A ∈ F n × n A \in \mathbb{F}^{n\times n} A ∈ F n × n לכסינה מעל F \mathbb{F} F אם היא דומה למטריצה אלכסונית.P ∈ F n × n P \in \mathbb{F}^{n\times n} P ∈ F n × n D = d i a g ( λ 1 , … , λ n ) ∈ F n × n D=diag(\lambda_{1},\dots,\lambda_{n})\in \mathbb{F}^{n\times n} D = d ia g ( λ 1 , … , λ n ) ∈ F n × n P − 1 A P = D P^{-1}AP=D P − 1 A P = D
דוגמה (מוטיבציה ללכסון(טריוויאלית))
P − 1 A P = D ⟹ A = P D P − 1 ⟹ A 3729 = ( P D P − 1 ) 3729 P^{-1}AP=D \implies A=PDP^{-1} \implies A^{3729}=(PDP^{-1})^{3729} P − 1 A P = D ⟹ A = P D P − 1 ⟹ A 3729 = ( P D P − 1 ) 3729 = P D P − 1 P D P − 1 … P D P − 1 = P D 3729 P − 1 =PDP^{-1}PDP^{-1}\dots PDP^{-1}=PD^{3729}P^{-1} = P D P − 1 P D P − 1 … P D P − 1 = P D 3729 P − 1 ⟺ \iff ⟺ P ∈ F n × n P \in \mathbb{F}^{n\times n} P ∈ F n × n D = d i a g ( λ 1 , … , λ n ) D=diag(\lambda_{1},\dots,\lambda_{n}) D = d ia g ( λ 1 , … , λ n ) A P = P D AP=PD A P = P D ⟺ \iff ⟺ v ‾ 1 , … , v ‾ n \underline{v}_{1},\dots,\underline{v}_{n} v 1 , … , v n λ 1 , … , λ n ∈ F \lambda_{1},\dots,\lambda_{n} \in \mathbb{F} λ 1 , … , λ n ∈ F
A ( v ‾ 1 … v ‾ n ) = ( v ‾ 1 … v ‾ n ) ( λ 1 ⋱ λ n ) A\begin{pmatrix}
\underline{v}_{1} & \dots & \underline{v}_{n}
\end{pmatrix}=\begin{pmatrix}
\underline{v}_{1} & \dots & \underline{v}_{n}
\end{pmatrix}\begin{pmatrix}
\lambda_{1} \\
& \ddots & \\
& & \lambda_{n}
\end{pmatrix} A ( v 1 … v n ) = ( v 1 … v n ) λ 1 ⋱ λ n ⟺ \iff ⟺ v ‾ 1 , … , v ‾ n \underline{v}_{1},\dots,\underline{v}_{n} v 1 , … , v n λ 1 , … , λ n ∈ F \lambda_{1},\dots,\lambda_{n} \in \mathbb{F} λ 1 , … , λ n ∈ F
( A v ‾ 1 … A v ‾ n ) = ( λ 1 v ‾ 1 λ 2 v ‾ n … λ n v ‾ n ) \begin{pmatrix}
A\underline{v}_{1} & \dots & A\underline{v}_{n}
\end{pmatrix}=\begin{pmatrix}
\lambda_{1}\underline{v}_{1} & \lambda_{2}\underline{v}_{n} & \dots & \lambda_{n}\underline{v}_{n}
\end{pmatrix} ( A v 1 … A v n ) = ( λ 1 v 1 λ 2 v n … λ n v n ) ⟺ \iff ⟺ v ‾ 1 , … , v ‾ n \underline{v}_{1},\dots,\underline{v}_{n} v 1 , … , v n λ 1 , … , λ n ∈ F \lambda_{1},\dots,\lambda_{n} \in \mathbb{F} λ 1 , … , λ n ∈ F
A v ‾ i = λ i v ‾ i i = 1 , … , n A\underline{v}_{i}=\lambda_{i}\underline{v}_{i} \qquad i=1,\dots,n A v i = λ i v i i = 1 , … , n ⟺ \iff ⟺ A A A ⟺ \iff ⟺ n n n
סקלר λ ∈ F \lambda \in \mathbb{F} λ ∈ F ערך עצמי(ע"ע) של A ∈ F n × n A \in \mathbb{F}^{n\times n} A ∈ F n × n 0 ≠ v ‾ ∈ F n 0 \neq \underline{v} \in \mathbb{F}^n 0 = v ∈ F n A v ‾ = λ v ‾ A\underline{v}=\lambda \underline{v} A v = λ v v ‾ \underline{v} v וקטור עצמי המתאים לע"ע λ \lambda λ
תהי A ∈ F n × n A \in \mathbb{F}^{n\times n} A ∈ F n × n A A A F \mathbb{F} F ⟺ \iff ⟺ B = { v ‾ 1 , … , v ‾ n } B=\left\{ \underline{v}_{1},\dots,\underline{v}_{n} \right\} B = { v 1 , … , v n } A A A A A A n n n
אם A A A P P P D D D P − 1 A P = D P^{-1}AP=D P − 1 A P = D P = ( v ‾ 1 , v ‾ 2 , … , v ‾ m ) P=(\underline{v}_{1},\underline{v}_{2},\dots,\underline{v}_{m}) P = ( v 1 , v 2 , … , v m ) D = ( λ 1 ⋱ λ n ) D=\begin{pmatrix}\lambda_{1}\\& \ddots \\ &&& \lambda_{n}\end{pmatrix} D = λ 1 ⋱ λ n
טכניקה למציאת D D D P P P
יהי λ ∈ F \lambda \in \mathbb{F} λ ∈ F ⟺ A ∈ F n × n \iff A \in \mathbb{F}^{n\times n} ⟺ A ∈ F n × n 0 ‾ ≠ v ‾ ∈ F n \underline{0}\neq \underline{v}\in \mathbb{F}^n 0 = v ∈ F n A v ‾ = λ v ‾ A\underline{v}=\lambda \underline{v} A v = λ v ⟺ \iff ⟺ v ‾ ≠ 0 ‾ \underline{v} \neq \underline{0} v = 0 λ v ‾ − A v ‾ = 0 ‾ \lambda \underline{v}-A\underline{v}=\underline{0} λ v − A v = 0 ⟺ \iff ⟺ v ‾ ≠ 0 ‾ \underline{v}\neq \underline{0} v = 0 ( λ I − A ) v ‾ = 0 ‾ (\lambda I-A)\underline{v}=\underline{0} ( λ I − A ) v = 0 ⟺ \iff ⟺ ( λ I − A ) x ‾ = 0 (\lambda I-A)\underline{x}=0 ( λ I − A ) x = 0 ⟺ \iff ⟺ λ I − A \lambda I-A λ I − A ⟺ \iff ⟺ ∣ λ I − A ∣ = 0 |\lambda I-A|=0 ∣ λ I − A ∣ = 0 דטרמיננטה )
P ( t ) = ∣ t I − A ∣ P(t)=|tI-A| P ( t ) = ∣ t I − A ∣ הפולניום האופייני של המטריצה A A A .
שלב 1
מוצאים את כל השורשים של הפ"א (פולינום אופייני).
P ( t ) = ∣ t I − A ∣ P(t)=|tI-A| P ( t ) = ∣ t I − A ∣ P ( t ) = ( t − λ 1 ) k 1 ( t − λ 2 ) k 2 … ( t − λ m ) k m P(t)=(t-\lambda_{1})^{k_{1}}(t-\lambda_{2})^{k_{2}}\dots(t-\lambda_{m})^{k_{m}} P ( t ) = ( t − λ 1 ) k 1 ( t − λ 2 ) k 2 … ( t − λ m ) k m אזי λ 1 , … , λ m \lambda_{1},\dots,\lambda_{m} λ 1 , … , λ m A A A k i k_{i} k i הריבוי האלגברי של הע"ע λ i \lambda_i λ i
שלב 2 - מציאת הו"ע
כדי למצוא את הו"ע פשוט פותרים את המערכת A v ‾ = λ i v ‾ A\underline{v}=\lambda _i\underline{v} A v = λ i v λ i \lambda_{i} λ i ( λ i I − A ) v ‾ = 0 ‾ (\lambda_{i}I-A)\underline{v}=\underline{0} ( λ i I − A ) v = 0 V λ i = { x ‾ ∈ F n ∣ ( λ i I − A ) x ‾ = 0 ‾ } V_{\lambda_{i}}=\left\{ \underline{x} \in \mathbb{F}^n|(\lambda_{i}I-A)\underline{x}=\underline{0} \right\} V λ i = { x ∈ F n ∣ ( λ i I − A ) x = 0 } המרחב העצמי של הע"ע λ i \lambda_{i} λ i
dim V λ i = n − r ( λ i I − A ) \dim V_{\lambda_{i}}=n-r(\lambda_{i}I-A) dim V λ i = n − r ( λ i I − A ) המימד של V λ i V_{\lambda_{i}} V λ i הריבוי הגיאומטרי של הע"ע λ i \lambda_{i} λ i
A A A ⟺ \iff ⟺ n n n n n n
תהא A = ( 3 1 1 − 3 5 − 3 − 1 1 1 ) A=\begin{pmatrix}3&1&1\\-3&5&-3\\-1&1&1\end{pmatrix} A = 3 − 3 − 1 1 5 1 1 − 3 1 A A A R \mathbb{R} R P P P D D D P − 1 A P = D P^{-1}AP=D P − 1 A P = D
נמצא ע"ע.
∣ t I − A ∣ = ∣ t − 3 − 1 − 1 3 t − 5 3 1 − 1 t − 1 ∣ = ↓ R 1 = R 1 + R 3 ∣ t − 2 0 t − 2 3 t − 5 3 1 − 1 t − 1 ∣ = ↓ C 3 = C 3 − C 1 ∣ t − 2 0 0 3 t − 5 0 1 − 1 t − 2 ∣ |tI-A|=\begin{vmatrix}
t-3 & -1 & -1 \\
3 & t-5 & 3 \\
1 & -1 & t-1
\end{vmatrix}\underset{\begin{matrix}
\downarrow \\
R_{1}=R_{1}+R_{3}
\end{matrix}}{=}\begin{vmatrix}
t-2 & 0 & t-2 \\
3 & t-5 & 3 \\
1 & -1 & t-1
\end{vmatrix}\underset{\begin{matrix}
\downarrow \\
C_{3}=C_{3}-C_{1}
\end{matrix}}{=}\begin{vmatrix}
t-2 & 0 & 0 \\
3 & t-5 & 0 \\
1 & -1 & t-2
\end{vmatrix} ∣ t I − A ∣ = t − 3 3 1 − 1 t − 5 − 1 − 1 3 t − 1 ↓ R 1 = R 1 + R 3 = t − 2 3 1 0 t − 5 − 1 t − 2 3 t − 1 ↓ C 3 = C 3 − C 1 = t − 2 3 1 0 t − 5 − 1 0 0 t − 2 = ( t − 2 ) 2 ( t − 5 ) =(t-2)^2(t-5) = ( t − 2 ) 2 ( t − 5 ) קיבלנו λ 1 = 2 \lambda_{1}=2 λ 1 = 2 2 2 2 λ 2 = 5 \lambda_{2}=5 λ 2 = 5 1 1 1 λ 1 = 2 \lambda_{1}=2 λ 1 = 2 V λ 1 = 2 V_{\lambda_{1}=2} V λ 1 = 2 ( 2 I − A ) x ‾ = 0 ‾ (2I-A)\underline{x}=\underline{0} ( 2 I − A ) x = 0
( − 1 1 − 1 3 − 3 3 1 − 1 1 ) → ( 1 − 1 1 0 0 0 0 0 0 ) ⟹ { x 1 − x 2 + x 3 = 0 x 1 = x 2 − x 3 = s − t x 2 = 5 x 3 = s \begin{pmatrix}
-1 & 1 & -1 \\
3 & -3 & 3 \\
1 & -1 & 1
\end{pmatrix}\to \begin{pmatrix}
1 & -1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix} \implies \begin{cases}
x_{1}-x_{2}+x_{3}=0 \\
x_{1}=x_{2}-x_{3}=s-t \\
x_{2}=5 \\
x_{3}=s
\end{cases} − 1 3 1 1 − 3 − 1 − 1 3 1 → 1 0 0 − 1 0 0 1 0 0 ⟹ ⎩ ⎨ ⎧ x 1 − x 2 + x 3 = 0 x 1 = x 2 − x 3 = s − t x 2 = 5 x 3 = s V λ 1 = 2 = { ( s − t s t ) s , t ∈ R } = s p { ( 1 1 0 ) , ( − 1 0 1 ) } V_{\lambda_{1}=2}=\left\{ \begin{array}{c|c}
\begin{pmatrix}
s-t \\
s \\
t
\end{pmatrix} & s,t \in \mathbb{R}
\end{array} \right\}=sp\left\{ \begin{pmatrix}
1 \\
1 \\
0
\end{pmatrix},\begin{pmatrix}
-1 \\
0 \\
1
\end{pmatrix} \right\} V λ 1 = 2 = ⎩ ⎨ ⎧ s − t s t s , t ∈ R ⎭ ⎬ ⎫ = s p ⎩ ⎨ ⎧ 1 1 0 , − 1 0 1 ⎭ ⎬ ⎫ הר"ג של λ 1 = 2 \lambda_{1}=2 λ 1 = 2 2 2 2 λ 2 = 5 \lambda_{2}=5 λ 2 = 5 ( 5 I − A ) x ‾ = 0 ‾ (5I-A)\underline{x}=\underline{0} ( 5 I − A ) x = 0
( 2 1 − 1 3 0 3 1 − 1 4 ) → ( 1 − 1 4 3 0 3 2 1 − 1 ) → ( 1 − 1 4 0 3 − 9 0 3 − 9 ) → ( 1 − 1 4 0 1 − 3 0 0 0 ) → ( 1 0 1 0 1 − 3 0 0 0 ) \begin{pmatrix}
2 & 1 & -1 \\
3 & 0 & 3 \\
1 & -1 & 4
\end{pmatrix}\to \begin{pmatrix}
1 & -1 & 4 \\
3 & 0 & 3 \\
2 & 1 & -1
\end{pmatrix}\to \begin{pmatrix}
1 & -1 & 4 \\
0 & 3 & -9 \\
0 & 3 & -9
\end{pmatrix}\to \begin{pmatrix}
1 & -1 & 4 \\
0 & 1 & -3 \\
0 & 0 & 0
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 1 \\
0 & 1 & -3 \\
0 & 0 & 0
\end{pmatrix} 2 3 1 1 0 − 1 − 1 3 4 → 1 3 2 − 1 0 1 4 3 − 1 → 1 0 0 − 1 3 3 4 − 9 − 9 → 1 0 0 − 1 1 0 4 − 3 0 → 1 0 0 0 1 0 1 − 3 0 { x 1 = − x 3 x 2 = 3 x 3 x 3 = t \begin{cases}
x_{1}=-x_{3} \\
x_{2}=3x_{3} \\
x_{3}=t
\end{cases} ⎩ ⎨ ⎧ x 1 = − x 3 x 2 = 3 x 3 x 3 = t V λ 2 = 5 = { ( − t 3 t t ) t ∈ R } = s p { ( − 1 3 1 ) } V_{\lambda_{2}=5}=\left\{ \begin{array}{c|c}
\begin{pmatrix}
-t \\
3t \\
t
\end{pmatrix} & t \in \mathbb{R}
\end{array} \right\}=sp\left\{ \begin{pmatrix}
-1 \\
3 \\
1
\end{pmatrix} \right\} V λ 2 = 5 = ⎩ ⎨ ⎧ − t 3 t t t ∈ R ⎭ ⎬ ⎫ = s p ⎩ ⎨ ⎧ − 1 3 1 ⎭ ⎬ ⎫ הר"ג של λ 2 = 5 \lambda_{2}=5 λ 2 = 5 { ( 1 1 0 ) , ( − 1 0 1 ) , ( − 1 3 1 ) } \left\{ \begin{pmatrix}1\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\1\end{pmatrix},\begin{pmatrix}-1\\3\\1\end{pmatrix} \right\} ⎩ ⎨ ⎧ 1 1 0 , − 1 0 1 , − 1 3 1 ⎭ ⎬ ⎫ R 3 \mathbb{R}^3 R 3
( 1 1 0 − 1 0 1 − 1 3 1 ) → ( 1 1 0 0 − 1 1 0 4 1 ) → ( 1 1 0 0 − 1 1 0 0 5 ) \begin{pmatrix}
1 & 1 & 0 \\
-1 & 0 & 1 \\
-1 & 3 & 1
\end{pmatrix}\to \begin{pmatrix}
1 & 1 & 0 \\
0 & -1 & 1 \\
0 & 4 & 1
\end{pmatrix}\to \begin{pmatrix}
1 & 1 & 0 \\
0 & -1 & 1 \\
0 & 0 & 5
\end{pmatrix} 1 − 1 − 1 1 0 3 0 1 1 → 1 0 0 1 − 1 4 0 1 1 → 1 0 0 1 − 1 0 0 1 5 ולכן המטריצה לכסינה מעל R \mathbb{R} R
P = ( 1 − 1 − 1 1 0 3 0 1 1 ) , D = ( 2 0 0 0 2 0 0 0 5 ) P=\begin{pmatrix}
1 & -1 & -1 \\
1 & 0 & 3 \\
0 & 1 & 1
\end{pmatrix},D=\begin{pmatrix}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 5
\end{pmatrix} P = 1 1 0 − 1 0 1 − 1 3 1 , D = 2 0 0 0 2 0 0 0 5
A ∈ F n × n A \in \mathbb{F}^{n\times n} A ∈ F n × n ⟺ F \iff\mathbb{F} ⟺ F
הפולינום האופייני(פ"א) של A A A F \mathbb{F} F
לכל ע"ע של A A A
יהא λ \lambda λ A A A ≤ \leq ≤ ≤ \leq ≤
אם ר"א=1 ⟸ \impliedby ⟸
הוכחה של טענה 1 : λ 1 \lambda_{1} λ 1 k k k { v ‾ 1 , … , v ‾ n } \left\{ \underline{v}_{1},\dots,\underline{v}_{n} \right\} { v 1 , … , v n } A v ‾ i = λ 1 v ‾ i A\underline{v}_{i}=\lambda_{1}\underline{v}_{i} A v i = λ 1 v i i = 1 , … , k i=1,\dots,k i = 1 , … , k { v ‾ 1 , … , v ‾ n } \left\{ \underline{v}_{1},\dots,\underline{v}_{n} \right\} { v 1 , … , v n } F n \mathbb{F}^n F n { v ‾ 1 , … , v ‾ k , v ‾ k + 1 , … , v ‾ n } \left\{ \underline{v}_{1},\dots,\underline{v}_{k},\underline{v}_{k+1},\dots,\underline{v}_{n} \right\} { v 1 , … , v k , v k + 1 , … , v n } F n \mathbb{F}^n F n
A v ‾ 1 = λ 1 v ‾ 1 A v ‾ 2 = λ 1 v ‾ 2 ⋮ A v ‾ k = λ 1 v ‾ k A v ‾ k + 1 = α 1 ( 1 ) v ‾ 1 + ⋯ + α n ( 1 ) v ‾ n ⋮ A v ‾ n = α 1 ( n − k ) v ‾ 1 + ⋯ + α n ( n − k ) v ‾ n \begin{align*}
& A\underline{v}_{1}=\lambda_{1}\underline{v}_{1} \\
& A\underline{v}_{2}=\lambda_{1}\underline{v}_{2} \\
& \vdots \\
& A\underline{v}_{k}=\lambda_{1}\underline{v}_{k} \\
& A\underline{v}_{k+1}=\alpha_{1}^{(1)}\underline{v}_{1}+\dots+\alpha_{n}^{(1)}\underline{v}_{n} \\
& \vdots \\
& A\underline{v}_{n}=\alpha_{1}^{(n-k)}\underline{v}_{1}+\dots+\alpha_{n}^{(n-k)}\underline{v}_{n}
\end{align*} A v 1 = λ 1 v 1 A v 2 = λ 1 v 2 ⋮ A v k = λ 1 v k A v k + 1 = α 1 ( 1 ) v 1 + ⋯ + α n ( 1 ) v n ⋮ A v n = α 1 ( n − k ) v 1 + ⋯ + α n ( n − k ) v n A ( v ‾ 1 v ‾ 2 … v ‾ k v ‾ k + 1 … v ‾ n ) ∣ ∣ P # = A \underset{\begin{matrix}
|| \\
P^{\#}
\end{matrix}}{\begin{pmatrix} \\
\underline{v}_{1} & \underline{v}_{2} & \dots & \underline{v}_{k} & \underline{v}_{k+1} & \dots & \underline{v}_{n}\\\
\end{pmatrix}}= A ∣∣ P # v 1 v 2 … v k v k + 1 … v n = ( v ‾ 1 v ‾ 2 … v ‾ k v ‾ k + 1 … v ‾ n ) ∣ ∣ P # ( λ 1 ⋱ λ 1 ∗ 0 ∗ ) ∣ ∣ D # \underset{\begin{matrix}
|| \\
P^{\#}
\end{matrix}}{\begin{pmatrix} \\
\underline{v}_{1} & \underline{v}_{2} & \dots & \underline{v}_{k} & \underline{v}_{k+1} & \dots & \underline{v}_{n}\\ \\
\end{pmatrix}}\underset{\begin{matrix}
|| \\
D^{\#}
\end{matrix}}{\begin{pmatrix}
\begin{array}{c|c}
\begin{matrix}
\lambda_{1} \\
& \ddots \\
& & \lambda_{1}
\end{matrix} & \qquad * \qquad \\
\hline \begin{matrix}
\\ 0\\\
\end{matrix} & \qquad * \qquad
\end{array}
\end{pmatrix}} ∣∣ P # v 1 v 2 … v k v k + 1 … v n ∣∣ D # λ 1 ⋱ λ 1 0 ∗ ∗ ( A v ‾ 1 … A v ‾ k ∣ A v ‾ k + 1 … A v ‾ n ) = ( λ v ‾ 1 … λ 1 v ‾ n ∣ ℵ k + 1 … ℵ n ) \begin{pmatrix}
A\underline{v}_{1}\dots A\underline{v}_{k} \vert A\underline{v}_{k+1}\dots A\underline{v}_{n}
\end{pmatrix}=(\lambda \underline{v}_{1}\dots\lambda_{1}\underline{v}_{n}\vert \aleph_{k+1}\dots \aleph_{n}) ( A v 1 … A v k ∣ A v k + 1 … A v n ) = ( λ v 1 … λ 1 v n ∣ ℵ k + 1 … ℵ n ) A P # = P # D # AP^{\#}=P^{\#}D^\# A P # = P # D # A = P # D # P # − 1 A=P^\#D^\#{P^\#}^{-1} A = P # D # P # − 1 ∣ t I − A ∣ = ∣ t I − P # D # P # − 1 ∣ |tI-A|=|tI-P^\#D^\#{P^\#}^{-1}| ∣ t I − A ∣ = ∣ t I − P # D # P # − 1 ∣ ∣ P # ( t I − D # ) P # − 1 ∣ = ∣ P # ∣ ∣ t I − D # ∣ ∣ P # − 1 ∣ \left|P^\#(tI-D^\#){P^\#}^{-1}\right|=\left| P^\# \right|\left| tI-D^\# \right|\left| {P^\#}^{-1} \right| P # ( t I − D # ) P # − 1 = P # t I − D # P # − 1 = ∣ P # ∣ ∣ t I − D # ∣ ∣ P # − 1 ∣ = ∣ t I − D # ∣ =\left| P^\# \right|\left| tI-D^\# \right|\left| {P^\#}^{-1} \right|=\left| tI-D^\# \right| = P # t I − D # P # − 1 = t I − D # A ( t I − B ) A − 1 = ( A t I − A B ) A − 1 = A t I A − 1 − A B A − 1 = t I − A B A − 1 A(tI-B)A^{-1}=(AtI-AB)A^{-1}=AtIA^{-1}-ABA^{-1}=tI-ABA^{-1} A ( t I − B ) A − 1 = ( A t I − A B ) A − 1 = A t I A − 1 − A B A − 1 = t I − A B A − 1
אם A ∼ B A \sim B A ∼ B
ז.א. קיבלנו כי
P ( t ) = ∣ t I − A ∣ = ∣ t I − D # ∣ = ∣ t − λ 1 ⋱ t − λ 1 ∗ 0 ∗ ∣ = ( t − λ 1 ) k ( ∗ ) P(t)=\left| tI-A \right|=\left| tI-D^\# \right|=\begin{vmatrix}
\begin{array}{c|c}
\begin{matrix}
t-\lambda_{1} \\
& \ddots \\
& & t-\lambda_{1}
\end{matrix} & \qquad * \qquad \\
\hline \begin{matrix}
\\ 0 \\\
\end{matrix} & *
\end{array}
\end{vmatrix}=(t-\lambda_{1})^k(*) P ( t ) = ∣ t I − A ∣ = t I − D # = t − λ 1 ⋱ t − λ 1 0 ∗ ∗ = ( t − λ 1 ) k ( ∗ ) ז.א. הר"א של λ \lambda λ k k k ≤ \leq ≤ ∎
יהיו λ 1 , … , λ k \lambda_{1},\dots,\lambda_{k} λ 1 , … , λ k A A A V λ 1 ⊕ V λ 2 ⊕ ⋯ ⊕ V λ k V_{\lambda_{1}}\oplus V_{\lambda_{2}}\oplus\dots \oplus V_{\lambda_{k}} V λ 1 ⊕ V λ 2 ⊕ ⋯ ⊕ V λ k
יהיו v ‾ 1 , v ‾ 2 , … , v ‾ k \underline{v}_{1},\underline{v}_{2},\dots,\underline{v}_{k} v 1 , v 2 , … , v k v ‾ 1 , … , v ‾ k \underline{v}_{1},\dots,\underline{v}_{k} v 1 , … , v k
הוכחה: באינדוקציה על מספר הוקטורים.בסיס: אם יש רק וקטור אחד, אזי הוא שונה מאפס(ו"ע) וברור שהוא בת"ל.
α v ‾ = 0 ⟹ α = 0 \alpha \underline{v}=0 \implies \alpha=0 α v = 0 ⟹ α = 0 צעד: נניח כי זה מתקיים עבור v ‾ 1 , … , v ‾ k \underline{v}_{1},\dots,\underline{v}_{k} v 1 , … , v k v ‾ 1 , … , v ‾ k , v ‾ k + 1 \underline{v}_{1},\dots,\underline{v}_{k},\underline{v}_{k+1} v 1 , … , v k , v k + 1
( ∗ ) α 1 v ‾ 1 + ⋯ + α k v ‾ k + 1 = 0 ‾ (*)\; \alpha_{1}\underline{v}_{1}+\dots+\alpha_{k}\underline{v}_{k+1}=\underline{0} ( ∗ ) α 1 v 1 + ⋯ + α k v k + 1 = 0 α 1 A v ‾ 1 + α 2 A v ‾ 2 + ⋯ + α k + 1 A v ‾ k + 1 = 0 ‾ \alpha_{1}A\underline{v}_{1}+\alpha_{2}A\underline{v}_{2}+\dots+\alpha_{k+1}A\underline{v}_{k+1}=\underline{0} α 1 A v 1 + α 2 A v 2 + ⋯ + α k + 1 A v k + 1 = 0 ( 1 ) α 1 λ 1 v ‾ 1 + α 2 λ 2 v ‾ 2 + ⋯ + α k + 1 λ k + 1 v ‾ k + 1 = 0 ‾ (1)\; \alpha_{1}\lambda_{1}\underline{v}_{1}+\alpha_{2}\lambda_{2}\underline{v}_{2}+\dots+\alpha_{k+1}\lambda_{k+1}\underline{v}_{k+1}=\underline{0} ( 1 ) α 1 λ 1 v 1 + α 2 λ 2 v 2 + ⋯ + α k + 1 λ k + 1 v k + 1 = 0 מאידרך, אם נכפול את ( ∗ ) (*) ( ∗ ) λ 1 \lambda_{1} λ 1
α 1 λ 1 v ‾ 1 + α 2 λ 1 v ‾ 2 + ⋯ + α k + 1 λ 1 v ‾ k + 1 = 0 ‾ \alpha_{1}\lambda_{1}\underline{v}_{1}+\alpha_{2}\lambda_{1}\underline{v}_{2}+\dots+\alpha_{k+1}\lambda_{1}\underline{v}_{k+1}=\underline{0} α 1 λ 1 v 1 + α 2 λ 1 v 2 + ⋯ + α k + 1 λ 1 v k + 1 = 0 נחסיר מ-( 1 ) (1) ( 1 )
α 2 ( λ 2 − λ 1 ) v ‾ 2 + α 3 ( λ 3 − λ 1 ) v ‾ 3 + ⋯ + α k + 1 ( λ k + 1 − λ 1 ) v ‾ k + 1 = 0 ‾ \alpha_{2}(\lambda_{2}-\lambda_{1})\underline{v}_{2}+\alpha_{3}(\lambda_{3}-\lambda_{1})\underline{v}_{3}+\dots+\alpha_{k+1}(\lambda_{k+1}-\lambda_{1})\underline{v}_{k+1}=\underline{0} α 2 ( λ 2 − λ 1 ) v 2 + α 3 ( λ 3 − λ 1 ) v 3 + ⋯ + α k + 1 ( λ k + 1 − λ 1 ) v k + 1 = 0 כעת, בצירוף הליניארי הזה יש k k k
α 2 ( λ 2 − λ 1 ) = 0 α 3 ( λ 3 − λ 1 ) = 0 ⋮ α k + 1 ( λ k + 1 − λ 1 ) = 0 ⟹ α 2 = α 3 = ⋯ = α k + 1 = 0 ⟹ α 1 = 0 \begin{matrix}
\alpha_{2}(\lambda_{2}-\lambda_{1})=0 \\
\alpha_{3}(\lambda_{3}-\lambda_{1})=0 \\
\vdots \\
\alpha_{k+1}(\lambda_{k+1}-\lambda_{1})=0
\end{matrix}\implies \alpha_{2}=\alpha_{3}=\dots=\alpha_{k+1}=0 \implies \alpha_{1}=0 α 2 ( λ 2 − λ 1 ) = 0 α 3 ( λ 3 − λ 1 ) = 0 ⋮ α k + 1 ( λ k + 1 − λ 1 ) = 0 ⟹ α 2 = α 3 = ⋯ = α k + 1 = 0 ⟹ α 1 = 0 ∎
הוכחה של הטענה :
v ‾ ∈ V λ i ∩ V λ 1 + V λ 2 + ⋯ + V ^ λ i + ⋯ + V λ k \underline{v}\in V_{\lambda_{i}} \cap V_{\lambda_{1}}+V_{\lambda_{2}}+\dots+\hat{V}_{\lambda_{i}}+\dots+V_{\lambda_{k}} v ∈ V λ i ∩ V λ 1 + V λ 2 + ⋯ + V ^ λ i + ⋯ + V λ k אזי:
v ‾ i = v ‾ = v ‾ 1 + v ‾ 2 + ⋯ + v ‾ i − 1 + v ‾ i + 1 + ⋯ + v ‾ k \underline{v}_{i}=\underline{v}=\underline{v}_{1}+\underline{v}_{2}+\dots+\underline{v}_{i-1}+\underline{v}_{i+1}+\dots+\underline{v}_{k} v i = v = v 1 + v 2 + ⋯ + v i − 1 + v i + 1 + ⋯ + v k ⟹ v ‾ 1 + v ‾ 2 + ⋯ − v ‾ i + ⋯ + v ‾ k = 0 ‾ \implies \underline{v}_{1}+\underline{v}_{2}+\dots-\underline{v}_{i}+\dots+\underline{v}_{k}=\underline{0} ⟹ v 1 + v 2 + ⋯ − v i + ⋯ + v k = 0 ⟹ v ‾ 1 = v ‾ 2 = ⋯ = v ‾ i = ⋯ = v ‾ k = 0 ‾ \implies \underline{v}_{1}=\underline{v}_{2}=\dots=\underline{v}_{i}=\dots=\underline{v}_{k}=\underline{0} ⟹ v 1 = v 2 = ⋯ = v i = ⋯ = v k = 0 וקטורים עצמיים של ע"ע שונים הם בת"ל.∎
איחוד הבסיסים של מ"ע שונים הוא בת"ל.
הוכחה:
v ‾ 1 , … , , v ‾ m 1 ∈ V λ 1 v ‾ m 1 + 1 , … , v ‾ m 1 + m 2 ∈ V λ 2 ⋮ v ‾ m k − 1 + 1 , … , v ‾ m k − 1 + m k ∈ V λ k \begin{matrix}
\underline{v}_{1},\dots,,\underline{v}_{m_{1}}\in V_{\lambda_{1}} \\
\underline{v}_{m_{1}+1},\dots,\underline{v}_{m_{1}+m_{2}}\in V_{\lambda_{2}} \\
\vdots \\
\underline{v}_{m_{k-1}+1},\dots,\underline{v}_{m_{k-1}+m_{k}}\in V_{\lambda_{k}}
\end{matrix} v 1 , … ,, v m 1 ∈ V λ 1 v m 1 + 1 , … , v m 1 + m 2 ∈ V λ 2 ⋮ v m k − 1 + 1 , … , v m k − 1 + m k ∈ V λ k בסיסים של המ"ע וכי
λ 1 v ‾ 1 + ⋯ + λ m 1 v ‾ m 1 ⏞ V λ 1 + λ m 1 + 1 v ‾ m 1 + 1 + ⋯ + λ m 1 + m 2 v ‾ m 1 + m 2 ⏞ V λ 2 + … \overbrace{\lambda_{1}\underline{v}_{1}+\dots+\lambda_{m_{1}}\underline{v}_{m_{1}}}^{V_{\lambda_{1}}}+\overbrace{\lambda_{m_{1}+1}\underline{v}_{m_{1}+1}+\dots+\lambda_{m_{1}+m_{2}}\underline{v}_{m_{1}+m_{2}}}^{V_{\lambda_{2}}}+\dots λ 1 v 1 + ⋯ + λ m 1 v m 1 V λ 1 + λ m 1 + 1 v m 1 + 1 + ⋯ + λ m 1 + m 2 v m 1 + m 2 V λ 2 + … + λ m k − 1 + 1 v ‾ m k − 1 + 1 + ⋯ + λ m k − 1 + m k v ‾ m k − 1 + m k ⏞ V λ k = 0 ‾ +\overbrace{\lambda_{m_{k-1}+1}\underline{v}_{m_{k-1}+1}+\dots+\lambda_{m_{k-1}+m_{k}}\underline{v}_{m_{k-1}+m_{k}}}^{V_{\lambda_{k}}}=\underline{0} + λ m k − 1 + 1 v m k − 1 + 1 + ⋯ + λ m k − 1 + m k v m k − 1 + m k V λ k = 0 כיוון שזה סכום ישר,
λ 1 v ‾ 1 + ⋯ + λ m 1 v ‾ m 1 = 0 ‾ ⟹ λ 1 = ⋯ = λ m 1 = 0 λ m 1 + 1 v ‾ m 1 + 1 + ⋯ + λ m 1 + m 2 v ‾ m 1 + m 2 = 0 ‾ ⟹ λ m 1 + 1 = ⋯ = λ m 1 + m 2 = 0 ⋮ λ m k − 1 + 1 v ‾ m k − 1 + 1 + ⋯ + λ m k − 1 + m k v ‾ m k − 1 + m k = 0 ‾ ⟹ λ m k − 1 + 1 = ⋯ = λ m k − 1 + m k = 0 \begin{matrix}
\lambda_{1}\underline{v}_{1}+\dots+\lambda_{m_{1}}\underline{v}_{m_{1}}=\underline{0} \implies \lambda_{1}=\dots=\lambda_{m_{1}}=0 \\
\lambda_{m_{1}+1}\underline{v}_{m_{1}+1}+\dots+\lambda_{m_{1}+m_{2}}\underline{v}_{m_{1}+m_{2}}=\underline{0}\implies \lambda_{m_{1}+1}=\dots=\lambda_{m_{1}+m_{2}}=0 \\
\vdots \\
\lambda_{m_{k-1}+1}\underline{v}_{m_{k-1}+1}+\dots+\lambda_{m_{k-1}+m_{k}}\underline{v}_{m_{k-1}+m_{k}}=\underline{0}\implies \lambda_{m_{k-1}+1}=\dots=\lambda_{m_{k-1}+m_{k}}=0
\end{matrix} λ 1 v 1 + ⋯ + λ m 1 v m 1 = 0 ⟹ λ 1 = ⋯ = λ m 1 = 0 λ m 1 + 1 v m 1 + 1 + ⋯ + λ m 1 + m 2 v m 1 + m 2 = 0 ⟹ λ m 1 + 1 = ⋯ = λ m 1 + m 2 = 0 ⋮ λ m k − 1 + 1 v m k − 1 + 1 + ⋯ + λ m k − 1 + m k v m k − 1 + m k = 0 ⟹ λ m k − 1 + 1 = ⋯ = λ m k − 1 + m k = 0 ∎
הוכחת המשפט המהולל : n n n n n n F \mathbb{F} F F \mathbb{F} F ≤ \leq ≤ n n n ר"ג = ר"א \R{\text{ר"ג}}=\R{\text{ר"א}} ר"ג = ר"א n n n ∎
תהא A = ( 1 0 0 2 − a 2 a a − a a ) A=\begin{pmatrix}1&0&0\\2&-a&2a\\a&-a&a\end{pmatrix} A = 1 2 a 0 − a − a 0 2 a a a a a A A A R \mathbb{R} R A A A C \mathbb{C} C
פתרון א':
P ( t ) = ∣ t I − A ∣ = ∣ t − 1 0 0 − 2 t + a − 2 a − a a t − a ∣ = ( t − 1 ) ( ( t + a ) ( t − a ) + 2 a 2 ) = ( t − 1 ) ( t 2 + a 2 ) P(t)=|tI-A|=\begin{vmatrix}
t-1 & 0 & 0 \\
-2 & t+a & -2a \\
-a & a & t-a
\end{vmatrix}=(t-1)((t+a)(t-a)+2a^2)=(t-1)(t^2+a^2) P ( t ) = ∣ t I − A ∣ = t − 1 − 2 − a 0 t + a a 0 − 2 a t − a = ( t − 1 ) (( t + a ) ( t − a ) + 2 a 2 ) = ( t − 1 ) ( t 2 + a 2 ) ראשית, הפ"א P ( t ) P(t) P ( t ) R \mathbb{R} R a = 0 ⟺ a=0\iff a = 0 ⟺ a = 0 a=0 a = 0 λ 1 = 1 \lambda_{1}=1 λ 1 = 1 1 1 1 λ 2 = 0 \lambda_{2}=0 λ 2 = 0
אם למטריצה A A A n n n A A A
dim V λ 2 = 0 = 3 − r ( − A ) = 3 − r ( A ) = 3 − 1 = 2 \dim V_{\lambda_{2}=0}=3-r(-A)=3-r(A)=3-1=2 dim V λ 2 = 0 = 3 − r ( − A ) = 3 − r ( A ) = 3 − 1 = 2 ולכן, ר"א=ר"ג=2 2 2
פתרון ב':
P ( t ) = ( t − 1 ) ( t − a i ) ( t + a i ) P(t)=(t-1)(t-ai)(t+ai) P ( t ) = ( t − 1 ) ( t − ai ) ( t + ai ) נחלק למקרים:− a i = a i -ai=ai − ai = ai a = 0 a=0 a = 0 R \mathbb{R} R C \mathbb{C} C a i = − 1 ai=-1 ai = − 1 a = i a=i a = i P ( t ) = ( t − 1 ) ( t + 1 ) ( t − 1 ) P(t)=(t-1)(t+1)(t-1) P ( t ) = ( t − 1 ) ( t + 1 ) ( t − 1 ) λ 1 = − 1 \lambda_{1}=-1 λ 1 = − 1 1 1 1 λ 2 = 1 \lambda_{2}=1 λ 2 = 1
dim V λ 2 = 3 − r ( 1 ⋅ I − A ) = 3 − r ( I − A ) = 3 − r ( ( 1 0 0 0 1 0 0 0 1 ) − ( 1 0 0 2 − i 2 i i − i i ) ) \dim V_{\lambda_{2}}=3-r(1\cdot I-A)=3-r(I-A)=3-r\left(\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmatrix}-\begin{pmatrix}
1 & 0 & 0 \\
2 & -i & 2i \\
i & -i & i
\end{pmatrix}\right) dim V λ 2 = 3 − r ( 1 ⋅ I − A ) = 3 − r ( I − A ) = 3 − r 1 0 0 0 1 0 0 0 1 − 1 2 i 0 − i − i 0 2 i i = 3 − r ( 0 0 0 − 2 1 + i − 2 i − i i 1 + i ) = 3 − 2 = 1 =3-r\begin{pmatrix}
0 & 0 & 0 \\
-2 & 1+i & -2i \\
-i & i & 1+i
\end{pmatrix}=3-2=1 = 3 − r 0 − 2 − i 0 1 + i i 0 − 2 i 1 + i = 3 − 2 = 1 ולכן המטריצה אינה לכסינה .a i = 1 ai=1 ai = 1 a = − i a=-i a = − i לא לכסינה .a ≠ 0 , i , − i a\neq0,i,-i a = 0 , i , − i 3 3 3
A A A C \mathbb{C} C ⟺ \iff ⟺ a = i ∨ a = − i a=i \lor a=-i a = i ∨ a = − i
הגדרה 4 (ערך עצמי ווקטור עצמי של טרנספורמציה)
תהא T : V → V T:V\to V T : V → V 0 ‾ ≠ v ‾ ∈ V \underline{0}\neq \underline{v}\in V 0 = v ∈ V T ( v ‾ ) = λ v ‾ T(\underline{v})=\lambda \underline{v} T ( v ) = λ v T T T λ \lambda λ v ‾ \underline{v} v
מציאת ע"ע וו"ע
T ( v ‾ ) = λ v ‾ T(\underline{v})=\lambda \underline{v} T ( v ) = λ v ⇕ \Updownarrow ⇕ [ T ( v ‾ ) ] B = [ λ v ‾ ] B [T(\underline{v})]_{B}=[\lambda \underline{v}]_{B} [ T ( v ) ] B = [ λ v ] B ⇓ \Downarrow ⇓ [ T ] B [ v ‾ ] B = λ [ v ‾ ] B [T]_{B}[\underline{v}]_{B}=\lambda[\underline{v}]_{B} [ T ] B [ v ] B = λ [ v ] B
λ \lambda λ T T T ⟺ \iff ⟺ λ \lambda λ [ T ] B [T]_{B} [ T ] B v ‾ \underline{v} v T T T ⟺ \iff ⟺ [ v ‾ ] B [\ \underline{v}\ ]_{B} [ v ] B [ T ] B [T]_{B} [ T ] B
T ( a x 2 + b x + c ) = ( a − b ) x 2 + ( b − c ) x T(ax^2+bx+c)=(a-b)x^2+(b-c)x T ( a x 2 + b x + c ) = ( a − b ) x 2 + ( b − c ) x נמצא את ע"ע וו"ע של T T T E = ( x 2 , x , 1 ) E=(x^2,x,1) E = ( x 2 , x , 1 ) T T T E E E
T ( x 2 ) = x 2 = 1 ⋅ x 2 + 0 ⋅ x + 0 ⋅ 1 T ( x ) = − x 2 + x = − 1 ⋅ x 2 + 1 ⋅ x + 0 ⋅ 1 T ( 1 ) = − x = 0 ⋅ x 2 + ( − 1 ) x + 0 ⋅ 1 \begin{matrix}
T(x^2)=x^2=1\cdot x^2+0\cdot x+0\cdot1 \\
T(x)=-x^2+x=-1\cdot x^2+1\cdot x+0\cdot1 \\
T(1)=-x=0\cdot x^2+(-1)x+0\cdot1
\end{matrix} T ( x 2 ) = x 2 = 1 ⋅ x 2 + 0 ⋅ x + 0 ⋅ 1 T ( x ) = − x 2 + x = − 1 ⋅ x 2 + 1 ⋅ x + 0 ⋅ 1 T ( 1 ) = − x = 0 ⋅ x 2 + ( − 1 ) x + 0 ⋅ 1 [ T ] E = ( 1 − 1 0 0 1 − 1 0 0 0 ) [T]_{E}=\begin{pmatrix}
1 & -1 & 0\\ 0 & 1 & -1\\ 0 & 0 & 0
\end{pmatrix} [ T ] E = 1 0 0 − 1 1 0 0 − 1 0 ∣ t I − [ T ] E ∣ = ( t − 1 ) 2 t = { 0 1 1 |tI-[T]_{E}|=(t-1)^2t =\begin{cases}
0 \\
1 \\
1
\end{cases} ∣ t I − [ T ] E ∣ = ( t − 1 ) 2 t = ⎩ ⎨ ⎧ 0 1 1 כעת, נמצא ו"ע עבור ע"ע λ 1 = 0 \lambda_{1}=0 λ 1 = 0
( 0 ⋅ I − A ) v ‾ = 0 ‾ (0\cdot I-A)\underline{v}=\underline{0} ( 0 ⋅ I − A ) v = 0 ( − 1 1 0 0 − 1 1 0 0 0 ) → ( 1 − 1 0 0 1 − 1 0 0 0 ) \begin{pmatrix}
-1 & 1 & 0 \\
0 & -1 & 1 \\
0 & 0 & 0
\end{pmatrix}\to \begin{pmatrix}
1 & -1 & 0 \\
0 & 1 & -1 \\
0 & 0 & 0
\end{pmatrix} − 1 0 0 1 − 1 0 0 1 0 → 1 0 0 − 1 1 0 0 − 1 0 כל פעם שיש ע"ע 0 המטריצה לא הפיכה .
[ V λ 1 ] E = s p { ( 1 1 1 ) } ⟹ V λ 1 = s p { x 2 + x + 1 } [V_{\lambda_{1}}]_{E}=sp\left\{ \begin{pmatrix}
1\\1\\1
\end{pmatrix} \right\}\implies V_{\lambda_{1}}=sp\left\{ x^2+x+1 \right\} [ V λ 1 ] E = s p ⎩ ⎨ ⎧ 1 1 1 ⎭ ⎬ ⎫ ⟹ V λ 1 = s p { x 2 + x + 1 } נמצא כעת מ"ע עבור ע"ע λ 2 = 1 \lambda_{2}=1 λ 2 = 1
( 1 ⋅ I − [ T ] E ) v ‾ = 0 (1\cdot I-[T]_{E})\underline{v}=0 ( 1 ⋅ I − [ T ] E ) v = 0 ( 0 1 0 0 0 1 0 0 1 ) → ( 0 1 0 0 0 1 0 0 0 ) \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 1
\end{pmatrix}\to \begin{pmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{pmatrix} 0 0 0 1 0 0 0 1 1 → 0 0 0 1 0 0 0 1 0 x 2 = x 3 = 0 x_{2}=x_{3}=0 x 2 = x 3 = 0 x 1 = t x_{1}=t x 1 = t ואז מקבלים:
V λ 2 = { ( t 0 0 ) } = s p { ( 1 0 0 ) } ⟹ V λ 2 = s p { x 2 } V_{\lambda_{2}}=\left\{ \begin{pmatrix}
t\\0\\0
\end{pmatrix} \right\}=sp\left\{ \begin{pmatrix}
1\\0\\0
\end{pmatrix} \right\} \implies V_{\lambda_{2}}=sp\left\{ x^2 \right\} V λ 2 = ⎩ ⎨ ⎧ t 0 0 ⎭ ⎬ ⎫ = s p ⎩ ⎨ ⎧ 1 0 0 ⎭ ⎬ ⎫ ⟹ V λ 2 = s p { x 2 }
תהא T : V → V T:V\to V T : V → V T T T B B B V V V [ T ] B [T]_{B} [ T ] B
[ T ] B [T]_{B} [ T ] B ⟺ \iff ⟺ V V V λ 1 , … , λ n \lambda_{1},\dots,\lambda_{n} λ 1 , … , λ n
[ T ] B = ( λ 1 0 λ 2 ⋱ 0 λ n ) [T]_{B}=\begin{pmatrix}
\lambda_{1} & & & 0\\
& \lambda_{2} \\
& & \ddots \\
0& & & \lambda_{n}
\end{pmatrix} [ T ] B = λ 1 0 λ 2 ⋱ 0 λ n
[ T ] B [T]_{B} [ T ] B ⟺ \iff ⟺ B = ( v ‾ 1 , … , v ‾ n ) B=(\underline{v}_{1},\dots,\underline{v}_{n}) B = ( v 1 , … , v n ) λ 1 , … , λ n ∈ F \lambda_{1},\dots,\lambda_{n} \in \mathbb{F} λ 1 , … , λ n ∈ F
[ T ( v ‾ 1 ) ] B = ( λ 1 0 ⋮ 0 ) , [ T ( v ‾ 2 ) ] B = ( 0 λ 2 0 ⋮ 0 ) , … , [ T ( v ‾ n ) ] B = ( 0 ⋮ 0 λ n ) [T(\underline{v}_{1})]_{B}=\begin{pmatrix}
\lambda_{1}\\0\\ \vdots\\0
\end{pmatrix}, [T(\underline{v}_{2})]_{B}=\begin{pmatrix}
0\\\lambda_{2}\\0\\ \vdots\\0
\end{pmatrix},\dots,[T(\underline{v}_{n})]_{B}=\begin{pmatrix}
0\\ \vdots\\ 0\\ \lambda_{n}
\end{pmatrix} [ T ( v 1 ) ] B = λ 1 0 ⋮ 0 , [ T ( v 2 ) ] B = 0 λ 2 0 ⋮ 0 , … , [ T ( v n ) ] B = 0 ⋮ 0 λ n
[ T ] B [T]_{B} [ T ] B ⟺ \iff ⟺ B = ( v ‾ 1 , … , v ‾ n ) B=(\underline{v}_{1},\dots,\underline{v}_{n}) B = ( v 1 , … , v n ) λ 1 , … , λ n ∈ F \lambda_{1},\dots,\lambda_{n}\in \mathbb{F} λ 1 , … , λ n ∈ F
T ( v ‾ 1 ) = λ 1 v ‾ 1 T ( v ‾ 2 ) = λ 2 v ‾ 2 ⋮ T ( v ‾ n ) = λ n v ‾ n \begin{align*}
& T(\underline{v}_{1})=\lambda_{1}\underline{v}_{1} \\
& T(\underline{v}_{2})=\lambda_{2}\underline{v}_{2} \\
& \vdots \\
& T(\underline{v}_{n})=\lambda_{n}\underline{v}_{n}
\end{align*} T ( v 1 ) = λ 1 v 1 T ( v 2 ) = λ 2 v 2 ⋮ T ( v n ) = λ n v n
[ T ] B [T]_{B} [ T ] B ⟺ \iff ⟺ B B B V V V T T T
T T T ⟺ \iff ⟺ B = ( v ‾ 1 , … , v ‾ n ) B=(\underline{v}_{1},\dots,\underline{v}_{n}) B = ( v 1 , … , v n ) V V V T T T [ T ] B [T]_{B} [ T ] B ⟺ \iff ⟺ C C C V V V [ T ] C [T]_{C} [ T ] C
T T T ⟺ \iff ⟺ C C C V V V [ T ] C [T]_{C} [ T ] C
תהא T : R 2 × 2 → R 2 × 2 T:\mathbb{R}^{2\times2}\to \mathbb{R}^{2\times2} T : R 2 × 2 → R 2 × 2
T ( A ) = A t T(A)=A^t T ( A ) = A t מצאו בסיס B B B [ T ] B [T]_{B} [ T ] B
T ( 1 0 0 0 ) = ( 1 0 0 0 ) = 1 ⋅ E 1 + 0 ⋅ E 2 + 0 ⋅ E 3 + 0 ⋅ E 4 4 T\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}=\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix}=1\cdot E_{1}+0\cdot E_{2}+0\cdot E_{3}+0\cdot E_{4}4 T ( 1 0 0 0 ) = ( 1 0 0 0 ) = 1 ⋅ E 1 + 0 ⋅ E 2 + 0 ⋅ E 3 + 0 ⋅ E 4 4 T ( 0 1 0 0 ) = ( 0 0 1 0 ) = 0 ⋅ E 1 + 0 ⋅ E 2 + 1 ⋅ E 3 + 0 ⋅ E 4 T\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}=\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}=0\cdot E_{1}+0\cdot E_{2}+1\cdot E_{3}+0\cdot E_{4} T ( 0 0 1 0 ) = ( 0 1 0 0 ) = 0 ⋅ E 1 + 0 ⋅ E 2 + 1 ⋅ E 3 + 0 ⋅ E 4 T ( 0 0 1 0 ) = ( 0 1 0 0 ) = 0 ⋅ E 1 + 1 ⋅ E 2 + 0 ⋅ E 3 + 0 ⋅ E 4 T\begin{pmatrix}
0 & 0 \\
1 & 0
\end{pmatrix}=\begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix}=0\cdot E_{1}+1\cdot E_{2}+0\cdot E_{3}+0\cdot E_{4} T ( 0 1 0 0 ) = ( 0 0 1 0 ) = 0 ⋅ E 1 + 1 ⋅ E 2 + 0 ⋅ E 3 + 0 ⋅ E 4 T ( 0 0 0 1 ) = ( 0 0 0 1 ) = 0 ⋅ E 1 + 0 ⋅ E 2 + 0 ⋅ E 3 + 1 ⋅ E 4 T\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}=\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix}=0\cdot E_{1}+0\cdot E_{2}+0\cdot E_{3}+1\cdot E_{4} T ( 0 0 0 1 ) = ( 0 0 0 1 ) = 0 ⋅ E 1 + 0 ⋅ E 2 + 0 ⋅ E 3 + 1 ⋅ E 4 [ T ] E = ( 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 ) [T]_{E}=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} [ T ] E = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 P ( t ) = ∣ t I − [ T ] E ∣ = ∣ t − 1 0 0 0 0 t − 1 0 0 − 1 t 0 0 0 0 t − 1 ∣ = ( t − 1 ) 2 ( t 2 − 1 ) = ( t − 1 ) 3 ( t + 1 ) P(t)=|tI-[T]_{E}|=\begin{vmatrix}
t-1 & 0 & 0 & 0 \\
0 & t & -1 & 0 \\
0 & -1 & t & 0 \\
0 & 0 & 0 & t-1
\end{vmatrix}=(t-1)^2(t^2-1)=(t-1)^3(t+1) P ( t ) = ∣ t I − [ T ] E ∣ = t − 1 0 0 0 0 t − 1 0 0 − 1 t 0 0 0 0 t − 1 = ( t − 1 ) 2 ( t 2 − 1 ) = ( t − 1 ) 3 ( t + 1 ) קיבלנו λ 1 = 1 \lambda_{1}=1 λ 1 = 1 λ 2 = − 1 \lambda_{2}=-1 λ 2 = − 1 1 1 1 λ 1 = 1 \lambda_{1}=1 λ 1 = 1
( 1 ⋅ I − [ T ] E ) v ‾ = 0 ‾ (1\cdot I-[T]_{E})\underline{v}=\underline{0} ( 1 ⋅ I − [ T ] E ) v = 0 ( 0 0 0 0 0 1 − 1 0 0 − 1 1 0 0 0 0 0 ) → ( 0 1 − 1 0 0 0 0 0 0 0 0 0 0 0 0 0 ) ⟹ { x 2 = x 3 = s x 1 = r x 4 = t \begin{pmatrix}
0 & 0 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & -1 & 1 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix}\to \begin{pmatrix}
0 & 1 & -1 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{pmatrix} \implies \begin{cases}
x_{2}=x_{3}=s \\
x_{1}=r \\
x_{4}=t
\end{cases} 0 0 0 0 0 1 − 1 0 0 − 1 1 0 0 0 0 0 → 0 0 0 0 1 0 0 0 − 1 0 0 0 0 0 0 0 ⟹ ⎩ ⎨ ⎧ x 2 = x 3 = s x 1 = r x 4 = t [ V λ 1 ] E = { ( r s s t ) r , s , t ∈ R } = s p { ( 1 0 0 0 ) , ( 0 1 1 0 ) , ( 0 0 0 1 ) } [V_{\lambda_{1}}]_{E}=\left\{ \begin{array}{c|c}
\begin{pmatrix}
r \\
s \\
s \\
t
\end{pmatrix} & r,s,t \in \mathbb{R}
\end{array} \right\}=sp\left\{ \begin{pmatrix}
1 \\
0 \\
0 \\
0
\end{pmatrix},\begin{pmatrix}
0 \\
1 \\
1 \\
0
\end{pmatrix},\begin{pmatrix}
0 \\
0 \\
0 \\
1
\end{pmatrix} \right\} [ V λ 1 ] E = ⎩ ⎨ ⎧ r s s t r , s , t ∈ R ⎭ ⎬ ⎫ = s p ⎩ ⎨ ⎧ 1 0 0 0 , 0 1 1 0 , 0 0 0 1 ⎭ ⎬ ⎫ קיבלנו ר"א=ר"ג=3.λ 2 = − 1 \lambda_{2}=-1 λ 2 = − 1
( − 1 ⋅ I − [ T ] E ) v ‾ = 0 ‾ (-1\cdot I-[T]_{E})\underline{v}=\underline{0} ( − 1 ⋅ I − [ T ] E ) v = 0 ( − 2 0 0 0 0 − 1 − 1 0 0 − 1 − 1 0 0 0 0 − 2 ) → ( 1 0 0 0 0 1 1 0 0 0 0 1 0 0 0 0 ) ⟹ { x 1 = 0 x 4 = 0 x 2 = − x 3 = − t ⟹ x 3 = t \begin{pmatrix}
-2 & 0 & 0 & 0 \\
0 & -1 & -1 & 0 \\
0 & -1 & -1 & 0 \\
0 & 0 & 0 & -2
\end{pmatrix}\to \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}\implies \begin{cases}
x_{1}=0 \\
x_{4}=0 \\
x_{2}=-x_{3}=-t \implies x_{3}=t
\end{cases} − 2 0 0 0 0 − 1 − 1 0 0 − 1 − 1 0 0 0 0 − 2 → 1 0 0 0 0 1 0 0 0 1 0 0 0 0 1 0 ⟹ ⎩ ⎨ ⎧ x 1 = 0 x 4 = 0 x 2 = − x 3 = − t ⟹ x 3 = t [ V λ 2 ] E = { ( 0 − t t 0 ) t ∈ R } = s p { ( 0 − 1 1 0 ) } [V_{\lambda_{2}}]_{E}=\left\{ \begin{array}{c|c}
\begin{pmatrix}
0 \\
-t \\
t \\
0
\end{pmatrix} & t \in \mathbb{R}
\end{array} \right\}=sp\left\{ \begin{pmatrix}
0 \\
-1 \\
1 \\
0
\end{pmatrix} \right\} [ V λ 2 ] E = ⎩ ⎨ ⎧ 0 − t t 0 t ∈ R ⎭ ⎬ ⎫ = s p ⎩ ⎨ ⎧ 0 − 1 1 0 ⎭ ⎬ ⎫ V λ 1 = s p { ( 1 0 0 0 ) , ( 0 1 1 0 ) , ( 0 0 0 1 ) } V λ 2 = s p { ( 0 − 1 1 0 ) } \boxed{V_{\lambda_{1}}=sp\left\{ \begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix},\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix},\begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix} \right\} \qquad V_{\lambda_{2}}=sp\left\{ \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix} \right\}} V λ 1 = s p { ( 1 0 0 0 ) , ( 0 1 1 0 ) , ( 0 0 0 1 ) } V λ 2 = s p { ( 0 1 − 1 0 ) } כעת, אנו רוצים למצוא את B B B
B = ( ( 1 0 0 0 ) ( 0 1 1 0 ) ( 0 0 0 1 ) ( 0 − 1 1 0 ) ) B=\begin{pmatrix}
\begin{pmatrix}
1 & 0 \\
0 & 0
\end{pmatrix} & \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix} & \begin{pmatrix}
0 & 0 \\
0 & 1
\end{pmatrix} & \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
\end{pmatrix} B = ( ( 1 0 0 0 ) ( 0 1 1 0 ) ( 0 0 0 1 ) ( 0 1 − 1 0 ) )
תהיינה A , B ∈ F n × n A,B \in \mathbb{F}^{n\times n} A , B ∈ F n × n A ∼ B A \sim B A ∼ B
r ( A ) = r ( B ) .1 ∣ A ∣ = ∣ B ∣ .2 t r ( A ) = t r ( B ) .3 יש להן אותו הפ"א .4 יש להן את אותם ע"ע עם אותו ר"א .5 לכל אחד מהע"ע אותו ר"ג .6 \begin{align*}
r(A)=r(B) & \quad .1 \\
|A|=|B| & \quad .2 \\
tr(A)=tr(B) & \quad .3 \\
\R{\text{יש להן אותו הפ"א}} & \quad .4 \\
\R{\text{יש להן את אותם ע"ע עם אותו ר"א}} & \quad .5 \\
\R{\text{לכל אחד מהע"ע אותו ר"ג}} & \quad .6
\end{align*} r ( A ) = r ( B ) ∣ A ∣ = ∣ B ∣ t r ( A ) = t r ( B ) יש להן אותו הפ"א יש להן את אותם ע"ע עם אותו ר"א לכל אחד מהע"ע אותו ר"ג .1 .2 .3 .4 .5 .6 הוכחה: (6)A = P B P − 1 A=PBP^{-1} A = PB P − 1 P P P
dim V λ 1 ( A ) = n − r ( λ 1 I − A ) = ↓ משפט n − r ( P − 1 ( λ 1 I − A ) P ) = n − r ( λ 1 I − P − 1 A P ) \dim V_{\lambda_{1}(A)}=n-r(\lambda_{1}I-A)\underset{\begin{matrix}
\downarrow \\
\R{\text{משפט}}
\end{matrix}}{=}n-r(P^{-1}(\lambda_{1}I-A)P)=n-r(\lambda_{1}I-P^{-1}AP) dim V λ 1 ( A ) = n − r ( λ 1 I − A ) ↓ משפט = n − r ( P − 1 ( λ 1 I − A ) P ) = n − r ( λ 1 I − P − 1 A P ) = n − r ( λ 1 I − B ) = dim V λ 1 ( B ) =n-r(\lambda_{1}I-B)=\dim V_{\lambda_{1}(B)} = n − r ( λ 1 I − B ) = dim V λ 1 ( B ) ∎
המשפט הזה הוא רק כיוון אחד ולא תקף לכיוון השני.
יהיו A , B ∈ F n × n A,B \in \mathbb{F}^{n\times n} A , B ∈ F n × n A , B A,B A , B
תהא A A A n × n n\times n n × n F \mathbb{F} F
המקדם החופשי של הפ"א הוא ( − 1 ) n ∣ A ∣ = ∣ − A ∣ (-1)^n|A|=|-A| ( − 1 ) n ∣ A ∣ = ∣ − A ∣
המקדם של x n x^n x n
המקדם של x n − 1 x^{n-1} x n − 1 T r ( A ) \mathrm{Tr}(A) Tr ( A )
אם A A A ∣ A ∣ |A| ∣ A ∣ T r ( A ) \mathrm{Tr}(A) Tr ( A )
ניתן להוכיח שסעיף 4 נכון אם הפ"א מתפרק לגורמים ליניאריים.
הוכחה: (1)
רורב רורב רורב p ( t ) = ∣ t I − A ∣ p(t)=|tI-A| p ( t ) = ∣ t I − A ∣ p ( 0 ) = ∣ 0 ⋅ I − A ∣ = ∣ − A ∣ = ( − 1 ) n ∣ A ∣ p(0)=|0\cdot I-A|=|-A|=(-1)^n|A| p ( 0 ) = ∣0 ⋅ I − A ∣ = ∣ − A ∣ = ( − 1 ) n ∣ A ∣ ∎
הוכחה: (2)
רורב רורב רורב ∎
הוכחה: (4)A A A D = ( λ 1 0 ⋱ 0 λ n ) D=\begin{pmatrix}\lambda_{1}&&0\\& \ddots\\0&&\lambda_{n}\end{pmatrix} D = λ 1 0 ⋱ 0 λ n
∣ A ∣ = ∣ D ∣ = λ 1 … λ n |A|=|D|=\lambda_{1}\dots\lambda_{n} ∣ A ∣ = ∣ D ∣ = λ 1 … λ n T r ( A ) = T r ( D ) = λ 1 + ⋯ + λ n \mathrm{Tr}(A)=\mathrm{Tr}(D)=\lambda_{1}+\dots+\lambda_{n} Tr ( A ) = Tr ( D ) = λ 1 + ⋯ + λ n ∎
אם λ \lambda λ A A A λ − 1 \lambda^{-1} λ − 1 A − 1 A^{-1} A − 1
** הוכחה:**
A v ‾ = λ v ‾ A\underline{v}=\lambda \underline{v} A v = λ v v ‾ = λ A − 1 v ‾ \underline{v}=\lambda A^{-1}\underline{v} v = λ A − 1 v A − 1 v ‾ = 1 λ v ‾ A^{-1}\underline{v}= \frac{1}{\lambda}\underline{v} A − 1 v = λ 1 v ∎
אם λ \lambda λ A A A λ k \lambda^k λ k A k A^k A k p ( x ) p(x) p ( x ) λ \lambda λ A A A p ( λ ) p(\lambda) p ( λ ) p ( A ) p(A) p ( A )
הוכחה:
A v ‾ = λ v ‾ A\underline{v}=\lambda \underline{v} A v = λ v A 2 v ‾ = λ A v ‾ = λ ( λ v ‾ ) = λ 2 v ‾ A^2\underline{v}=\lambda A\underline{v}=\lambda(\lambda \underline{v})=\lambda^2\underline{v} A 2 v = λ A v = λ ( λ v ) = λ 2 v ⋮ \vdots ⋮ באופן כללי - ברור, עבור פולינום.∎
כמה הגדרות לסיכום
מתקיים ∣ T ∣ = ∣ [ T ] B ∣ |T|=|[T]_{B}| ∣ T ∣ = ∣ [ T ] B ∣
הפ"א של T T T ∣ λ I − T ∣ |\lambda I-T| ∣ λ I − T ∣
∣ λ I − T ∣ = ∣ [ λ I − T ] B ∣ = ∣ [ λ I ] B − [ T ] B ∣ = ∣ λ I − [ T ] B ∣ |\lambda I-T|=|[\lambda I-T]_{B}|=|[\lambda I]_{B}-[T]_{B}|=|\lambda I-[T]_{B}| ∣ λ I − T ∣ = ∣ [ λ I − T ] B ∣ = ∣ [ λ I ] B − [ T ] B ∣ = ∣ λ I − [ T ] B ∣
הע"ע של T T T
אם α \alpha α T T T v ‾ \underline{v} v α \alpha α ⟺ \iff ⟺ v ‾ ∈ ker ( α I − T ) \underline{v} \in \ker(\alpha I-T) v ∈ ker ( α I − T ) v ‾ ≠ 0 ‾ \underline{v}\neq \underline{0} v = 0
V α = ker ( α I − T ) V_{\alpha}=\ker(\alpha I-T) V α = ker ( α I − T ) V V V V α V_{\alpha} V α α \alpha α
λ \lambda λ A A A ⟺ \iff ⟺ λ \lambda λ A t A^t A t ל-A B AB A B B A BA B A
הוכחה: (2)
∣ t I − A ∣ = ∣ ( t I − A ) t ∣ = ∣ ( t I ) t − A t ∣ = ∣ t I − A t ∣ |tI-A|=|(tI-A)^t|=|(tI)^t-A^t|=|tI-A^t| ∣ t I − A ∣ = ∣ ( t I − A ) t ∣ = ∣ ( t I ) t − A t ∣ = ∣ t I − A t ∣ ∎
אם סכום השורות(או העמודות) במטריצה ריבועית A A A c c c c c c A A A ( 1 1 1 1 ) \begin{pmatrix}1\\1\\1\\1\end{pmatrix} 1 1 1 1
הוכחה:
רורב רורב רורב ∎
האם המטריצה ( 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 ) \begin{pmatrix}1&2&3&4&5\\1&2&3&4&5\\1&2&3&4&5\\1&2&3&4&5\\1&2&3&4&5\end{pmatrix} 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5 P P P D D D A = P D P − 1 A=PDP^{-1} A = P D P − 1
15 ע"ע עם ר"א לפחות 1.
dim V λ = 0 = 5 − r ( 0 ⋅ I − A ) = 5 − 1 = 4 \dim V_{\lambda=0}=5-r(0\cdot I-A)=5-1=4 dim V λ = 0 = 5 − r ( 0 ⋅ I − A ) = 5 − 1 = 4 ולכן, עם ר"א לפחות 4.
15 ע"ע עם ר"א=ר"ג=1
D = ( 15 0 0 0 ) D=\begin{pmatrix}
15 & 0 \\
0 & 0
\end{pmatrix} D = ( 15 0 0 0 ) את P P P
תהא A ∈ R 4 × 4 A \in \mathbb{R}^{4\times4} A ∈ R 4 × 4
∣ A − 2 I ∣ = 0 ∧ r ( 3 I − A ) = 2 |A-2I|=0 \land r(3I-A)=2 ∣ A − 2 I ∣ = 0 ∧ r ( 3 I − A ) = 2 א. מצאו את הפ"א של A A A I − A 2 I-A^2 I − A 2 I − A 2 I-A^2 I − A 2