Question

Find the basis of the space Vof all skew-symmetric 3x3 matrices, and thus determine the dimension of V.

Short answer

The dimension of a $nxn$skew-symmetric matrices is $\frac{n^2-n}{2}$which is spanned by .

$Span\left\{\left[\begin{array}{cccc}0& 1& \dots & 0\\ -1& 0& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 0\end{array}\right],...,\left[\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & 1\\ 0& -1& 0\end{array}\right]\right\}$.

Step-by-step solution

Determine the basis.

Consider the matrix $A=\left[\begin{array}{cccc}0& a_{12}& \dots & a_{1n}\\ -a_{12}& 0& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& a_{2n}& \dots & 0\end{array}\right]$where all$a_{ij}$are real.

The matrix A is skew-symmetric if ${\mathit{A}}^{\mathbf{T}}\mathbf{=}\mathbf{-}\mathit{A}$and the general form of any skew-symmetric matrix is $\left[\begin{array}{cccc}0& a_{12}& \dots & a_{1n}\\ -a_{12}& 0& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& a_{2n}& \dots & 0\end{array}\right]$.

The total number of elements in the strictly upper triangular matrix is $\frac{{\mathbf{n}}^{\mathbf{2}}\mathbf{-}\mathbf{n}}{\mathbf{2}}$.

Simplify the equation localid="1660129735066" $A=\left[\begin{array}{cccc}0& a_{12}& \dots & a_{1n}\\ -a_{12}& 0& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& -a_{2n}& \dots & 0\end{array}\right]$as follows:

$$\begin{gathered} A=\left[\begin{array}{cccc}0& a_{12}& \dots & a_{1n}\\ -a_{12}& 0& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& -a_{2n}& \dots & 0\end{array}\right] \\ A=\left[\begin{array}{cccc}0& a_{12}& \dots & a_{1n}\\ -a_{12}& 0& \dots & a_{2n}\\ ⋮& ⋮& \ddots & ⋮\\ -a_{1n}& -a_{2n}& \dots & 0\end{array}\right]+...+\left[\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & a_{\left(n-1\right)n}\\ 0& a_{\left(n-1\right)n}& 0\end{array}\right] \\ A=a_{12}\left[\begin{array}{cccc}0& 1& \dots & 0\\ -1& 0& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 0\end{array}\right]+...+a_{\left(n-1\right)n}\left[\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & 1\\ 0& -1& 0\end{array}\right] \\ \mathrm{where}\left[\begin{array}{cccc}0& 1& \dots & 0\\ -1& 0& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 0\end{array}\right],...,\left[\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & 1\\ 0& -1& 0\end{array}\right]\mathrm{are}\mathrm{linear}\mathrm{independent} \end{gathered}$$

Therefore, the matrix A is spanned by localid="1660130418420" $Span\left\{\left[\begin{array}{cccc}0& 1& \dots & 0\\ -1& 0& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 0\end{array}\right],...,\left[\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & 1\\ 0& -1& 0\end{array}\right]\right\}$.

By the definition, the total number of elements in the set is .

$\left\{\left[\begin{array}{cccc}0& 1& \dots & 0\\ -1& 0& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 0\end{array}\right],...,\left[\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & 1\\ 0& -1& 0\end{array}\right]\right\}\mathrm{is}\frac{{\mathrm{n}}^2-\mathrm{n}}{2}$

Hence, the dimension of A is which is $\frac{n^2-n}{2}$spanned by $Span\left\{\left[\begin{array}{cccc}0& 1& \dots & 0\\ -1& 0& \dots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 0\end{array}\right],...,\left[\begin{array}{ccc}0& \dots & 0\\ ⋮& \ddots & 1\\ 0& -1& 0\end{array}\right]\right\}$width="263">Span01…0-10…0⋮⋮⋱⋮00…0,...,0…0⋮⋱10-10