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Square root of a 0 matrix

  1. Dec 2, 2012 #1
    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)?

  2. jcsd
  3. Dec 2, 2012 #2
    The square of that matrix is the same matrix, not the zero matrix. Did you accidentally multiply when you should've added?
  4. Dec 2, 2012 #3

    I like Serena

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    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
    $$\begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix}$$
  5. Dec 2, 2012 #4


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    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.
  6. Dec 2, 2012 #5


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    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.
  7. Dec 2, 2012 #6
    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?

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