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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

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    $\begingroup$ If $A=A A^T$, can you express $A^T$? $\endgroup$ Commented Jun 30, 2014 at 17:10

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

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  • $\begingroup$ I follow the A^T til =A, the first part.Then A^2=AA but how does AA=AA^T? $\endgroup$
    – cele
    Commented Jun 30, 2014 at 17:48
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    $\begingroup$ Use the fact $A^T = A$ to substitute $A^T$ for the second $A$ in $AA$. $\endgroup$
    – David K
    Commented Jun 30, 2014 at 17:54
  • $\begingroup$ oh, ok. I figured it was something plain as day. Thanks $\endgroup$
    – cele
    Commented Jun 30, 2014 at 17:59
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get transpose from $A=AA^T$ thus we have $A^T=AA^T$ and thus we have $A=A^T$ and it proved that $A=A^2$

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$AA^T=A$

$A$ 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 $det \ A\neq 0$

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    $\begingroup$ Consider $A=0$, which has no inverse. $\endgroup$
    – abnry
    Commented Jun 30, 2014 at 17:41
  • $\begingroup$ Thank you for your hint. I didn´t think of this case. $\endgroup$ Commented Jun 30, 2014 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
    Commented Jun 30, 2014 at 17:56

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