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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...

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  • $\begingroup$ $B$ should be symmetric, so it is probably meant that the entry $(12)$ is equal to the entry $(21)$ which is $\begin{pmatrix}2\\1\end{pmatrix}$, which is defined. $\endgroup$ – Julian Kuelshammer Jun 22 '13 at 7:06
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    $\begingroup$ You should either drop the "symmetric" qualification for $B$, or change the definition of its entries. Setting $B_{i,j}=\binom{i+j}i$ would be symmetric for instance. For the record, $\binom12=0$ and in general $\binom nk=0$ whenever $0\leq n<k$; these are perfectly well defined, even by the combinatorial definition (there are $0$ ways to choose $k$ among $n$ objects in this case). The matrix $B$ as specified in the question is lower triangular with diagonal entries equal to$~1$. $\endgroup$ – Marc van Leeuwen Jun 22 '13 at 7:11
  • $\begingroup$ @ Julian Kuelshammer sir,I get $$B= \begin{pmatrix}1& 2 & 3 & 4 & 5 \\ 2& 1 & 3 & 6 & 10\\ 3 & 3 & 1 & 4 & 10 \\ 4 & 6 & 4 & 1 & 5\\ 5 & 10 & 10 &5 & 1 \end{pmatrix}$$ $\endgroup$ – user45799 Jun 22 '13 at 7:16
  • $\begingroup$ @ Julian Kuelshammer sir, then how can I find trace C=? det C=? $\endgroup$ – user45799 Jun 22 '13 at 7:25
  • $\begingroup$ @ Julian Kuelshammer sir,since A be a 5*5 skew symmetric matrix with entries in $\mathbb R$, trace A =0 and trace B=5,so trce C=5 ....am I right... what about det C=? $\endgroup$ – user45799 Jun 22 '13 at 7:33
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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 $n$.

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  • $\begingroup$ @ user1551 please provide me prove that det(A)=0 When n is odd and Why is the oddness of n essential? $\endgroup$ – user45799 Jun 22 '13 at 13:29
  • $\begingroup$ @Prasanta If $k$ is a real number and $A$ is $n\times n$, can you express $\det(kA)$ in terms of $\det(A)$? $\endgroup$ – user1551 Jun 22 '13 at 13:31
  • $\begingroup$ @ user1551 det(kA) = $k^n$ det(A) $\endgroup$ – user45799 Jun 22 '13 at 13:37
  • $\begingroup$ @Prasanta Yes. So, what if $k=-1$ and $n$ is odd? $\endgroup$ – user1551 Jun 22 '13 at 13:54
  • $\begingroup$ det(−A) = - det(A) so, det(A)=0 thanks @ user1551... $\endgroup$ – user45799 Jun 22 '13 at 13:57

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