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

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 $A$?

• The eigenvalues are $5$ and $-2$. Check your work again. In particular, have you got the characteristic polynomial wrong? It should be $(x-1)(x-2) - 3 \times 4 = 0$. – астон вілла олоф мэллбэрг Apr 21 '18 at 23:54
• @астонвіллаолофмэллбэрг nice catch! I forgot to include the -12 in the characteristic equation. Thanks! – adamcasey Apr 22 '18 at 0:03
• You are welcome. I hope you managed to find the eigenvectors as well. – астон вілла олоф мэллбэрг Apr 22 '18 at 0:42