Say I have matrix A:
$$A = \begin{bmatrix} -1 & -2 \\ -2 & 2 \end{bmatrix}$$
First lets solve for the $\lambda$ in:
$| \lambda * I – A | = \begin{vmatrix} \lambda + 1 & 2 \\ 2 & \lambda – 2 \end{vmatrix} = \lambda^2 – \lambda – 2 – 4 = (\lambda – 3)*(\lambda + 2)$
$$ \lambda = {3 , -2 }$$
solve x in the equation $(\lambda * I – A)x = 0$ for each lambda:
For the eigenvector 3:
$$\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} – \begin{bmatrix} -1 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} 4 & -2 \\ 2 & 1 \end{bmatrix} -> \begin{bmatrix} 1 & \frac{1}{2} \\ 0 & 0 \end{bmatrix}$$
So $$\begin{bmatrix} t_1 \\ t_2 \end{bmatrix} = \begin{bmatrix} -t \\ 2t \end{bmatrix} = t * \begin{bmatrix} -1 \\ 2 \end{bmatrix}$$
For eigenvector -2:
$$\begin{bmatrix} -2 & 0 \\ 0 & -2 \end{bmatrix} – \begin{bmatrix} -1 & -2 \\ -2 & 2 \end{bmatrix} = \begin{bmatrix} -1 & 2 \\ 2 & -4 \end{bmatrix} -> \begin{bmatrix} 1 & -2 \\ 0 & 0 \end{bmatrix}$$
So $$\begin{bmatrix} t_1 \\ t_2 \end{bmatrix} = \begin{bmatrix} 2t \\ t \end{bmatrix} = t * \begin{bmatrix} 2 \\ 1 \end{bmatrix}$$
$(-1,2) \cdot (2,1) = 0$ so orthogonal.
Is this right?
Last related question. Why is this statement true:
I think I get most of it… but why does an nxn must have n eigenvectors? Is this true that the number of eigenvectors is = number of rows/columns in a square matrix? Why? I think I get why it has n linearly independent eigenvectors
Similarly…does an nxn matrix always have n eigenvalues?
Best Answer
The matrix doesn't always have $n$ linearly independent eigenvectors; but when it does, it's diagonalizable. (Symmetric matrices have this property.)
Your work above looks correct.