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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?

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The only thing that remains to be observed is that the diagonal of a skew-symmetric matrix is made of $0$'s. – 1015 Feb 2 '13 at 19:25
What kind of question title is "Let A be a 5x5 skew-symmetric matrix"? – Henning Makholm Feb 3 '13 at 2:03
I have edited the title. – user52976 Feb 3 '13 at 2:43
up vote 3 down vote accepted

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 $Tr\;C=Tr\;A+Tr\;B=0+5=5$.

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In a skew-symmetric matrix the diagonal entries are zero implying the trace is zero.

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