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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 $ \begin{pmatrix} i \\ j \end{pmatrix} $ for $ 1 \leq i,j \leq 5 $.
Consider the $ 10 \times 10$ matrix , given in the block form by $$ C =\begin{pmatrix} A & A+B \\ 0 & B \end{pmatrix} .$$ Then,

(a) $\det C=1$ or $-1$,

(b) $\det C=0$,

(c) trace of $C$ is $0$,

(d) trace of $C$ is $5$.

Since $A$ is skew-symmetric, we have $\det A=0$. Hence $\det C=\det(A) \det(B)=0$. But how to find the trace of $C$ ? I need help

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  • $\begingroup$ Skew symmetric matrices do not have to have $det = 0$. $\endgroup$ – Paul Jun 5 '17 at 11:48
  • $\begingroup$ @Paul The precise result is that skew symmetric $n \times n$ matrices with $n$ odd have det=0. We are here in this case. $\endgroup$ – Jean Marie Jun 5 '17 at 12:47
  • $\begingroup$ It will be $ 1\leq i\leq j\leq 5 $ ,instead of $ 1\leq i ,j\leq 5 $ , otherwise the matrix $B $ wouldn't be symmetric. and hence $ B $ is the $ 5\times 5 $ identity matrix. $\endgroup$ – suchanda adhikari May 7 at 17:30
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I do not have enough reputation to comment, so I have to give an answer. Unless I am mistaken, the trace of $C$ is simply $Tr(C) = Tr(A) + Tr(B)$. Since $A$ is skew-symmetric, we have $A=-A^T$ and from this can deduce that $Tr(A)=0$ (can you figure this out?)

In his comment, Paul says that the determinant of a skew-symmetric is not necessarily $0$. But, if I am not mistaken, for matrices of odd dimension it is necessarily true.

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  • $\begingroup$ I think you are right. $\endgroup$ – астон вілла олоф мэллбэрг Jun 5 '17 at 11:49
  • $\begingroup$ No, the trace is given $ \ 5 $ $\endgroup$ – M. A. SARKAR Jun 5 '17 at 14:02
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    $\begingroup$ @mabmath Sure. The trace of $C$ is $5$, but the trace of $A$ is zero, since $Tr(A) = Tr(-A^T)$ or $Tr(A) = - Tr(A)$ and finally $Tr(A) = 0$. In fact, one can prove the stronger claim that the diagonal of $A$ is zero, since $A=-A^T$ implies $a_{ii} = - a_{ii}$. $\endgroup$ – Eli Bashwinger Jun 5 '17 at 14:05
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Hints

  1. The eigenvalues of a real skew-symmetric matrix are imaginary, and so its nonzero eigenvalues come in pairs.
  2. By skew-symmetry, the diagonal entries of any skew-symmetric matrix are all zero.
  3. Since ${i \choose j} = 0$ for $j > i$, the matrix $B$ is lower triangular. (in fact, by definition this matrix is essentially the first five rows of Pascal's Triangle, padded with zeros).
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  • $\begingroup$ One could show that skew-symmetric matrices are singular (for odd sizes) and of trace zero without using or knowing about eigenvalues. $\endgroup$ – Omnomnomnom Jun 5 '17 at 12:09
  • $\begingroup$ @Omnomnomnom That's a good point, and this is easy: All one needs to know is the definition of skew-symmetric, that $\det A = \det (A^{\top})$, and that $\det(\lambda A) = \lambda^n \det A$, where $A$ has size $n \times n$. $\endgroup$ – Travis Jun 5 '17 at 12:14

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