Answer to Question #162957 in Linear Algebra for Ra

Question

To determine the eigenvalues and eigenvectors of the following matrix :

2 -2 0

P = -2 1 -2

0 -2 0

Answer 1

$P=\left(\begin{array}{rrr}2&-2&0\\-2&1&-2\\0&-2&0\end{array}\right)$

Eigenvalues $\lambda_1$, $\lambda_2$, $\lambda_3$ are the roots of the characteristic polynomial of $P$ .

Obtain the characteristic polynomial taking the determinant

$\left|\begin{array}{rrr}2-\lambda&-2&0\\-2&1-\lambda&-2\\0&-2&-\lambda\end{array}\right|=$

$(2-\lambda)(1-\lambda)(-\lambda)+(-2)\cdot(-2)\cdot0+$ $0\cdot(-2)\cdot(-2)-$

$0\cdot(1-\lambda)\cdot0-(-2)\cdot(-2)\cdot(-\lambda)-$ $(2-\lambda)\cdot(-2)\cdot(-2)=$

$(2-\lambda-2\lambda+\lambda^2)(-\lambda)+0+0-0+4\lambda-(2-\lambda)\cdot4=$

$(2-3\lambda+\lambda^2)(-\lambda)+4\lambda-8+4\lambda=$

$-2\lambda+3\lambda^2-\lambda^3+8\lambda-8=$

$-\lambda^3+3\lambda^2+6\lambda-8$

Find the roots factoring the polynomial

$-(\lambda^3-3\lambda^2-6\lambda+8)=$

$-(\lambda^3+8-3\lambda^2-6\lambda)=$

$-((\lambda^3+8)-(3\lambda^2+6\lambda))=$

$-((\lambda^3+2^3)-3\lambda(\lambda+2))=$

$-((\lambda+2)(\lambda^2-2\lambda+2^2)-3\lambda(\lambda+2))=$

$-((\lambda+2)(\lambda^2-2\lambda+4)-3\lambda(\lambda+2))=$

$-(\lambda+2)(\lambda^2-2\lambda+4-3\lambda)=$

$-(\lambda+2)(\lambda^2-5\lambda+4)=$

$-(\lambda+2)(\lambda^2-4\lambda-\lambda+4)=$

$-(\lambda+2)((\lambda^2-4\lambda)-(\lambda-4))=$

$-(\lambda+2)(\lambda(\lambda-4)-(\lambda-4))=$

$-(\lambda+2)(\lambda-1)(\lambda-4)$

So the roots of polynomial are

$\lambda_1=-2$,

$\lambda_2=1$,

$\lambda_3=4$.

Find eigenvectors solving the homogeneous system substituting $\lambda$ by $-2$, $1$, $4$

$\left\{\begin{array}{rrrr}(2-\lambda)x_1&-2x_2&&=0\\-2x_1&+(1-\lambda)x_2&-2x_3&=0\\&-2x_2&-\lambda x_3&=0\end{array}\right.$

$\lambda=\lambda_1=-2$

$\left\{\begin{array}{rrrr}4x_1&-2x_2&&=0\\-2x_1&+3x_2&-2x_3&=0\\&-2x_2&+2x_3&=0\end{array}\right.$

Use Gaussian elimination modifying the augmented matrix

$\left(\begin{array}{rrr}4&-2&0\\-2&3&-2\\0&-2&2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim$ $L_2+\frac12L_1\to L_2\sim$

$\left(\begin{array}{lll}4&-2&0\\-2+\frac12\cdot4&3+\frac12\cdot(-2)&-2+\frac12\cdot0\\0&-2&2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=$

$\left(\begin{array}{rrr}4&-2&0\\0&2&-2\\0&-2&2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim$ $L_3+L_2\to L_3\sim$

$\left(\begin{array}{rrr}4&-2&0\\0&2&-2\\0+0&-2+2&2-2\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=$

$\left(\begin{array}{rrr}4&-2&0\\0&2&-2\\0&0&0\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)$.

System represented as the matryx is

$\left\{\begin{array}{rrrr}4x_1&-2x_2&&=0\\&+2x_2&-2x_3&=0\end{array}\right.$

Let $x_3$ be free while $x_1$ and $x_2$ be dependent on $x_3$ . Then

$\left\{\begin{array}{rrrr}4x_1&&&=2x_2\\&2x_2&&=2x_3\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rrrr}4x_1&&&=2x_3\\&x_2&&=x_3\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rrrr}x_1&&&=\frac12x_3\\&x_2&&=x_3\end{array}\right.$

Thus, the eigenvector corresponding to the eigenvalue $\lambda_1=-2$ is

$\bold{v}_1=\left(\begin{array}{r}x_1\\x_2\\x_3\end{array}\right)=$ $\left(\begin{array}{r}\frac12x_3\\x_3\\x_3\end{array}\right)$.

Here $x_3$ is allowed to take any value. In particular, if $x_3=1$

$\bold{v}_1=\left(\begin{array}{r}\frac12\\1\\1\end{array}\right)$.

$\lambda=\lambda_2=1$

$\left\{\begin{array}{rrrr}x_1&-2x_2&&=0\\-2x_1&&-2x_3&=0\\&-2x_2&-x_3&=0\end{array}\right.$

Use Gaussian elimination modifying the augmented matrix

$\left(\begin{array}{rrr}1&-2&0\\-2&0&-2\\0&-2&-1\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim$ $L_2+2L_1\to L_2\sim$

$\left(\begin{array}{lll}1&-2&0\\-2+2\cdot1&0+2\cdot(-2)&-2+ 2\cdot0\\0&-2&-1\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=$

$\left(\begin{array}{rrr}1&-2&0\\0&-4&-2\\0&-2&-1\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim$ $L_3-\frac12L_2\to L_3\sim$

$\left(\begin{array}{lll}1&-2&0\\0&-4&-2\\0-\frac12\cdot0&-2-\frac12\cdot(-4)&-1-\frac12\cdot(-2)\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=$

$\left(\begin{array}{rrr}1&-2&0\\0&-4&-2\\0&0&0\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)$

System represented as the matryx is

$\left\{\begin{array}{rrrr}x_1&-2x_2&&=0\\&-4x_2&-2x_3&=0\end{array}\right.$

Let $x_3$ be free while $x_1$ and $x_2$ be dependent on $x_3$ . Then

$\left\{\begin{array}{rrrr}x_1&&&=2x_2\\&-4x_2&&=2x_3\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rrrl}x_1&&&=2\left(-\frac12x_3\right)\\&x_2&&=-\frac12x_3\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rrrl}x_1&&&=-x_3\\&x_2&&=-\frac12x_3\end{array}\right.$

Thus, the eigenvector corresponding to the eigenvalue $\lambda_2=1$ is

$\bold{v}_2=\left(\begin{array}{r}x_1\\x_2\\x_3\end{array}\right)=$ $\left(\begin{array}{r}-x_3\\-\frac12x_3\\x_3\end{array}\right)$.

Here $x_3$ is allowed to take any value. In particular, if $x_3=1$

$\bold{v}_2=\left(\begin{array}{r}-1\\-\frac12\\1\end{array}\right)$.

$\lambda=\lambda_3=4$

$\left\{\begin{array}{rrrr}-2x_1&-2x_2&&=0\\-2x_1&-3x_2&-2x_3&=0\\&-2x_2&-4x_3&=0\end{array}\right.$

Use Gaussian elimination modifying the augmented matrix

$\left(\begin{array}{rrr}-2&-2&0\\-2&-3&-2\\0&-2&-4\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim$ $L_2-L_1\to L_2\sim$

$\left(\begin{array}{lll}-2&-2&0\\-2-(-2)&-3-(-2)&-2-0\\0&-2&-4\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=$

$\left(\begin{array}{rrr}-2&-2&0\\0&-1&-2\\0&-2&-4\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)\sim$ $L_3-2L_2\to L_3\sim$

$\left(\begin{array}{lll}-2&-2&0\\0&-1&-2\\0-2\cdot0&-2-2\cdot(-1)&-4-2\cdot(-2)\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)=$

$\left(\begin{array}{rrr}-2&-2&0\\0&-1&-2\\0&0&0\end{array}\right|\left.\begin{array}{r}0\\0\\0\end{array}\right)$ .

System represented as the matryx is

$\left\{\begin{array}{rrrr}-2x_1&-2x_2&&=0\\&-x_2&-2x_3&=0\end{array}\right.$

Let $x_3$ be free while $x_1$ and $x_2$ be dependent on $x_3$ . Then

$\left\{\begin{array}{rrrr}-2x_1&&&=2x_2\\&-x_2&&=2x_3\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rrrl}x_1&&&=-x_2\\&x_2&&=-2x_3\end{array}\right.\Rightarrow$ $\left\{\begin{array}{rrrl}x_1&&&=2x_3\\&x_2&&=-2x_3\end{array}\right.$

Thus, the eigenvector corresponding to the eigenvalue $\lambda_2=1$ is

$\bold{v}_3=\left(\begin{array}{r}x_1\\x_2\\x_3\end{array}\right)=$ $\left(\begin{array}{r}2x_3\\-2x_3\\x_3\end{array}\right)$.

Here $x_3$ is allowed to take any value. In particular, if $x_3=1$

$\bold{v}_3=\left(\begin{array}{r}2\\-2\\1\end{array}\right)$.

Answer:

Eigenvalues

$\lambda_1=-2$, $\lambda_2=1$, $\lambda_3=4$.

Eigenvectors

$\bold{v}_1=\left(\begin{array}{r}\frac12\\1\\1\end{array}\right)t_1$, $\bold{v}_2=\left(\begin{array}{r}-1\\-\frac12\\1\end{array}\right)t_2$, $\bold{v}_3=\left(\begin{array}{r}2\\-2\\1\end{array}\right)t_3$,

where factors $t_1$ , $t_2$ , $t_3$ may take any value.