4
$\begingroup$

I've been working trying to understand the following question:

Let n be a positive integer, let $F$ be a field, and let $A \in \mathrm{Mat}(n,F)$ satisfy the condition $A=AA^{\top}$. Show that $A^2=A$.

I haven't made much progress since my knowledge is pretty basic but I ran across this link and was wondering if this example was essentially the same?

Example

$\endgroup$
  • 3
    $\begingroup$ If $A=A A^T$, can you express $A^T$? $\endgroup$ – Peter Franek Jun 30 '14 at 17:10
20
$\begingroup$

Since $A^T = (AA^T)^T = AA^T = A$, You have $A^2 = AA = AA^T = A$

$\endgroup$
  • $\begingroup$ I follow the A^T til =A, the first part.Then A^2=AA but how does AA=AA^T? $\endgroup$ – cele Jun 30 '14 at 17:48
  • 1
    $\begingroup$ Use the fact $A^T = A$ to substitute $A^T$ for the second $A$ in $AA$. $\endgroup$ – David K Jun 30 '14 at 17:54
  • $\begingroup$ oh, ok. I figured it was something plain as day. Thanks $\endgroup$ – cele Jun 30 '14 at 17:59
6
$\begingroup$

get transpose from $A=AA^T$ thus we have $A^T=AA^T$ and thus we have $A=A^T$ and it proved that $A=A^2$

$\endgroup$
-1
$\begingroup$

$AA^T=A$

$A$ from the LHS to the RHS

$A^T=AA^{-1} \Rightarrow A^T=I\Rightarrow A=I$

$I^2=I \Rightarrow A^2=A$

Edit: It works only, if $det \ A\neq 0$

$\endgroup$
  • 1
    $\begingroup$ Consider $A=0$, which has no inverse. $\endgroup$ – abnry Jun 30 '14 at 17:41
  • $\begingroup$ Thank you for your hint. I didn´t think of this case. $\endgroup$ – callculus Jun 30 '14 at 17:46
  • $\begingroup$ And $A^2 = A$ in the special case $A = 0$ too, although not via the exact same reasoning. $\endgroup$ – David K Jun 30 '14 at 17:56

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.