# Computing the determinant and trace of a $10 \times 10$ matrix with a particular block form

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

• Skew symmetric matrices do not have to have $det = 0$. – Paul Jun 5 '17 at 11:48
• @Paul The precise result is that skew symmetric $n \times n$ matrices with $n$ odd have det=0. We are here in this case. – Jean Marie Jun 5 '17 at 12:47
• 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. – suchanda adhikari May 7 at 17:30

## 2 Answers

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.

• I think you are right. – астон вілла олоф мэллбэрг Jun 5 '17 at 11:49
• No, the trace is given $\ 5$ – M. A. SARKAR Jun 5 '17 at 14:02
• @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 $a_{ii} = - a_{ii}$. – Eli Bashwinger Jun 5 '17 at 14:05

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).
• One could show that skew-symmetric matrices are singular (for odd sizes) and of trace zero without using or knowing about eigenvalues. – Omnomnomnom Jun 5 '17 at 12:09
• @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 $n \times n$. – Travis Jun 5 '17 at 12:14