How to find eigenvectors of this matrix

Question

I want to find eigenvectors of the following matrix manually.

$$ A = \begin{bmatrix} 300 & 100 & 75 \\ 100 & 200 & 50 \\ 75 & 50 & 100 \end{bmatrix} $$

I found eigenvalues as $$ A1 = 389.809 \\ A2 = 71.864 \\ A3 = 138.325 $$

I substitute A1, in the equation of form

$$ (A - \lambda I ) \overline x = 0 $$ $ \overline x $ are eigenvectors.

When I do this substitution, I get following matrix:

A= \begin{bmatrix} -89.809 & 100 & 75 \\ 100 & -189.809 & 50 \\ 75 & 50 & -289.809 \end{bmatrix}

X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}

B = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

$ AX=B $

My problem is to solve this system of equation. It seems that I can only get a trivial solution. If I assume, say $ x_3 = 1 $ still I am going wrong, if I compare my answer with matlab's.

Can someone help me how to get eigenvectors?

Answer 1

You have made an error somewhere: the eigenvalues for that matrix are, roughly: $ 516.65, 41.68 \pm 14.09î$ (or you have written down the wrong matrix).

The rest of your process and thinking is correct.

Answer 2

The point here is that if $\lambda A$ is an $n \times n$ matrix (with $\lambda \in \mathbb{C}$ or $\mathbb{R}$) then $$\sigma(\lambda A) = \lambda^n \sigma(A)$$

Where $\sigma$ is the spectrum with eigenvalues $\zeta$

So do yourself a favor and factor out $100$:

Then solving the characteristic equation becomes

$$100^3 [ (3-\zeta) ((2-\zeta)*(1-\zeta)-1*1) - 2(2*(1-\zeta)-1*1)+ 1.5 (2*1-2*1)]=0$$

perhaps you can get it from here?