Calculating Eigenvectors of a Diagonal Matrix

Question

I'm feeling dumb even asking this. But there might be a definition for this somewhat like why $1$ is not a prime number. Therefor this might be the right place to ask this question anyway.

Given the matrix $$\begin{bmatrix} 4 & 0 \\ 0 & 4 \\ \end{bmatrix}$$

One sees immediately that the eigenvalues are $4$ and $4$ and the corresponding eigenvectors $$\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$$ and $$\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$$

Assuming one doesn't see that or one tries to program this he would use $(A-\lambda_i E)v_i=0$ to calculate the eigenvectors. But using this in this really simple example leads to $$\begin{gather} \begin{bmatrix} 0 & 0 \\ 0 & 0 \\ \end{bmatrix}v=0 \end{gather}$$ So every solution would be correct for $v$.

Where is my mistake? I hope it is something obvious. I really hate it when there are special cases and one can not always use one scheme for all related problems.

Answer 1

There is nothing wrong with your calculations. Note that if $v_1$ and $v_2$ are eigenvectors corresponding to an eigenvalue $\lambda $, so is $c_1v_1+c_2v_2$. In your case, note that $e_1$ and $e_2$ are basis elements for $\mathbb R^2$.

Answer 2

This happens for any $n \times n$ identity matrix since the eigenvectors are always orthogonal and hence they span the entire $\mathbb{R}^n$ space. Thus, any vector in the space is an eigenvector. Therefore, there is no mistake in your solution.