Square Enix has an
If so, would you e
Inhibitory effect
Tales from The Loo
Q: Python: How to
/* * Licensed to
Oliver, Wisconsin
Q: What is the us
The present invent
Q: How can I chan

Introduction {#Sec
Q: Find a value i
Q: Is there a way
In this article we
Synthetic lethalit
The goal of this p
Q: How to add a f
Q: D3 force graph
package org.eclips
We use cookies to
<<      >>
Q: Square root (square) of a matrix and the matrix This is regarding the square root of a matrix. I would like to know how can I calculate the matrix of matrix of $(A^{t}A)^{1/2}$. A: $(A^TA)^{1/2}=((A^T A)^T)^{1/2}=((A^T)^T(A^T))^T=A^T=AA^T$. A: Let $A$ be an $m \times n$ matrix and let $C=(A^TA)^{1/2}$ be its square root. If $U$ is $n\times n$ unitary, then $U^T$ is also unitary. Therefore, $$ \text{trace}(C)=\text{trace}(A^TA)^{1/2}=\text{trace}(A^TU^{-1}U^TA)=\text{trace}(AA^T)=\|A\|_F^2. $$ As noted by Bill and Yoni, $$ A^TA=AA^T. $$ So, trace$(C)$ is equal to the square of the Frobenius norm of the matrix $A$, $$ \|A\|_F^2=\text{trace}(C). $$ Since the square of the Frobenius norm is an invariant under unitary transformation, we get $$ \text{trace}(C)=\|A\|_F^2. $$ A: $$AA^T=(A^TA)^{1/2}\implies \operatorname{trace}(AA^T)=\operatorname{trace}(A^TA)^{1/2}=\operatorname{trace}(A^TA)=\operatorname{trace}(A^T A)=\|A\|_F^2$$ $$A^TA=AA^T\implies A^TA=(AA^T)^{1/2}$$ Edit: The question was changed after this answer (and I don't have edit rights). $$AA^T=(A^TA)^{1/2}\implies AA^T=(A^TA)^{1/2}(A^TA)^{1/2}=(A^TA)^{1/2}(A^TA)^{1/2}=A^TA$$ The answer to the question "I would like to know how can I calculate the matrix of matrix of $(A^TA)^{1/2}$" is trivial: it's $(A^TA)^{1/2}$ (that's it). Edit 2: This is a more generalized result. Let $A$ be any matrix, and $D$ be a square diagonal matrix (possibly not square). If $D$ is invertible, then $D^{-1}AD=(D^{-1}AD)^{1/2}=(D^{1/2}AD^{1/2})^{1/2}=(AD^{1/2})^{1/2}=A^{1/2}D^{1/2}$ (notice that $A^{1/2}D^{1/2}$ is $A$ times the square root of $D$). To see this for square matrices: $A^{1/2}=A^{1/2}I=A^{1/2}(A^{-1}A)=(A^{-1}A)^{1/2}=I^{1/2}A^{1/2}=I^{1/2}A^{1/2}I$. Now, the square root of a scalar times a diagonal matrix is still a scalar times a diagonal matrix. And a scalar times the identity matrix is still the same identity matrix. So, if $A$ is $m\times n$, and $D$ is $k\times k$, then $A^{1/2}D^{1/2}$ is $mk\times kn$ (which is also equal to $m\times n$). But this matrix is $A$ times the square root of the diagonal matrix $D$. The same formula is valid for every $D$. And the fact that $(AA^T)^{1/2}=A^{1/2}A^{1/2}=A$ is trivial, of course.