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Somewhere in a proof I used the fact that for a vector $a\in\mathbb R^n$, the matrix $a a^T$ is symmetric and can thus be diagonalised into $a a^T = Q^T D Q$, with $Q$ orthogonal and $D$ diagonal. I needed this to prove that, given $x\in \mathbb R^n$, there exist orthogonal matrix $Q$ and diagonal matrix $D$ s.t. $x^T a a^T x = (Qx)^T D Qx$.

Now I need to say something about the reverse of this, so my question is: given diagonal matrix $D$, under what conditions does there exist a vector $a$ and orthogonal matrix $Q$ s.t. $(Qx)^T D Qx$ can be written as $x^T a a^T x$, with $a_1,\ldots ,a_n \neq 0$ ?

I think the condition is that $D$ has positive elements but I don't know how to prove this.

Lawrencelot
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    What rank has $aa^T$? That yields a necessary condition. Is it also sufficient? – Daniel Fischer Jul 30 '13 at 10:13
  • Good question. I want $a a^T$ to have rank at least $1$, but where I use this, $D$ actually has full rank. $a a^T$ does not have to have full rank, but I do want the vector $a$ to have nonzero elements. – Lawrencelot Jul 30 '13 at 10:42
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    However, $n \times n$ matrix of the form $aa^T$ for $a \in \mathbb{R}^n$ a column vector either has rank $0$ (if $a = 0$) or rank $1$ (if $a \neq 0$), so... – Branimir Ćaćić Jul 30 '13 at 10:47
  • I don't see that. But I did notice a mistake in my comment: $D$ does not necessarily have full rank – Lawrencelot Jul 30 '13 at 11:19

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A couple of hints: First of all, there exists a vector $a$ and orthogonal matrix $Q$ s.t. $(Qx)^T D Qx$ can be written as $x^T a a^T x$ iff there exists $a, Q$, where $Q$ is orthogonal, such that $D = Qaa^TQ^T$. That is maybe a simpler way to look at the problem.

Second, the diagonal entries of $D$ must be the eigenvalues of $aa^T$. Some of the commenters have hinted that those can be found. (What do you think an eigenvector would be?) If $aa^T$ has rank one, as the commenters have suggested, what can we say about its eigenvalues?

Eric Auld
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