For any square matrix $A$, if $A$ commutes with $A^{\ast}A$ then is $A$ normal? I solved for $2 \times 2$ matrices and it is true that $A$ is normal. For other order I can't prove or can't find counter example.
Asked
Active
Viewed 144 times
2
-
Is $A$ invertible? – wgrenard Oct 20 '16 at 07:33
-
The question is for any matrix. – Hemant Oct 20 '16 at 08:13
-
@daw you assumed, without loss of generality, that $A$ is upper triangular and $A^A$ diagonal. Perhaps it is useful, then, to note that the off-diagonal entries of $A^A$ are dot-products of columns from $A$. – Ben Grossmann Oct 20 '16 at 13:05
-
@daw another bit to notice is that if $A^A$ is diagonal with no repeated eigenvalues, then $A^A$ commutes only with other diagonal matrices. – Ben Grossmann Oct 20 '16 at 13:06
1 Answers
1
The answer is yes.
We begin by noting that $$ (AA^*)A = A(A^*A) = (A^*A)A $$ Thus, for every vector $x$, we have $(A^*A)(Ax) = (AA^*)(Ax)$. That is, for every vector $y$ in the image of $A$, we have $AA^* y = A^*Ay$. Note, however, that $$ \ker(AA^*) = \ker(A^*) = im(A)^\perp $$ So we may conclude from the above that $im(A^*A) \subseteq im(AA^*) = im(A)$. In fact, since these matrices have the same rank, we have $im(A^*A) = im(AA^*)$. Since these matrices are self-adjoint, we also have $\ker(A^*A) = \ker(AA^*)$.
Thus, $AA^*$ and $A^*A$ have the same kernel $K$, and they are the same transformation over $K^\perp$. It follows that $AA^* = A^*A$, as desired.
Ben Grossmann
- 225,327
-
-
Every element of $im(A^A)$ has the form $A^Ay$ for some $y\in \ker(A^*A)^\perp$. – Ben Grossmann Oct 20 '16 at 13:45
-
This seems trivial. But where does $AA^$ come into play? Did you mix up the order of $A$ and $A^$ in the comment? – daw Oct 20 '16 at 13:56
-
Sorry, that should be $AA^*$ in the above instead. Yes, I mixed the order – Ben Grossmann Oct 20 '16 at 14:02
-
I still can't figure out how $im(A^{\ast}A) \subset im(AA^{\ast})$. Also, is it a result that if two self adjoint matrices have same kernel then they are equal? How did you get the second part of the proof? – Hemant Oct 20 '16 at 14:21
-
$$ im(AA^) = AA^(\Bbb C^n) = AA^(im(A)^\perp) = A^A(im(A)^\perp) \subseteq A^*A(\Bbb C^n) $$ – Ben Grossmann Oct 20 '16 at 14:27
-
@Hemant the self-adjoint matrices have the same kernel and they are the same transformation over its orthogonal complement. – Ben Grossmann Oct 20 '16 at 14:28