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Help plz with this problem

Let a $3 × 3$ matrix $A$ be such that for any column vector $v ∈ \Bbb R^3$ the vectors $Av$ and $v$ are orthogonal. Prove that $A^t + A = 0$, where $A^t$ is the transpose matrix.

I've read all about orthogonal matrices but not have yet results.

Marc van Leeuwen
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heiverjust
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2 Answers2

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The conclusion is NOT true for real matrices. Let $A$ be an $n \times n$ real matrix, with $n=3$. Suppose it is skew-symmetric, that is, $A^T = -A$. Then, $A^T A = -A^2$. Since the matrix is orthogonal, $I = -A^2$. Then we have $\det(A)^2 + 1 = 0.$ That cannot happen.

Marc van Leeuwen
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Matha Mota
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  • Where in the question does it say that $A$ is an orthogonal matrix? Also, if you are arguing that it is impossible to satisfy the hypotheses, then the implication would be true (vacuously). I cannot see however exactly what question you think you were answering. – Marc van Leeuwen Apr 30 '21 at 08:35
  • My bad I misunderstood the problem statement. I read that A is orthogonal. Sorry, and thanks. – Matha Mota Apr 30 '21 at 09:04
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We have here $v^TAv=0$.

Also we have $v^TA^Tv=0$ and consequently $v^T(A + A^T)v=0$ where $A+A^T$ is symmetric matrix.
Symmetric matrix is orthogonally diagonalizable so in some orthogonally changed basis for every vector $u$ in $ {R^3}$ we have $ u^TDu=0$.
Now place standard basis vectors $e_1, ...e_n$ into formula above and you will see that all diagonal entries of $D$ have to be $0$ what means that $A+A^T=0$

Widawensen
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  • I think "in some basis for every vector $u$ in $\Bbb R^3$ we have $uTDu=0$" could be expressed more elegantly, especially since the statement does not refer to a basis at all. (Of course there is some relation between $D$ and $A$ that depends on a basis, so that could be made explicit.) Probably you can do without $D$ altogether if you don't insist on using the standard basis in the next sentence. – Marc van Leeuwen Apr 30 '21 at 09:21
  • $D$ is a diagonal form of the matrix $S=A+A^T$ in other orthogonally changed basis. If $D$ is zero matrix the same can be said about $S$. Of course vectors $v$ and $Av$ are orthogonal independently of orthogonal change of basis. I should add that the change of basis where $S$ becomes $D$ is orthogonal change. – Widawensen Apr 30 '21 at 09:38
  • By vectors of standard basis I mean simply vectors of the form with single $1$ and rest of $0$ for example $e_2=[ \ \ 0 \ \ 1 \ \ 0 \ \ \dots \ \ 0]^T$. For such vectors $e_i^TDe_i$ gives $i^{th}$ diagonal entry of $D$. – Widawensen Apr 30 '21 at 09:47
  • I know what a standard basis is. I just think that under the basis change you have in mind, the standard basis vectors would correspond to the eigenvectors of $A$, so you could argue directly in terms of those. Also I don't think orthogonality of the eigenspaces (or well chosen eigenvectors) is really necessary, just having a basis of eigenvectors, at least one of which has nonzero eigenvalue. – Marc van Leeuwen Apr 30 '21 at 09:50
  • Yes, the standard basis vectors where the matrix became diagonal correspond to eigenvectors of $S$. Sorry, I, for a some time, have to return to work. – Widawensen Apr 30 '21 at 09:58
  • @Widawensen please tell me why you substitute the basis vectors? After all, these vectors can be anything. – heiverjust Jun 10 '21 at 13:46
  • @heiverjust Calculations with standard basis vectors are very easy (easier than others) and lead fast to the result, so I've chosen them. Please see comment from 30April 9:47. – Widawensen Jun 10 '21 at 14:07
  • @heiverjust Hey heiverjust. Now my proof is fully clear for you? I simply used the fact that zero matrix is zero matrix in any frame. – Widawensen Jun 11 '21 at 09:15