
#1
Dec212, 03:19 AM

P: 783

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 



#2
Dec212, 04:05 AM

P: 166





#3
Dec212, 04:31 AM

HW Helper
P: 6,189

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 $$\begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix}$$ 



#4
Dec212, 07:00 AM

Math
Emeritus
Sci Advisor
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PF Gold
P: 38,879

Square root of a 0 matrix
IF A^{2}= 0 and A is invertible, then we could multiply both sides by A^{1} and get A= 0. However, the ring of matrices as "noninvertible" matrices. It is quite possible to have AB= 0 with neither A nor B 0 and, in particular, nonzero A such that A^{2}= 0.




#5
Dec212, 11:48 AM

Engineering
Sci Advisor
HW Helper
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P: 6,340

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. 



#6
Dec212, 12:37 PM

P: 783

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 


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