Can you row reduce and then diagonalize a matrix?

Question

I'm studying for a midterm and I've got a recurring question:

If we have an nxn matrix A, can we row reduce A and then diagonalize? I seem to be getting conflicting answers from 2 different TAs.

Also, if the determinant of A is 0, does that mean A is not diagonalizable? Thanks!

Answer 1

Suppose an $n\times n$ matrix has rank $n$; then its row reduced echelon form is the identity. If you could first row reduce and then diagonalize, then every invertible matrix would be similar to the identity, which is clearly false.

A diagonal matrix can have zero determinant, but it is obviously diagonalizable.

Answer 2

No, performing row reduction on a matrix changes its eigenvalues, so changes its diagonalization.

Consider the matrix $A = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}$. By subtracting the first row from the second, we get a matrix in row echelon form: $$ A \sim \begin{bmatrix} 1 & 2 \\ 0 & -1 \end{bmatrix} $$ The eigenvalues of the matrix on the right are $1$ and $-1$.

But the eigenvalues of $A$ are the roots of $(\lambda-1)^2 - 2 = 0$. That is, $\lambda = 1 \pm \sqrt{2}$. So the diagonalization of $A$ is going to be something else entirely.

As for your second question, the matrix $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$ is already diagonal (hence, diagonalizable), but has determinant zero. So invertible and diagonalizable are different things entirely.