3
$\begingroup$

I got the above problem in an exam, the problem stated to show there is at most $2^n$ such matrices.

What I did was to show the matrix is similar to a matrix with a diagonal matrix with distinct complex numbers, and now if $A$ is such a matrix and $B$ is a square root. Then $B$ is diagonalizable so since $A$ and $B$ commute there is a matrix $P$ such that that $P^{-1}BP$ and $P^{-1}AP$ are both are diagonal.

So my argument was that to consider the diagonal case (w.l.g) and show that each diagonal entry of $P^{-1}BP$ should be root of the corresponding diganoal entry of $P^{-1}AP$. So we can have atmost 2 options for each and overall we get $2^n$.

Now the problem I just realized is that I just saw that for any invertible diagonalizable matrix $A^2$, $A$ is also diagonalizable. So then we can apply the same logic as what I just did to any such invertible matrix. But I also know the identity has infinitely many roots. So where did I go wrong?

Thank You

edit: added non zero

$\endgroup$
  • 1
    $\begingroup$ The identity doesn't have distinct eigenvalues ... $\endgroup$ – Michael Burr May 4 '18 at 17:14
  • $\begingroup$ Yes but my argument doesn't use the distinct eigenvalues property that's what I was worried $\endgroup$ – user68099 May 4 '18 at 17:16
  • $\begingroup$ If the eigenvalues are different, then the base change matrix is determined (up to a permutation matrix). $\endgroup$ – dan_fulea May 4 '18 at 17:21
2
$\begingroup$

It seems to me that you missed one subtlety when you wrote "w.l.g". If $A=PDP^{-1}$ and $E^2=D$, then $(PEP^{-1})^2=A$. But who says that there are no other invertible $Q$ with $(QEQ^{-1})^2=A$? That's precisely what happens here.

When you know that all diagonal elements in $D$ are distinct, the following happens: if $(QEQ^{-1})^2=A=(PEP^{-1})^2$, we can rewrite this as $$ (Q^{-1}P) D=D(Q^{-1}P). $$ Because all diagonal entries of $D$ are distinct, this implies that $Q^{-1}P=I$, that is $Q=P$. That's why you get precisely $2^n$ square roots.

When eigenvalues are repeated, this goes off the window. For $I$, you can take the $2^n$ matrices $D_k$ with $1$ and $-1$ in the diagonal, so $D_k^2=I$. But now, for any invertible $P$, the matrixi $PD_kP^{-1}$ is a root, and distinct $P$ will give us mostly distinct matrices; that's how we get infinitely many roots.

$\endgroup$

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.