Is it possible to diagonalize a singular matrix?

Question

I have not seen anywhere written that it is impossible, but it seems impossible, so I want to check if I missed something.

According to a theorem, an $n\times n$ matrix is diagonalizable if it has $n$ independent eigenvectors.

Let's say, the matrix has $1$ row with only zeros (worst singular case). As it has one row with only zeros, it will zero out the corresponding row of any vector it is multiplied by. That means, the corresponding rows of its eigenvectors have to be zero. And if one of the rows is fixed, there cannot be n independent eigenvectors. Am I correct?

Did I miss anything?

Answer 1

A matrix is singular if and only if $0$ is one of its eigenvalues. A singular matrix can be either diagonalizable or not diagonalizable. For example, $$ \pmatrix{ 1&0\\0&0 } $$ Is diagonalizable (since it is diagonal), whereas $$ \pmatrix{ 0&1\\0&0 } $$ is not diagonalizable.

Answer 2

Yes, diagonalize the zero matrix.

THM Let $A$ be a matrix associated to a linear transformation $T:V\to V$, $V$ a $k$-vector space. Let $\lambda_1,\ldots,\lambda_l$ be it's distinct eigenvalues. Then $A$ is diagonalizable iff $$V=E(\lambda_1)\oplus\cdots \oplus E(\lambda_l)$$ where $E(\lambda_i)=\ker(A-\lambda_i1)$.

The sum is always direct, but it might not comprise all of $V$. This is equivalent to the easier claim that $V$ admits a basis of eigenvectors of $A$, also equivalent to the fact that the minimal polinomial splits over $k$ and has all simple roots. In turn, this is equivalent to the fact that $\chi_A$ splits and the multiplicity of $\lambda_i$ as a root equals the dimension of $E(\lambda_i)$. We always have $\dim E(\lambda_i)\leqslant {\rm mult}\;(\chi_A,\lambda_i)$, and this is in fact a good thing to use when checking for diagonalizability, in some cases. For example, $$\begin{pmatrix}1&1\\0&1\end{pmatrix}$$ is not diagonalizable, since $\chi=(X-1)^2$ but $\dim E(1)=1<2$.