Question

TRUE OR FALSE

There exist a linear transformation L from ${\mathit{R}}^{\mathbf{3}\mathbf{\times }\mathbf{3}}$to${\mathit{R}}^{\mathbf{2}\mathbf{\times }\mathbf{2}}$ whose kernel is the space of all skew-$\mathbf{3}\mathbf{\times }\mathbf{3}$symmetric matrices.

Short answer

The given statement is false.

Step-by-step solution

Given Information

Consider the statement as follows:

There exists a linear transformation from $R^{3\times 3}to{R}^{2\times 2}$ whose kernel is the space of all skew-symmetric$3\times 3$ matrices.

Explanation of the solution

The formula is as follows:

$$\begin{gathered} dim\left(R^{n\times n}\right)=n^2 \\ dim\left(R^{3\times 3}\right)=9 \\ dim\left(R^{2\times 2}\right)=4 \end{gathered}$$

Use the above formula for $3\times 3$and $2\times 2$matrices to find the dimension.

$$\begin{gathered} dim\left(R^{3\times 3}\right)=9 \\ dim\left(R^{2\times 2}\right)=4 \end{gathered}$$

Therefore, $dim\left(ker\left(L\right)\right)$should be at least 5.

But it is given that kernel of L is the space of all skew-symmetric $3\times 3$matrices whose dimension is 3.

That is, the basis for the space of set all skew-symmetric $3\times 3$matrices are given by as follows.

$\left[\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right],\left[\begin{array}{ccc}0& 0& -1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]$and $\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& -1\\ 0& 1& 0\end{array}\right]$

Therefore, its dimension is 3.

Thus, the given statement “There exists a linear transformation from $R^{3\times 3}to{R}^{2\times 2}$ whose kernel is the space of all skew-symmetric $3\times 3$matrices” is false.