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Background: I'm a chemistry major so I'm sorry if this seems obviously wrong...

This question states a lower bound for $\text{Trace}(B^TB)$ in terms of $\text{Trace}(B)$ derived via the Cauchy-Schwarz inequality.

Is it possible to instead find an upper bound for $\text{Trace}(B^TB)$ in terms of $\text{Trace}(B)$?

I have in the past seen lower and upper bounds derived for sums of square roots using the Cauchy-Schwarz and Minkwoski inequalities respectively but haven't been able to figure it out. I am aware that $\text{Trace}(B^TB) \leq \text{Trace}(B)^2$ when $B$ is semi-positive definite but I am interested in the case of a general square matrix with real entries.

My interest in this problem stems from a practical problem involving the Frobenius norm so I am sorry if it seems out of place. I know the trace of the matrix so it would be incredibly useful if I could relate it via an inequality.

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"I am aware that $\text{Trace}(B^TB) \leq \text{Trace}(B)^2$ when 𝐵 is semi-positive definite but I am interested in the case of a general square matrix with real entries."

You should prove to yourself that in reals,
$\text{Trace}(B^TB) = \big\Vert B \big \Vert_F^2 \geq 0$ with equality iff $B = \mathbf 0$.

Now pick some general $B$ that is traceless. It could for example have bipartite like structure like e.g.

$B:= \begin{bmatrix} 0 & A \\ C & 0 \\ \end{bmatrix} \quad$ for some $A \neq \mathbf 0$ Your desired inequality can never be true here.

Also: consider permutation matrices that don't have fixed points.

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  • $\begingroup$ Beautiful answer. Disappointing result. Thank you! $\endgroup$ – Jack Holmes Feb 8 '20 at 21:51
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Let $A=\begin{bmatrix} 0 & n\\ 0& 0\\ \end{bmatrix}$.Then $\mbox{tr}(A)=0$ but $A^TA=\begin{bmatrix}0 &0 \\0 &n^2 \\ \end{bmatrix}$ and hence $$\mbox{trace}(A^TA)=n^2$$

Basically, what you are asking is the following question

Question Can I bound the sum of squares of all elements in $A$ (this is exactly $tr(A^TA)$) by the sum of diagonal elements in $A$?

The answer is obviously no, since the first sum increases when we increase the non diagonal entries of $A$, while teh second stays unchanged.

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