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We have a theorem says that "ODD-SIZED SKEW-SYMMETRIC MATRICES ARE SINGULAR" . Proof link is given here if needed. Now let us assume we have a $3\times 3$ skew symmetric matrices of the form $ \begin{bmatrix}\,0&\!-a_3&\,\,a_2\\ \,\,a_3&0&\!-a_1\\-a_2&\,\,a_1&\,0\end{bmatrix}$ and an Identity matrix $I_{3\times3}$

Question

  1. Can we say determinant of $I_{3\times3}+\begin{bmatrix}\,0&\!-a_3&\,\,a_2\\ \,\,a_3&0&\!-a_1\\-a_2&\,\,a_1&\,0\end{bmatrix} \tag 1$

    is not zero always? if so how can we prove it mathematically ?

NB:: $a_1,a_2,a_3$ cant be zero together at a time

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3 Answers 3

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Another way: if $A$ is skew-symmetric, then for any vector $v$ we have $v^T A v = - v^T A^T v = - v^T A v$ so $v^T A v = 0$. Now for any real vector $v$, if $(I + A) v = 0$ we have $$0 = v^T (I+A) v = v^T v + v^T A v = v^T v$$ so $v = 0$. This works whether $n$ is odd or even.

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Yes, it is indeed never zero. We can expand the determinant along the first column, and find: $$ \det(I_{3\times 3} + A) = 1(1 + a_1^2) + a_3(a_1a_2 + a_3) - a_2(a_1a+3 - a_2) = 1 + a_1^2 + a_2^2 + a_3^2 \neq 0 $$

We can even allow $a_1 = a_2 = a_3$ since then we would just have the identity matrix.

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Yes, and moreover for any $n\times n$ skew-symmetric matrix $A$ with $n$ odd, $I_n+A$ is non-singular. This is because the eigenvalues of $A$ are $0$ or purely imaginary (coming in pairs of the form $\pm\lambda i$ for $\lambda\in\mathbb{R}$). Hence $I_n+A$ has eigenvalues $1$ and $1\pm\lambda i.$

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