N x N matrix has N nonzero eigenvalues but has the zero-vector as eigenvector

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I’m trying to diagonalize the following matrix:
$$ A = \left[ \begin{matrix} 1&3\\ 4&2 \end{matrix} \right] $$

I’ve found the eigenvalues to be:
$$ \lambda = 2$$ $$ \lambda = 1$$

For $\text{$\lambda=2$}$, the eigenvector I have is
$$\left[ \begin{matrix} 3\\ 1 \end{matrix} \right]$$

But for $\text{$\lambda=1$}$, I am getting an eigenvector of:
$$\left[ \begin{matrix} 0\\ 0 \end{matrix} \right]$$

Are my eigenvalues wrong or does this just mean that I can’t diagonalize the matrix $A$?