Square root of a 0 matrix

Bipolarity

At first I thought that there is no square matrix whose square is the 0 matrix. But I found a counterexample to this. My counterexample is:
[tex]\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)[/tex]

However it appears that my counterexample has a 0 row. I'm curious, must a square root of the 0 matrix necessarily have at least one 0 row (or 0 column)?

BiP

justsomeguy

The square of that matrix is the same matrix, not the zero matrix. Did you accidentally multiply when you should've added?

I like Serena

I suspect you intended the following matrix?
$$\begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$$
Square it and you get the zero matrix.

The same holds for
$$\begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix}$$

HallsofIvy

IF A²= 0 and A is invertible, then we could multiply both sides by A^-1 and get A= 0. However, the ring of matrices as "non-invertible" matrices. It is quite possible to have AB= 0 with neither A nor B 0 and, in particular, non-zero A such that A²= 0.

AlephZero

The "square root of a matrix" isn't a very useful idea for general matrices, because it is hardly ever unique. See http://en.wikipedia.org/wiki/Square_root_of_a_matrix for the sort of (probably unexpected) things that can happen.

However the positive definite square root of a positive definite matrix (called its "principal square root") is unique, and sometimes useful.

If A is a symmetric matrix, finding B such that A = BB^T, is even more useful. B has most of the useful properties of the "square root or A", even when it is not a symmetric matrix.

Bipolarity

Thank you all for your replies! Sorry for my mistake but I get it now!!

HallsofIvy, does your post essentially prove that square roots of the 0 matrix must be singular?

BiP