Let $A$ be a $5 \times 5$ skew-symmetric matrix with entries in $\mathbb{R}$ and $B$ be the $5 \times 5$ symmetric matrix whose $(i, j)^{th}$ entry is the binomial coefficient $\binom{i}{j}$ for $1 \le i \le j \le 5$. Consider the $10 \times 10$ matrix, given in block form by $$ C = \left (
\begin{matrix}
A & A+B \\
0 & B
\end{matrix} \right ).$$ Then show that trace of $C$ is $5$.
I came across the above problem. I see that $\text{trace} C = \text{trace} A + \text{trace} B$. Now we see that $\text{trace}B=1+1+1+1+1=5$ (by the given condition, since $\binom{i}{i}=1$). Now the only thing that remains to prove that $\text{trace}A=0$. Can someone point me in the right direction?
Best Answer
Since $A$ is skew-symmetric, $a_{i,i}=-a_{i,i}$ so $a_{i,i}=0$. Hence $Tr\;A=0$.
Now you have seen that $Tr\;B=5$.
So $Tr\;C=Tr\;A+Tr\;B=0+5=5$.