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Is there square matrix $A$ of size $3$ with real entries such that $$ \operatorname{tr}(A)=0\text{ and }A^2+A^T=I. $$ I have proved that there is not with size $2$ using definition of "trace", but for size $3$ it becomes complicated.

Here is the sketch of the proof for $2$. $$ a_{11}+a_{22}=0,\\ a_{11}^2+a_{12}a_{21}+a_{11}=1,\\ a_{11}a_{12}+a_{12}a_{22}+a_{21}=0,\\ a_{21}a_{11}+a_{21}a_{22}+a_{12}=0,\\ a_{12}a_{21}+a_{22}^2+a_{22}=1. $$ Putting $a_{11}=-a_{22}$ it is easy to see that above inequalities can't be true together.

Thanks!

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  • $\begingroup$ Can you post your solution for$n=2$ so we can mimic it or something? $\endgroup$ – PenasRaul Jun 25 '14 at 17:30
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The answer is no.

If we substitute $A^T=I-A^2$ into the transpose of $A^2+A^T=I$, namely $(A^T)^2+A=I$, we get $$(I-A^2)^2+A=I$$ Hence $I-2A^2+A^4+A=I$, or $$A^4-2A^2+A=0 ~~~~~~(\star)$$ must be satisfied by our matrix $A$. This has just four roots: $0,1,\frac{-1+\sqrt{5}}{2},\frac{-1-\sqrt{5}}{2}$.

Now, consider $B=QAQ^{-1}$, which is $A$ in Jordan canonical form. By multiplying $(\star)$ on the left and right by $Q, Q^{-1}$ respectively, we must also have $$B^4-2B^2+B=0 ~~~~(\dagger)$$ $B$ has the same eigenvalues as $A$, which must sum to $0$ (counted by multiplicity) by the first condition. Each block of $B$ must separately verify $(\dagger)$, hence the only possible nonzero $B$, up to reordering of the three blocks, is $$B=\left(\begin{smallmatrix}1&0&0\\0&\frac{-1+\sqrt{5}}{2}&0\\0&0&\frac{-1-\sqrt{5}}{2}\end{smallmatrix}\right)$$ Now, we plug $A=Q^{-1}BQ$ into $A^2+A^T=I$, to get $$Q^{-1}B^2Q+Q^TB^T(Q^{-1})^T=I$$ Multiplying on the left and right by $Q$ and $Q^{-1}$ respectively, and set $R=QQ^T$, to get $$B^2+RBR^{-1}=I$$ It turns out that $$I-B^2=\left(\begin{smallmatrix}0&0&0\\0&\frac{-1+\sqrt{5}}{2}&0\\0&0&\frac{-1-\sqrt{5}}{2}\end{smallmatrix}\right)$$ But since $RBR^{-1}=(I-B^2)$, $B$ and $I-B^2$ must have the same eigenvalues. They have two in common, but not all three -- $B$ has $1$ while $I-B^2$ has $0$. Hence no such $A$ exists.

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    $\begingroup$ Thank you. Is following proof correct. We have $$t=1^2+\left(\frac{-1-\sqrt{5}}{2}\right)^2+\left(\frac{-1+\sqrt{5}}{2}\right)^2 = 4$$, but on other hand we have $$t=tr(A^2)=3$$, which is contradiction? $\endgroup$ – pointer Jun 25 '14 at 18:42
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    $\begingroup$ The proof was great ! Thanks @vadim123 $\endgroup$ – the8thone Jun 25 '14 at 18:44
  • $\begingroup$ why do I know that $B$ has no of-diagonal elements? $\endgroup$ – Michael Jun 25 '14 at 18:47
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    $\begingroup$ @user159543 Because an annihilating polynomial of A has distinct roots, therefore the minimal polynomial of A is a product of distinct linear factors, i.e. it is diagonalizable.(the Jordan form has to be diagonal with eigenvalues on the diagonal) $\endgroup$ – the8thone Jun 25 '14 at 18:51
  • $\begingroup$ thanks, I overread that he just chooses the nonzero ones $\endgroup$ – Michael Jun 25 '14 at 18:59
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  1. The matrix $A$ is normal (and hence diagonalizable), since $$AA^T=A(I-A^2)=(I-A^2)A=A^TA.$$

  2. Further it is easy to show (see vadim123's answer) that $A$ satisfies the equation $$A^4-2A^2+A=0,$$ hence its eigenvalues $\lambda_{1,2,3}$ satisfy it as well.

  3. The possibilities to get zero trace are then $\lambda_1=\lambda_2=\lambda_3=0$ (immediately excluded as $A$ cannot be zero) and $\lambda_1=1$, $\lambda_2=\frac{-1+\sqrt5}{2}$, $\lambda_3=\frac{-1-\sqrt5}{2}$. In the latter case $A^T=I-A^2$ has zero eigenvalue which leads to a contradiction.

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Thanks vadim123 for idea. Here is another proof. As we saw $A$ has three different eigenvalues $a=1,~b=\frac{-1-\sqrt{5}}{2},~c=\frac{-1+\sqrt{5}}{2}$. We have $$a^2+b^2+c^2=4$$,on other hand we have $$3=tr(A^2)=a^2+b^2+c^2,$$which is contradiction.

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