Does $\sum_{i = 1}^{r} \Sigma_{ii}^2 -2\text{tr}(\Sigma C) + \text{tr}(C^{T} C) \geq 0$ hold for positive diagonal $ \Sigma$ and singular $C$?

Question

Let $n>1$ be a natural number, which is strictly bigger than $1$.

Let $\Sigma$ be a $n$-dimensional square diagonal matrix with strictly positive diagonal entries $\Sigma_{ii}>0$.

Let $C$ be an $n$-dimensional square matrix with rank $r \in \{1,...,n-1\}$.

I strongly suspect, that

$$\sum_{i = 1}^{r} \Sigma_{ii}^2 -2\text{tr}(\Sigma C) + \text{tr}(C^{T} C) \geq 0.$$

Is this true?

Answer 1

Let $\Sigma = \begin{bmatrix} 1 & 0 \\ 0 & 100 \end{bmatrix},$ $C= \begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} \Rightarrow C^TC=\begin{bmatrix} 2 & 4 \\ 4 & 8 \end{bmatrix} \hspace{.2cm} \wedge \Sigma C= \begin{bmatrix} 1 & 2 \\ 100 & 200 \end{bmatrix}.$

$tr(C^TC) = 10, tr(\Sigma C) = 201$ and your expression evaluates to $1- 2\times 201 + 10 = -391$. It is all because the first sum goes only to $r$ and not to $n$, which makes it possible to exclude the large diagonal element from the first sum.