Diagonalization by left multiplication Given a $N\times N$ matrix $B$. Is there a matrix $A$ which I can multiply from the left to make $A B$ diagonal?
I know the solution that with Eigen decomposition I get $A^{-1} B A$. However, I need to multiple my original matrix from left and right to get diagonal form.
Assuming it is not possible, is there a (possibly numerical) method to make a matrix $B$ at least more diagonal by left multiplying with a matrix $A$? (i.e., reducing the off diagonal components)
Assuming $B$ is already close to diagonal, is there a matrix $A$ to multiply from the left to make the off diagonal elements smaller?
 A: The problem is interesting only if the matrix $A$ is assumed to be invertible, otherwise choosing $A=0$ solves the problem in a trivial way.
The multiplication on the left by an invertible matrix doesn't change the linear relations between the columns. More precisely, if $\{b_{i_1},\dots,b_{i_k}\}$ is a set of columns of $B$, with $1\le i_1<i_2<\dots<i_k\le n$ (where $B$ is $m\times n$), and we denote by $b'_i$ the columns of $AB$ ($A$ invertible), then


*

*$\{b_{i_1},\dots,b_{i_k}\}$ is linearly independent if and only if $\{b'_{i_1},\dots,b'_{i_k}\}$ is linearly independent;

*$b_i=\alpha_1b_{i_1}+\dots+\alpha_kb_{i_k}$ if and only if $b'_i=\alpha_1b'_{i_1}+\dots+\alpha_kb'_{i_k}$.
Assume, for instance, that $b_2=\gamma b_1$, with $b_1\ne0$ and that $AB$ is diagonal. Then we must have $b_2'=0$, so also $b_2=0$.
Similarly, if a column of $B$ is a linear combination of the columns to the left of it, it must be $0$.
Thus a matrix can be “left diagonalized with an invertible matrix” if and only if its nonzero columns are linearly independent. One direction has been shown above. For the other direction, use the reduced row echelon form and swap rows until you get diagonal form.
