【Linear Algebra】Eigen Vectors and Eigen Values

Introduction

In the last post, we require you should have knowledge about Linear algebra, like eigen vectors and eigen values. So now, we can learn what eigen vectors and eigen values is.

Eigen Vectors and Eigen Values

For a arbitary square matirx $A$, it is exist an eigen vector and eigen value make this equation true.
$Av = \lambda v$
where the $v$ is the eigen vector and the $\lambda$ is the eigen value.

So, how do we find these things?

Firstly, we know this equation must be true.
$Av=\lambda v$
Now, let us put an identity matrix in right side.
$Av=\lambda I v$
Then, bring all to left side.
$Av-\lambda Iv=0$
If $v$ is non-zero then we can change the thing of left side to the determinant.
$|A-\lambda I|=0$
Then we can solve for $\lambda$ using this determinant.

Example: Solve for $\lambda$
Start with $|A-\lambda I|=0$ :
$\begin{vmatrix} \begin {bmatrix} -6 & 3\\ 4 & 5\\ \end{bmatrix} -\lambda \begin {bmatrix} 1 & 0\\ 0 & 1\\ \end{bmatrix}\end{vmatrix} =0$
Then:
$\begin{vmatrix} -6-\lambda & 3\\ 4 & 5-\lambda \end{vmatrix} =0$
Gets:
$(-6-\lambda)(5-\lambda)-3\times4=0$
Then get this quadratic equation:
$\lambda^2+\lambda-42=0$
And solve for $\lambda$:
$\lambda_1=-7 \quad \lambda_2=6$
Finally, there two possible eigen values.

Now we get the eigenvalues, then we can solve for their matching eigen vectors.

Start with:
$Av=\lambda v$
Put in the value we know:
$\begin {bmatrix} -6 & 3\\ 4 & 5\\ \end{bmatrix} \begin {bmatrix} x\\ y\\ \end{bmatrix} = 6 \begin {bmatrix} x\\ y\\ \end{bmatrix}$
Then we can two equations:
$\begin{aligned} -6x+3y =&\ 6x\\ 4x+5y =&\ 6y\\ \end{aligned}$
Then we get $y=4x$, so the eigen vector is any non-zero multiple of this:
$\begin {bmatrix} 1\\ 4\\ \end{bmatrix}$
So now we can get the solution:
$Av = \begin {bmatrix} -6 & 3\\ 4 & 5\\ \end{bmatrix}\begin {bmatrix} 1\\ 4\\ \end{bmatrix} = \begin {bmatrix} 6\\ 24\\ \end{bmatrix}$
And:
$\lambda v = 6 \begin {bmatrix} 1\\ 4\\ \end{bmatrix} = \begin {bmatrix} 6\\ 24\\ \end{bmatrix}$
So, get $Av=\lambda v$

Finally, we get the solution of eigen values and eigen vector.
Now, we know how to solve for eigen values and eigen vector of a arbitary matrix.
Try to find the eigen vector for the other eigenvalue of $-7$.

Renferences

Eigenvector and Eigenvalue