# upper bound for $\operatorname{trace}(A^TA)$ in terms of $\operatorname{trace}(A)$

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.

"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.

• Beautiful answer. Disappointing result. Thank you! – Jack Holmes Feb 8 '20 at 21:51

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.