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Consider

$$ E = V\Lambda V^{\top} $$ $$ (m\times m) = (m\times p) (p\times p) (p\times m)$$

where $p<m$, $E$ is symmetric and singular, $\Lambda$ is diagonal, $V$ has orthonormal columns, and

$$ \begin{align} V^{\top}V &= I \\ V V^{\top} &\neq I \end{align} $$

What can we say about the trace of $E$? Normally I would say $E$ and $\Lambda$ are similar, therefore their traces must be equal, but since they have different sizes I don't know.

Could it be that the missing $(m-p)$ eigenvalues are all zero? In this case I suppose the traces would be equal.

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  • $\begingroup$ The matrix $VV^\top$ behaves in one respect like an identity matrix: If $x$ is in the column space of $V$ then $VV^\top x=x$. But if $x$ is orthogonal to that column space, then $VV^\top x=0$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Dec 19 '15 at 1:10
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Could it be that the missing $(m−p)$ eigenvalues are all zero?

Yes. They are indeed zero. Note that $\text{rank}(E) \leq p$ and $E$ has at most $p$ (not necessarily distinct) non-zero eigenvalues.

Normally I would say $E$ and $V$ are similar, therefore their traces must be equal, but since they have different sizes I don't know.

I suppose you mean $E$ and $\Lambda$ are similar. Note that the commutativity of trace does not depend on the matrix size, as long as the product is a square matrix, namely:

Suppose $A$ is a $m \times n$ matrix and $B$ is a $n \times m$ matrix. Then $\text{trace}(AB) = \text{trace}(BA)$.

You can therefore check the desired result directly from this fact:

$$\text{trace}(E) = \text{trace}(V \Lambda V^T) = \text{trace}(V^T V \Lambda) = \text{trace}(\Lambda).$$

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  • $\begingroup$ Thanks. Yes I meant $E$ and $\Lambda$. Corrected. $\endgroup$ – Ernest A Dec 19 '15 at 0:44

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