I have been set some work to do over the holidays, and one of the questions gives a hint that is as follows: $A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^TA\text{ is diagonalisable}$.

I know that

  • $A\in\mathrm{M}_{n\times n}(\mathbb{R})\implies A^TA\text{ is diagonalisable}$

  • $A\in\mathrm{M}_{n\times n}(\mathbb{C})\implies A^*A\text{ is diagonalisable}$

where the former can be thought of as a particular case of the latter. Both those statements are true because $A^*A$ is self adjoint, and we can then apply the Spectral Theorem for normal operators.

But is the statement at the top of my question true, or has the lecturer simply mistyped one of the two facts that I've written? If it is true, I can't see how to prove it, so any hints would be appreciated.

| cite | improve this question | | | | |
  • $\begingroup$ Note that, " a matrix is normal if and only if it is unitarily similar to a diagonal matrix ". $\endgroup$ – Mhenni Benghorbal Dec 18 '13 at 14:20
  • $\begingroup$ The first case corresponds to matrices with real entries which is true. $\endgroup$ – Mhenni Benghorbal Dec 18 '13 at 14:26

I think that's a typo. Your lecturer should write $A^\ast A$ instead of $A^TA$.

In general, every complex square matrix $C$ is $\mathbb C$-similar to a complex symmetric matrix $S$. Since a symmetric matrix represents a symmetric bilinear form, and every symmetric bilinear form is diagonalisable (via $T$-congruence), $S$ can always be written as $A^TA$ for some complex matrix $A$. So, if what your lecturer wrote is correct, that means every complex square matrix $C$ is diagonalisable. Yet this is obviously untrue.

For a concrete example, consider $$ A=\pmatrix{1&\tfrac{i+1}2\\ 1&\tfrac{i-1}2},\ A^TA=\pmatrix{2&i\\ i&0}= \pmatrix{-i&-i\\ 1&0}\pmatrix{1&1\\ 0&1}\pmatrix{0&1\\ i&-1}. $$ Since the Jordan form of $A^TA$ is $\pmatrix{1&1\\ 0&1}$, $A^TA$ is nondiagonalisable.

| cite | improve this answer | | | | |
  • 1
    $\begingroup$ @Tim I didn't notice that julien posted his answer when I was typing mine. Since the two examples are essentially identical, and julien's answer appeared first, would you please unaccept mine and accept his instead? Thanks. $\endgroup$ – user1551 Dec 18 '13 at 15:12
  • $\begingroup$ No problem, I'm fine with that :-) Thank you, though. $\endgroup$ – Julien Dec 18 '13 at 15:27

This is false.

$$ A=\pmatrix{\frac{1}{2}+i&1\\-1-\frac{i}{2}&i}\qquad A^TA=\pmatrix{2i&1\\1&0}\qquad \mathrm{Spectrum}(A^TA)=\{i\} $$

| cite | improve this answer | | | | |
  • $\begingroup$ I think for the case $A^T A$ means a matrix with real entries. $\endgroup$ – Mhenni Benghorbal Dec 18 '13 at 14:24
  • $\begingroup$ @MhenniBenghorbal Read the title of the thread again. Obviously $A^TA$ is diagonalizable if $A$ is real square. $\endgroup$ – Julien Dec 18 '13 at 14:26
  • $\begingroup$ Just read under " I know that..." $\endgroup$ – Mhenni Benghorbal Dec 18 '13 at 14:29
  • 2
    $\begingroup$ @MhenniBenghorbal Yes, the OP knows these two facts. And is asking about a different question. And was given two counterexamples in the two posted answers. All is well. Read the question. Please... $\endgroup$ – Julien Dec 18 '13 at 14:32

As the counter-examples show, $A^TA$ is does not need to be diagonalisable for a complex $A$. However, $A^TA$ is "almost diagonalisable". This is known as Takagi's decomposition: for a complex symmetric $B$, there is a unitary $U$ and diagonal $D$ such that $B=UDU^T$. Note that $A^TA$ is (complex) symmetric.

| cite | improve this answer | | | | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.