Let $A$ be a $5\times5$ real skew symmetric matrix with entries and $B$ be a $5\times5$ real symmetric matrix whose $(i,j)$-th entery is the binomial coefficient ${i \choose j}$ when $i\ge j$. Now, if
$$C= \begin{pmatrix}A & A+B \\ 0 & B\\ \end{pmatrix},$$
then $\mathrm{trace}(C)$=? $\det(C)$=?
Since $A$ is real and skew symmetric, $\mathrm{trace}(A)=0$; also, $\mathrm{trace}(B)=5$, so $\mathrm{trace}(C)=5$. Am I right?
And given $\det(C)=0$, how can I get $\det(C)=0$?
Can anyone help me please…
Best Answer
Your answer that $\mathrm{trace}(C)=5$ and your reasoning are correct. In general, if $A$ and $B$ are both $n\times n$, since $C$ is block upper triangular, we have $\mathrm{trace}(C)=\mathrm{trace}(A)+\mathrm{trace}(B)=0+n=n$.
We also have $\det(C)=\det(A)\det(B)$. When $n$ is odd, the determinant of an $n\times n$ skew symmetric matrix $A$ is zero. (Hint: For any matrix -- skew symmetric or not -- we have $\det(A)=\det(A^T)$. For skew symmetric matrix, we also have $\det(A^T)=\det(-A)$. Therefore $\det(A)=\det(-A)$. How to prove that $\det(A)=0$ from here? Why is the oddness of $n$ essential?) Therefore $\det(C)=0$.
For even $n$, the determinant of a skew symmetric matrix is not necessarily zero. Since the entries of $A$ are unspecified, there is no way to find its determinant. Also, the determinants of $B$ from $n=1$ to $5$ are resp. $1,-3,15,-97,628$. According to the OEIS database (A079689), there apparently isn't any explicit formula for $\det(B)$ in terms of $n$.