trace of a 2x2 matrix with itself transposed $\operatorname{tr}(uu^T)=\sum a_{ij}^2$

Question

I have noticed that if I have a 2x2 matrix the following is true: $$\text{tr}(uu^T)=\sum a_{ij}^2$$ by which i mean if we have: $$u=\begin{pmatrix}a&b\\c&d\end{pmatrix},u^T=\begin{pmatrix}a&c\\b&d\end{pmatrix}$$ Then: $$uu^T=\begin{pmatrix}a^2+b^2&ac+bd\\ac+bd&c^2+d^2\end{pmatrix}\Rightarrow \operatorname{tr}(uu^T)=a^2+b^2+c^2+d^2$$ I have also noticed that this shows: $$uu^T=(uu^T)^T$$ However I am not sure if this is true for larger square matrices and a reason why this is true using known rules

Answer 1

This is true in general and if you take square-roots of both sides, you'll get the definition of the Frobenius norm.

To prove it in full generality, notice that:

$$\operatorname{tr}(uu^T)=\sum_{i=1}^n [uu^T]_{ii}=\sum_{i=1}^n\sum_{k=1}^nu_{ik}[u^T]_{ki}=\sum_{i=1}^n\sum^n_{k=1}u_{ik}^2$$

Concerning $uu^T=(uu^T)^T$, the proof is similar, and easier if you know or derive that $(ab)^T=b^Ta^T$