Question

5 4 2 1. Give A 4 5 2 (1) Numerically prove that A has an orthonormal basis of eigenvectors. (10%) (2) Find A5 by stating a proper similarity transformation (13%)

Please answer both questions.. thank you! :)

Answer 1

5 4 2 A-4 5 2

(A-xI) is

001 010 100 2 452 542

3 2 4-2 一42 5

take determinant=0

45-0 2 (5 - x) det2 2-r 4 det 2 2+2 det

(5-2) (2.2-7x+6) _ 4 (-4x + 4) + 2 (2.r-2)=0

12r241r 30+16x-16+4.x-40

$-x^3+12x^2-21x+10=0$

(z-1) (2-1)(x-10) = 0

(2-1)2 (2-10)-0

${\color{Red} x=1,\:x=10}$..............eigenvalues

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for x=1, (A-xI) is

$\begin{pmatrix}5&4&2\\ 4&5&2\\ 2&2&2\end{pmatrix}-1\cdot \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$

$=\begin{pmatrix}4&4&2\\ 4&4&2\\ 2&2&1\end{pmatrix}$

reduced this matrix

1200 100 100

reduced system is

$\begin{pmatrix}1&1&\frac{1}{2}\\ 0&0&0\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$

$x+y+\frac{z}{2}=0...................x=-y-\frac{z}{2}$

$y=y$.....................free

$z=z$.....................free

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general solution is

12 0 1 10

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eigenvectors are

${\color{Red} \begin{pmatrix}-1\\ \:1\\ \:0\end{pmatrix},\:\begin{pmatrix}-\frac{1}{2}\\ \:0\\ \:1\end{pmatrix}}$

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for x=10, (A-xI) is

$\begin{pmatrix}5&4&2\\ 4&5&2\\ 2&2&2\end{pmatrix}-10\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$

-5 42 4-5 2 2 2 -8

reduce the matrix

$\begin{pmatrix}1&0&-2\\ 0&1&-2\\ 0&0&0\end{pmatrix}$

reduced system is

$\begin{pmatrix}1&0&-2\\ 0&1&-2\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\end{pmatrix}$

$x-2z=0...................x=2z$

$y-2z=0...................y=2z$

$z=z$....................free

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general solution is

$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}z$

eigenvector is

${\color{Red} \begin{pmatrix}2\\ 2\\ 1\end{pmatrix}}$

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total 3 eigenvectors are

$v_1=\begin{pmatrix}-1\\ \:1\\ \:0\end{pmatrix},v_2=\begin{pmatrix}-\frac{1}{2}\\ \:0\\ \:1\end{pmatrix},\:v_3=\begin{pmatrix}2\\ \:2\\ \:1\end{pmatrix}$

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${\color{Blue} v_1=u_1=\begin{pmatrix}-1\\ \:1\\ \:0\end{pmatrix} }$

orthonormal vector is

$u_1'=\frac{1}{\sqrt{\left(-1\right)^2+1^2+0^0}}\begin{pmatrix}-1\\ \:1\\ \:0\end{pmatrix}$

$u_1'=\frac{1}{\sqrt{2}}\begin{pmatrix}-1\\ \:1\\ \:0\end{pmatrix}$

${\color{Red} u_1'= \begin{pmatrix}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\\ 0\end{pmatrix}}$

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$u_2=v_2-\frac{v_2 \cdot u_1}{u_1 \cdot u_1}u_1$

$u_2=\begin{pmatrix}-\frac{1}{2}\\ \:\:0\\ \:\:1\end{pmatrix}-\frac{\begin{pmatrix}-\frac{1}{2}\\ \:\:0\\ \:\:1\end{pmatrix}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}}{\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}$

$u_2=\begin{pmatrix}-\frac{1}{2}\\ \:\:0\\ \:\:1\end{pmatrix}-\frac{\frac{1}{2}+0+0}{1+1}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}$

$u_2=\begin{pmatrix}-\frac{1}{2}\\ \:\:0\\ \:\:1\end{pmatrix}-\frac{\frac{1}{2}}{2}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}$

$u_2=\begin{pmatrix}-\frac{1}{2}\\ \:\:0\\ \:\:1\end{pmatrix}-\frac{1}{4}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}$

${\color{Blue} u_2=\begin{pmatrix}-\frac{1}{4}\\ -\frac{1}{4}\\ 1\end{pmatrix}}$

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orthonormal vector is

$u_2'= \frac{1}{\sqrt{\left(-\frac{1}{4}\right)^2+\left(-\frac{1}{4}\right)^2+1^2}}\begin{pmatrix}-\frac{1}{4}\\ -\frac{1}{4}\\ 1\end{pmatrix}$

${\color{Red} u_2'= \begin{pmatrix}-\frac{1}{3\sqrt{2}}\\ -\frac{1}{3\sqrt{2}}\\ \frac{2\sqrt{2}}{3}\end{pmatrix}}$

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$u_3=\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}-\frac{\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}\:\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}}{\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}-\frac{\begin{pmatrix}2\\ \:2\\ \:1\end{pmatrix}\begin{pmatrix}-\frac{1}{4}\\ \:-\frac{1}{4}\\ \:1\end{pmatrix}}{\begin{pmatrix}-\frac{1}{4}\\ \:-\frac{1}{4}\\ \:1\end{pmatrix}\begin{pmatrix}-\frac{1}{4}\\ \:-\frac{1}{4}\\ \:1\end{pmatrix}}\begin{pmatrix}-\frac{1}{4}\\ \:-\frac{1}{4}\\ \:1\end{pmatrix}$

$u_3=\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}-\frac{-2+2+0}{-1+1+0}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}-\frac{-\frac{1}{2}-\frac{1}{2}+1}{\frac{1}{16}+\frac{1}{16}+1}\begin{pmatrix}-\frac{1}{4}\\ \:-\frac{1}{4}\\ \:1\end{pmatrix}$

$u_3=\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}-\frac{0}{-1+1+0}\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}-\frac{0}{\frac{1}{16}+\frac{1}{16}+1}\begin{pmatrix}-\frac{1}{4}\\ \:-\frac{1}{4}\\ \:1\end{pmatrix}$

$u_3=\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}-0\begin{pmatrix}-1\\ \:\:1\\ \:\:0\end{pmatrix}-0\begin{pmatrix}-\frac{1}{4}\\ \:-\frac{1}{4}\\ \:1\end{pmatrix}$

${\color{Blue} u_3=\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}}$

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orthonormal vector is

$u_3 '=\frac{1}{\sqrt{2^2+2^2+1^2}}\begin{pmatrix}2\\ 2\\ 1\end{pmatrix}$

${\color{Red} u_3 '= \begin{pmatrix}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{pmatrix}}$

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orthonormal basis are

${\color{Red} u_1'= \begin{pmatrix}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\\ 0\end{pmatrix}}$${\color{Red} u_2'= \begin{pmatrix}-\frac{1}{3\sqrt{2}}\\ -\frac{1}{3\sqrt{2}}\\ \frac{2\sqrt{2}}{3}\end{pmatrix}}$${\color{Red} u_3 '= \begin{pmatrix}\frac{2}{3}\\ \frac{2}{3}\\ \frac{1}{3}\end{pmatrix}}$

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D=eigenvalue matrix .

${\color{Red} D=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&10\end{pmatrix}}$

P=eigenvector matrix

${\color{Red} P=\begin{pmatrix}-1&-\frac{1}{2}&2\\ 1&0&2\\ 0&1&1\end{pmatrix}}$

inverse of P is

${\color{Red} P^{-1}=\begin{pmatrix}-\frac{4}{9}&\frac{5}{9}&-\frac{2}{9}\\ -\frac{2}{9}&-\frac{2}{9}&\frac{8}{9}\\ \frac{2}{9}&\frac{2}{9}&\frac{1}{9}\end{pmatrix}}$

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as we know

$A^n=PD^nP^{-1}$

to A⁵ put n=5

$A^5=PD^5P^{-1}$

$A^5= \begin{pmatrix}-1&-\frac{1}{2}&2\\ \:1&0&2\\ \:0&1&1\end{pmatrix}\begin{pmatrix}1^5&0&0\\ \:0&1^5&0\\ \:0&0&10^5\end{pmatrix}\begin{pmatrix}-\frac{4}{9}&\frac{5}{9}&-\frac{2}{9}\\ \:-\frac{2}{9}&-\frac{2}{9}&\frac{8}{9}\\ \:\frac{2}{9}&\frac{2}{9}&\frac{1}{9}\end{pmatrix}$

$A^5= \begin{pmatrix}-1&-\frac{1}{2}&2\\ \:1&0&2\\ \:0&1&1\end{pmatrix}\begin{pmatrix}1&0&0\\ \:0&1&0\\ \:0&0&100000\end{pmatrix}\begin{pmatrix}-\frac{4}{9}&\frac{5}{9}&-\frac{2}{9}\\ \:-\frac{2}{9}&-\frac{2}{9}&\frac{8}{9}\\ \:\frac{2}{9}&\frac{2}{9}&\frac{1}{9}\end{pmatrix}$

$A^5= \begin{pmatrix}-1&-\frac{1}{2}&200000\\ 1&0&200000\\ 0&1&100000\end{pmatrix}\begin{pmatrix}-\frac{4}{9}&\frac{5}{9}&-\frac{2}{9}\\ -\frac{2}{9}&-\frac{2}{9}&\frac{8}{9}\\ \frac{2}{9}&\frac{2}{9}&\frac{1}{9}\end{pmatrix}$

44445 44444 22222 A"= 44444 44445 22222 22222 22222 11112