Is a Zero Row Necessary in the Square Root of a Zero Matrix?

Click For Summary
A square root of the zero matrix must be singular, as demonstrated by the discussion surrounding the counterexample provided. The initial matrix proposed did not yield the zero matrix when squared, indicating a misunderstanding of matrix multiplication. It was clarified that non-invertible matrices can produce a product of zero without either being zero themselves. The concept of square roots for matrices is generally not unique, but the principal square root of a positive definite matrix is an exception. The conversation highlights the necessity of at least one zero row or column in any square root of the zero matrix.
Bipolarity
Messages
773
Reaction score
2
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:
\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)

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
 
Physics news on Phys.org
Bipolarity said:
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:
\left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array} \right)

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

The square of that matrix is the same matrix, not the zero matrix. Did you accidentally multiply when you should've added?
 
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
$$\begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix}$$
 
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.
 
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.
 
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
 
I am studying the mathematical formalism behind non-commutative geometry approach to quantum gravity. I was reading about Hopf algebras and their Drinfeld twist with a specific example of the Moyal-Weyl twist defined as F=exp(-iλ/2θ^(μν)∂_μ⊗∂_ν) where λ is a constant parametar and θ antisymmetric constant tensor. {∂_μ} is the basis of the tangent vector space over the underlying spacetime Now, from my understanding the enveloping algebra which appears in the definition of the Hopf algebra...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K