Eigenvectors of the $2\times2$ zero matrix

Question

I have been given a problem that involves the following matrix:

$$\begin{bmatrix}-2 & 0\\0 & -2\end{bmatrix}$$

I calculated the eigenvalues to be $\lambda_{1,2} = -2$

When I go to calculate the eigenvectors I get the following system:

$$\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$

The eigenvectors are clearly $\begin{bmatrix}1 \\ 0\end{bmatrix}$ and $\begin{bmatrix}0 \\ 1\end{bmatrix}$ and any multiple of these

But why is this? Is there a formal reasoning or method why it's only these two? Why doesn't $\begin{bmatrix}1 \\ 1\end{bmatrix}$ exist as an eigenvector?

Thank you

Answer 1

The eigenvalues are not $\pm1$. You might want to have another go at that.
The procedure for eigenvectors gives those two vectors. They form a basis for the eigenspace. Since they have the same eigenvalue, any linear combination of them is also an eigenvector. Including $[1,1]^T$.

Answer 2

A number $\lambda$ is an eigenvalue of a matrix $A$ iff $\det (A - \lambda I) = 0$. If $A$ denotes the matrix under consideration, we have $\det (A - \lambda I) = 0$ iff $\lambda = -2$, so $\lambda := -2$ is the eigenvalue of $A$. By definition, a nonzero vector $v$ is an eigenvector of $A$ corresponding to the eigenvalue $\lambda$ iff $Av = \lambda v$, i.e. iff $(-2v_{1},-2v_{2}) = (-2v_{1},-2v_{2})$, so every eigenvector corresponding to $\lambda$ is a linear combination of the vectors $(1,0)$, $(0,1)$.

Answer 3

Suppose $A$ is a square matrix. Then $A$ is a multiple of the identity iff every vector is an eigenvector of $A$.

If $A$ is a multiple of the identity, then $A = \alpha I$ for some $\alpha$ and hence $Av = \alpha v$ for any $v$.

Now suppose for any non zero $v$ there is some $\lambda_v$ such that we have $Av = \lambda_v v$. Choose $u,v$ to be linearly independent, then $A u = \lambda_u u, Av = \lambda_v v, A(u+v) = \lambda_{u+v} (u+v)=\lambda_u u + \lambda_v v$. It follows from linearity and independence that $\lambda_v = \lambda_{u+v} = \lambda_u$, and hence that $Av = \lambda_u v$ for all $v$.