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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.

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up vote 2 down vote accepted

Answer is the zero matrix obviously.

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.

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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 $A = P^{-1}0P = 0$.

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Another way to look at the problem.

Consider the geometric multiplicity of the matrix. It has two linearly independent eigenvectors corresponding to zero and so the geometric multiplicity is equal to its algebraic multiplicity. Therefore the matrix is diagonalizable. But the diagonal form is the zero matrix and any vector similar to the zero matrix is still just the zero matrix.

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