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I am reading Boyd's Convex Optimization text, and I am looking at a relation between the trace and the eigenvalue of a matrix. It is on page 92, example 3.23, line 7.

The matrix $Y$ (apparently) has only one eigenvector $v$, $v^T v = 1$, and it (apparently) has only one eigenvalue $\lambda$. Also, $Y$ is not positive semidefinite, nor is it positive definite.

$ tr(Y v v^T) = \lambda$

Why is this so? I know the trace is equal to the sum of the eigenvalues.

Thanks.

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    $Yv$ is $\lambda v$, so $Yvv^T$ is $\lambda v v^T$... Can you continue?2011-04-18

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Note that $Yv = \lambda v$, so you are trying to compute the trace of $\lambda vv^T$.

Let $v = \left(\begin{array}{c}a_1\\a_2\\\vdots\\a_n\end{array}\right).$ The relation $v^tv = 1$ tells you that $a_1^2+\cdots + a_n^2 = 1$.

The diagonal entries of $vv^T$ are precisely $a_1^2$, $a_2^2,\ldots,a_n^2$. So the diagonal entries of $\lambda vv^T$ are $\lambda a_1^2,\ldots,\lambda a_n^2$. Adding the diagonal entries to get the trace gives: $\mathrm{Tr}(Yvv^t) = \mathrm{Tr}(\lambda vv^T) = \lambda\mathrm{Tr}(vv^T) = \lambda(a_1^2+\cdots+a_n^2) = \lambda$ since $a_1^2+\cdots + a_n^2 = v^Tv = 1$.

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    @Mariano Suárez-Alvarez: Yes, with all the tricks and properties, I missed the first line.2011-04-18