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Which of the following matrices are non-singular?

  1. $I + A$ where $A$ not equal to $0$ is a skew-symmetric real $n\times n$ matrix, $n\geq 2$.
  2. Every skew-symmetric non-zero real $5 \times 5$ matrix.
  3. Every skew-symmetric non-zero real $2 \times 2$ matrix.
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  • $\begingroup$ b & c are not true. what about a $\endgroup$ – poton Aug 7 '12 at 10:47
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We will use the properties $\det A=\det(A^t)$ and $\det(-A)=(-1)^d\det A$, where $d$ is the dimension of the matrix.

  1. We have $\det(A+I)=\det(A^t+I)=\det(I-A)$. Let $x$ be such that $(A-I)x=0$. Then $$\langle x,x\rangle=\langle x,Ax\rangle=\langle A^tx,x\rangle=-\langle Ax,x\rangle=-\langle x,x\rangle,$$ which proves that $x=0$. We have seen that if $(A-I)x=0$ then $x=0$, hence $A-I$ is invertible. This implies that $\det(I-A)\neq 0 $. Since $\det(I-A)=\det(A+I)$ we get that $\det(A+I)\neq 0$, which proves that $A+I$ is invertible.

  2. Using the mentioned properties, for a skew-symmetrix matrix of odd dimension the determinant is $0$.

  3. In dimension 2, a non-zero skew-symmetric matrix is of the form $A:=\pmatrix{0&x\\-x&0}$ where $x\neq 0$. Its determinant is $x^2\neq 0$ hence $A$ is necessarily invertible. For the other even dimensions (say $d=2N$), we can construct a non-zero skew-symmetric matrix which is not invertible. Indeed, consider the block matrix defined by $$ \pmatrix{A&0_{2,2N-2}\\ 0_{2N-2,2}& 0_{2N-2,2N-2} } $$ where $0_{i,j}$ denoted the matrix with $i$ rows and $j$ columns and all its entries are zero as $A= \pmatrix{0&1\\-1&0}$.

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  • $\begingroup$ How did you write the statement '$x$ such that $Ax=x$'? Please explain. $\endgroup$ – Unknown x Oct 21 '17 at 16:44
  • $\begingroup$ Sir for Matrix $A_{n\times n}$, case where $n$ odd, I got the answer. for the even case, how can i do? $\endgroup$ – Unknown x Oct 21 '17 at 17:42
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    $\begingroup$ The answer to 3. is a bit confusing. A nonzero skew-symmetric $2\times 2$ matrix must be of the form $\begin{bmatrix} 0 & a\\ -a & 0\end{bmatrix}$ for some $a\ne 0$, and is therefore nonsingular. For $n\times n$ with $n$ even and at least $4$, the matrix can be either singular or nonsingular. Use $2\times 2$ blocks of these with at least one block of zeroes (down the diagonal) and it will be singular. $\endgroup$ – Ted Shifrin Oct 21 '17 at 18:09
  • $\begingroup$ @TedShifrin You are right. It needs to be rewritten. $\endgroup$ – Davide Giraudo Oct 21 '17 at 19:11
  • $\begingroup$ @DavideGiraudo Please explain (1). How did you get $Ax=x$? $\endgroup$ – Unknown x Oct 22 '17 at 2:04

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