Matrix of Linear Transformation

[skylit]

Let L : M22 → M22 be given by L(X) = XA − AX, where
A =
[ 1 2 ]
[ 2 1 ]

Find the representation of L with respect to the basis S given below:
[ 1 0 ] [ 0 1 ] [ 0 0 ] [ 0 0 ]
[ 0 0 ],[ 0 0 ],[ 1 0 ],[ 0 1 ]

and with respect to the basis U=
[ 1 0 ] [ 1 0 ] [ 0 1 ] [ 0 1 ]
[ 0 1 ],[ 0 -1],[ 1 0 ],[-1 0 ]
and with respect to the usual basis.

All matrices shown are 2x2 square matrices.

I'm not sure what he means by representation and how to approach this problem?

 

[renegade05]

Let L : M22 → M22 be given by L(X) = XA − AX, where
A =
[ 1 2 ]
[ 2 1 ]

Find the representation of L with respect to the basis S given below:
[ 1 0 ] [ 0 1 ] [ 0 0 ] [ 0 0 ]
[ 0 0 ],[ 0 0 ],[ 1 0 ],[ 0 1 ]

and with respect to the basis U=
[ 1 0 ] [ 1 0 ] [ 0 1 ] [ 0 1 ]
[ 0 1 ],[ 0 -1],[ 1 0 ],[-1 0 ]
and with respect to the usual basis.

All matrices shown are 2x2 square matrices.

I'm not sure what he means by representation and how to approach this problem?

I believe what is being asked is to plug in the values and perform matrix multiplication and subtraction.

For example :
\(\displaystyle L(\textbf{X}) = \textbf{XA} − \textbf{AX} \)

Where X = \(\displaystyle \left( \begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right)\) This is the first standard basis in your list.

And A = \(\displaystyle \left( \begin{array}{cc}1 & 2 \\2 & 1 \end{array} \right)\) This is just the matrix A

We Have \(\displaystyle L(\textbf{X}) = \left( \begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right) \left( \begin{array}{cc}1 & 2 \\2 & 1 \end{array} \right) − \left( \begin{array}{cc}1 & 2 \\2 & 1 \end{array} \right) \left( \begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right) = \left( \begin{array}{cc}0 & 2 \\-2 & 0 \end{array} \right)\)

 

[HallsofIvy]

I believe what is being asked is to plug in the values and perform matrix multiplication and subtraction.

For example :
\(\displaystyle L(\textbf{X}) = \textbf{XA} − \textbf{AX} \)

Where X = \(\displaystyle \left( \begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right)\) This is the first standard basis in your list.

And A = \(\displaystyle \left( \begin{array}{cc}1 & 2 \\2 & 1 \end{array} \right)\) This is just the matrix A

We Have \(\displaystyle L(\textbf{X}) = \left( \begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right) \left( \begin{array}{cc}1 & 2 \\2 & 1 \end{array} \right) − \left( \begin{array}{cc}1 & 2 \\2 & 1 \end{array} \right) \left( \begin{array}{cc}1 & 0 \\0 & 0 \end{array} \right) = \left( \begin{array}{cc}0 & 2 \\-2 & 0 \end{array} \right)\)

\(\displaystyle \begin{pmatrix}0 & 2 \\ -2 & 0 \end{pmatrix}= 2\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}\)
So applying the linear transformation to the first of the given basis matrices for U gives 2 times the third of the given basis matrices for V. That means that the first column of the matrix representation of the linear transformation is \(\displaystyle \begin{pmatrix}0 \\ 0 \\ 2 \\ 0 \end{pmatrix}\).

Get the other columns in the same way- apply the linear transformation to each basis matrix, of U, write the result as a linear combination of the basis matrices of V. The coefficients are the numbers in the column of the matrix representing L.