# Two linearly independent eigenvectors with eigenvalue zero

What is the only $2\times 2$ matrix that only has eigenvalue zero but does have two linearly independent eigenvectors?

I know there is only one such matrix, but I'm not sure how to find it.

EDIT, here is a simple reason: let the matrix be $(c_1\ c_2)$, where $c_1$ and $c_2$ are both $2\times1$ column vectors. For any eigenvector $(a_1 \ a_2)^T$ with eigenvalue $0$, $a_1c_1 + a_2c_2 = 0$. Similarly, for another eigenvector $(b_1 \ b_2)^T$, $b_1c_1 + b_2c_2 = 0$. So $(a_2b_2 - a_1b_2)c_1 = 0$, therefore $c_1=0$ as the eigenvectors are linearly independent. From this, $c_2=0$ also.
Let $A$ be any such matrix. Let $\beta=[\mathbf{v}_1,\mathbf{v}_2]$ be a basis made up of eigenvectors of $A$. If $P$ is the matrix that has $\beta$ in the columns, then $P^{-1}AP$ is diagonal, with the eigenvalues of $A$ in the diagonals. But such a matrix is $$\left(\begin{array}{cc} 0&0\\0&0\end{array}\right).$$ So $PAP^{-1}=0$. Multiplying on the left by $P^{-1}$ and on the right by $P$, we get $A = P^{-1}0P = 0$.