Question

Q-Declare a matrix
1 2 3
0 4 5
0 0 3
Find eigen values and eigen vector.​

Answer 1

We have the matrix,

$\longrightarrow A=\left[\begin{array}{ccc}1&2&3\\0&4&5\\0&0&3\end{array}\right]$

We need to find its eigenvalues and eigenvectors.

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Eigenvalues of a square matrix $A$ of order n are the possible values of a complex number $\lambda,$ such that the determinant of the matrix $A-\lambda I$ equals zero, i.e., $|A-\lambda I|=0,$ where I is the identity matrix of order n.

Eigenvector of a square matrix $A$ of order n, for a particular eigenvalue $\lambda,$ is any non - zero complex multiple of an n × 1 matrix $X$ such that $(A-\lambda I)X=O,$ where $O$ is a zero matrix of order n × 1.

[Note:- Never misinterpret 'complex number' as only numbers with non - zero imaginary part, as complex numbers include every real numbers also.]

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The matrix $A-\lambda I$ is simply the matrix $A$ but $\lambda$ is subtracted from each element of main diagonal, i.e.,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}1-\lambda&2&3\\0&4-\lambda&5\\0&0&3-\lambda\end{array}\right]$

Now,

$\longrightarrow|A-\lambda I|=0,$

$\longrightarrow\left|\begin{array}{ccc}1-\lambda&2&3\\0&4-\lambda&5\\0&0&3-\lambda\end{array}\right|=0$

Expanding along $C_1,$

$\longrightarrow(1-\lambda)\big((4-\lambda)(3-\lambda)-5\cdot0\big)=0$

$\longrightarrow(1-\lambda)(4-\lambda)(3-\lambda)=0$

From this we get,

$\longrightarrow\underline{\underline{\lambda\in\left\{1,\ 3,\ 4\right\}}}$

These are the eigenvalues.

Let the eigenvector, the 3×1 matrix $X,$ be,

$\longrightarrow X=\left[\begin{array}{c}a\\b\\c\end{array}\right]$

Take $\lambda=1.$ Then,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}0&2&3\\0&3&5\\0&0&2\end{array}\right]$

Now,

$\longrightarrow(A-\lambda I)X=O$

$\longrightarrow\left[\begin{array}{ccc}0&2&3\\0&3&5\\0&0&2\end{array}\right]\cdot\left[\begin{array}{c}a\\b\\c\end{array}\right]=0$

$\longrightarrow\left[\begin{array}{c}2b+3c\\3b+5c\\2c\end{array}\right]=0$

Then we get,

• $2b+3c=0$

• $3b+5c=0$

• $2c=0$

Solving them we get,

• $b=0$

• $c=0$

$a$ is free variable here. Take $a=1.$

Hence,

$\longrightarrow\underline{\underline{X=k\left[\begin{array}{c}1\\0\\0\end{array}\right],\quad k\in\mathbb{C}-\{0\}}}$

This is the corresponding eigenvector for $\lambda=1.$

Take $\lambda=3.$ Then,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}-2&2&3\\0&1&5\\0&0&0\end{array}\right]$

Now,

$\longrightarrow(A-\lambda I)X=O$

$\longrightarrow\left[\begin{array}{ccc}-2&2&3\\0&1&5\\0&0&0\end{array}\right]\cdot\left[\begin{array}{c}a\\b\\c\end{array}\right]=0$

$\longrightarrow\left[\begin{array}{c}-2a+2b+3c\\b+5c\\0\end{array}\right]=0$

Then we get,

• $-2a+2b+3c=0$

• $b+5c=0$

Solving them we get,

• $b:c=-5:1$

• $a:c=7:-2$

Combining both we get,

• $a:b:c=7:10:-2$

Hence,

$\longrightarrow\underline{\underline{X=k\left[\begin{array}{c}7\\10\\-2\end{array}\right],\quad k\in\mathbb{C}-\{0\}}}$

This is the corresponding eigenvector for $\lambda=3.$

Take $\lambda=4.$ Then,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}-3&2&3\\0&0&5\\0&0&-1\end{array}\right]$

Now,

$\longrightarrow(A-\lambda I)X=O$

$\longrightarrow\left[\begin{array}{ccc}-3&2&3\\0&0&5\\0&0&-1\end{array}\right]\cdot\left[\begin{array}{c}a\\b\\c\end{array}\right]=0$

$\longrightarrow\left[\begin{array}{c}-3a+2b+3c\\5c\\-c\end{array}\right]=0$

Then we get,

• $-3a+2b+3c=0$

• $5c=0$

• $-c=0$

Solving them we get,

• $c=0$

• $a:b=2:3$

Hence,

$\longrightarrow\underline{\underline{X=k\left[\begin{array}{c}2\\3\\0\end{array}\right],\quad k\in\mathbb{C}-\{0\}}}$

This is the corresponding eigenvector for $\lambda=4.$

Answer 2

$\huge{\underline{\underline{\mathrm{\red{AnswEr}}}}}$

We have the matrix,

$\longrightarrow A=\left[\begin{array}{ccc}1&2&3\\0&4&5\\0&0&3\end{array}\right]$

We need to find its eigenvalues and eigenvectors.

____________________________________

Eigenvalues of a square matrix $A$ of order n are the possible values of a complex number $\lambda,$ such that the determinant of the matrix $A-\lambda I$ equals zero, i.e., $|A-\lambda I|=0,$ where I is the identity matrix of order n.

Eigenvector of a square matrix $A$ of order n, for a particular eigenvalue $\lambda,$ is any non - zero complex multiple of an n × 1 matrix $X$ such that $(A-\lambda I)X=O,$ where $O$ is a zero matrix of order n × 1.

[Note:- Never misinterpret 'complex number' as only numbers with non - zero imaginary part, as complex numbers include every real numbers also.]

____________________________________

The matrix $A-\lambda I$ is simply the matrix $A$ but $\lambda$ is subtracted from each element of main diagonal, i.e.,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}1-\lambda&2&3\\0&4-\lambda&5\\0&0&3-\lambda\end{array}\right]$

Now,

$\longrightarrow|A-\lambda I|=0,$

$\longrightarrow\left|\begin{array}{ccc}1-\lambda&2&3\\0&4-\lambda&5\\0&0&3-\lambda\end{array}\right|=0$

Expanding along $C_1,$

$\longrightarrow(1-\lambda)\big((4-\lambda)(3-\lambda)-5\cdot0\big)=0$

$\longrightarrow(1-\lambda)(4-\lambda)(3-\lambda)=0$

From this we get,

$\longrightarrow\underline{\underline{\lambda\in\left\{1,\ 3,\ 4\right\}}}$

These are the eigenvalues.

Let the eigenvector, the 3×1 matrix $X,$ be,

$\longrightarrow X=\left[\begin{array}{c}a\\b\\c\end{array}\right]$

Take $\lambda=1.$ Then,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}0&2&3\\0&3&5\\0&0&2\end{array}\right]$

Now,

$\longrightarrow(A-\lambda I)X=O$

$\longrightarrow\left[\begin{array}{ccc}0&2&3\\0&3&5\\0&0&2\end{array}\right]\cdot\left[\begin{array}{c}a\\b\\c\end{array}\right]=0$

$\longrightarrow\left[\begin{array}{c}2b+3c\\3b+5c\\2c\end{array}\right]=0$

Then we get,

$2b+3c=0$

$3b+5c=0$

$2c=0$

Solving them we get,

$b=0$

$c=0$

$a$ is free variable here. Take $a=1.$

Hence,

$\longrightarrow\underline{\underline{X=k\left[\begin{array}{c}1\\0\\0\end{array}\right],\quad k\in\mathbb{C}-\{0\}}}$

This is the corresponding eigenvector for $\lambda=1.$

Take $\lambda=3.$ Then,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}-2&2&3\\0&1&5\\0&0&0\end{array}\right]$

Now,

$\longrightarrow(A-\lambda I)X=O$

$\longrightarrow\left[\begin{array}{ccc}-2&2&3\\0&1&5\\0&0&0\end{array}\right]\cdot\left[\begin{array}{c}a\\b\\c\end{array}\right]=0$

$\longrightarrow\left[\begin{array}{c}-2a+2b+3c\\b+5c\\0\end{array}\right]=0$

Then we get,

$-2a+2b+3c=0$

$b+5c=0$

Solving them we get,

$b:c=-5:1$

$a:c=7:-2$

Combining both we get,

$a:b:c=7:10:-2$

Hence,

$\longrightarrow\underline{\underline{X=k\left[\begin{array}{c}7\\10\\-2\end{array}\right],\quad k\in\mathbb{C}-\{0\}}}$

This is the corresponding eigenvector for $\lambda=3.$

Take $\lambda=4.$ Then,

$\longrightarrow A-\lambda I=\left[\begin{array}{ccc}-3&2&3\\0&0&5\\0&0&-1\end{array}\right]$

Now,

$\longrightarrow(A-\lambda I)X=O$

$\longrightarrow\left[\begin{array}{ccc}-3&2&3\\0&0&5\\0&0&-1\end{array}\right]\cdot\left[\begin{array}{c}a\\b\\c\end{array}\right]=0$

$\longrightarrow\left[\begin{array}{c}-3a+2b+3c\\5c\\-c\end{array}\right]=0$

Then we get,

$-3a+2b+3c=0$

$5c=0$

$-c=0$

Solving them we get,

$c=0$

$a:b=2:3$

Hence,

$\longrightarrow\underline{\underline{X=k\left[\begin{array}{c}2\\3\\0\end{array}\right],\quad k\in\mathbb{C}-\{0\}}}$

This is the corresponding eigenvector for $\lambda=4.$