GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 10

Question

Which of the following is/are correct? (Mark all the appropriate choices)

• A. The determinant of a skew-symmetric matrix over real numbers is non-negative.
• B. When the identity matrix is added to a real skew-symmetric matrix, the resultant matrix is invertible.
• C. For two matrices $A$ and $B,$ the equation $AB = 0$ implies that either $A$ or $B$ is a zero matrix.
• D. For two matrices $A$ and $B,$ if $AB = 0,$ then $BA = 0.$

Answer 1

For option $(A):$ A matrix $A_{n \times n}$ is called skew-symmetric if $A^{T} = -A.$ Here $A^{T}$ is the transpose of $A.$

We know that $\mid A \mid = \mid A^{T} \mid $

$\implies \mid A \mid = \mid -A \mid $

$\implies \mid A \mid =  (-1)^{n} \mid A \mid$

If $n$ is odd, then $\mid A \mid = - \mid A \mid \implies  \mid A \mid = 0$

$\therefore$ Every odd degree skew-symmetric matrix is not invertible, or equivalently singular.

If $n$ is even, the determinant of a skew-symmetric matrix can always be written as the square of a polynomial in the matrix entries, a polynomial with integer coefficients that only depend on the size of the matrix.

Ref: https://en.wikipedia.org/wiki/Pfaffian

For even order, let's $A  = \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix} \implies \mid A \mid  = 0 - (-a\cdot a) = a^{2}$

So, the determinant of a skew-symmetric matrix over real numbers is non-negative.

For option $(B):$ The diagonal of skew-symmetric matrix consists of zero elements and therefore the sum of elements in the main diagonals is equal to zero. When identity matrix is added to skew-symmetric matrix then all the diagonal elements become $1$ and the resultant matrix is invertible.

Let $A  = \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix} \implies \mid A \mid  = 0 - (-a\cdot a) = a^{2} \neq 0$

Now, $A + I = \begin{bmatrix} 0 & a \\ -a & 0 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\implies A + I = \begin{bmatrix} 1 & a \\ -a & 1 \end{bmatrix} \implies \mid A + I \mid = 1 + a^{2} \neq 0$

So, the matrix $A + I$ is invertible.

For option $(C),$ a counter example is enough.

$A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}_{2 \times 2}, B = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$

Now, $AB = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$

For option $(D),$ a counter example is enough.

$A = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}_{2 \times 2}, B = \begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$

Now, $AB = \begin{bmatrix}
0 & 0 \\
0 & 0
\end{bmatrix}_{2 \times 2}$

and, $BA = \begin{bmatrix}
0 & 1 \\
0 & 0
\end{bmatrix}_{2 \times 2}$

So, the correct answer is $A;B.$

Comments

$\begin{bmatrix} 0 &i \\ -i &0 \end{bmatrix}$

for this matrix how is A valid as det value is -1 ?

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$i$ is complex number, not real number, and $i^{2}=-1$.

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for
$A=\begin{bmatrix} 0 & i\\ -i& 0 \end{bmatrix}$

how is (B) valid ?

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for option B proof refer to this: https://math.stackexchange.com/questions/3018066/prove-that-if-a-is-a-real-skew-symmetric-matrix-then-i-a-is-invertible-where