Eigenvalue and Eigenvector of $\small\pmatrix{0 & 0 \\ 0 & -7}$

Question

I need help working out the eigenvectors for this matrix.

$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $

The original matrix is $ \begin {pmatrix} 5 & 0 \\ 0 & -2 \end{pmatrix} $ , eigenvalues are 5,-2,

but I am not sure how to about the eigenvectors,

as for 5

$ \begin {pmatrix} 0 & 0 \\ 0 & -7 \end{pmatrix} $ $ \begin{pmatrix} x \\ y \end{pmatrix}$ = $ \begin{pmatrix} 0 \\ 0 \end{pmatrix}$

from the first equation, $x$ and $y$ are both zero, but from the second equation $y = 0$, so what is the eigenvector?

Answer 1

From first equation you deduce whatever is x and y the equation holds

$$0x+0y=0$$ From second equation you deduce that $y=0$ $$0x-7y=0 \implies -7y=0 \implies y=0$$ So $$(x,y)=(x,0)=x(1,0)$$

Answer 2

No, from the first equation, $x$ and $y$ are free. From the second equation, $y=0$. So your eigenvector is $$ \begin{bmatrix}1\\0 \end{bmatrix} $$ as you can check, the equation is satisfied.

Answer 3

To find the eigenvectors, you need to solve the linear system :

$$(A-\lambda I)v = 0$$

For the case of $\lambda = 5$, you have :

$$(A-5I)v_5=0 \Rightarrow\begin{bmatrix} 0 & 0 \\ 0 & -7 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2\end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \Rightarrow \{ 0v_1 + 0v_2 = 0 |v_2 = 0 \}$$

This means that the eigenspace gets generated by $\{(v_1,v_2) \in \mathbb R^2 | v_1\in \mathbb R \space \text{and} \space v_2=0\}$, thus $v_1$ can take any value over the reals since $0v_1 = 0$ is true for all $v_1 \in \mathbb R$. Simply letting $v_1=1$, you yield the eigenvector :

$$v_5 = \begin{bmatrix} 1 \\ 0\end{bmatrix}$$