Trace of $U'U$ with $U$ upper-triangular

Question

I have found the following formula for $U$ $(m\times m)$ upper-triangular matrix. \begin{equation}\operatorname{tr}[U'U]=\sum_{i\leq j}^m U_{ij}^2\end{equation} where $\operatorname{tr}[\cdot]$ is the trace operator, $'$ denotes the transposition, $U_{ij}$ the $ij$-th entry of $U$. I can't show this formula, I need an help.

Answer 1

For a general $m\times m$matrix $A$ the trace of $A^TA$ is simply the sum of squares of all elements of $A$:

$$\operatorname{Tr}( A^TA) =\sum_{i=1}^m (A^TA)_{ii} = \sum_{i=1}^m \sum_{j=1}^m (A^T)_{ij}A_{ji} = \sum_{i=1}^m \sum_{j=1}^m A_{ji}A_{ji} = \sum_{i,j=1}^m A_{ij}^2.$$

Now, if your matrix $U$ is upper-triangular, this means that $U_{ij} = 0$ whenever $i > j$ so the sum reduces to $$\operatorname{Tr}( U^TU) = \sum_{i,j=1}^m U_{ij}^2 = \sum_{i \le j}^m U_{ij}^2.$$