I don't understand this linear transformation problem.

Question

I am learning linear algebra from brilliant.org, and I don't understand what this problem is asking for or why the answer is $24$,can anyone help me?

This is the problem and the screenshot

Answer 1

The standard basis for the vector space of cubic polynomials is given by the $4$ polynomials: $$ \mathbf{p}_{1}=x^3,\quad\mathbf{p}_{2}=x^2,\quad\mathbf{p}_{3}=x,\quad\mathbf{p}_{4}=1 $$ The standard basis for the vector space of $2\times2$ matrices is given by the $4$ matrices: $$ \mathbf {M}_{11}=\begin{pmatrix}1&0\\0&0\end{pmatrix},\quad \mathbf {M}_{12}=\begin{pmatrix}0&1\\0&0\end{pmatrix},\quad \mathbf {M}_{21}=\begin{pmatrix}0&0\\1&0\end{pmatrix},\quad \mathbf {M}_{22}=\begin{pmatrix}0&0\\0&1\end{pmatrix} $$ Now each of these vector spaces has dimension $4$. Thus we can map $\mathbf{p}_{1}$ to one of $\mathbf {M}_{11}$, $\mathbf {M}_{12}$, $\mathbf {M}_{21}$, $\mathbf {M}_{22}$, that is we initially have $4$ choices. Say we map it to $\mathbf {M}_{11}$, then we have to map $\mathbf{p}_{2}$ to one of the remaining matrices $\mathbf {M}_{12}$, $\mathbf {M}_{21}$, $\mathbf {M}_{22}$, that is we have $3$ choices. Going on like this we find we have $4\cdot3\cdot2\cdot1=4!=24$ choices in total for mapping the basis polynomials to the basis matrices.

Answer 2

You want to form a $2\times 2 $ matrix with four elements {a,b,c,d} without repeating any element.

For the first spot you have$ 4$ options.

After your first spot is filled then for your second spot you have only $3$ options.

For the third spot you have only two options left and finally for the last spot you only have one option.

That makes a total of $$4\times 3\times 2\times1=24 $$options.