Find the change of co-ordinate matrix for the following bases for $M(2\times 2)$

Question

I have the following question that I am having trouble solving. My teacher has given the answer, and it roughly makes sense, but it skips a bunch of steps. Can someone walk though a solution to this question?

Let $$\mathcal{B} = \bigg\{ E_{11} = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} , E_{21} = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix} , E_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}, E_{22} = \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} \bigg\} $$ and $$\mathcal{C} = \bigg\{ A = \begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix} , B = \begin{pmatrix} 1 & 1 \\ 0 & 0\end{pmatrix} , C = \begin{pmatrix} 1 & 1 \\ 1 & 0\end{pmatrix}, D = \begin{pmatrix} 1 & 1 \\ 1 & 1\end{pmatrix} \bigg\} $$ be bases for $M_{22}(\mathbb{R})$. Note that $\mathcal{B}$ is not the standard basis for $M_{22}(\mathbb{R})$

A) Find the change-of-coordinate matrices $P_{\mathcal{C} \leftarrow \mathcal{B}}$ and $P_{\mathcal{B} \leftarrow \mathcal{C}}$

B) Use part A) to find $\bigg[\begin{pmatrix} 1 & 2 \\ 3 & 4\end{pmatrix} \bigg]_{\mathcal{C}} $

How do I solve this given that the matrices are represented in this manner? I understand what the question is asking but am uncertain of how to find the solution

Answer 1

to find change of basis matrix from $B \rightarrow C$

$A = 1E_{11}$

$B = 1E_{11}+1E_{12}$

$C = 1E_{11}+1E_{12}+1E_{21}+0E_{22}$

$D = 1E_{11}+1E_{12}+1E_{21}+0E_{22}$

therefore change of basis matrix from $B \rightarrow C$ can be written as \begin{bmatrix} 1 & 1 & 1 &1 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

other change of matrix from $C \rightarrow B$ can be obtained by taking inverse of above matrix or follow same procedure as above

$E_{11} = A$

$E_{12}=-A+B+0C+0D$

$E_{21}=-A+B+0C+0D$

$E_{22} = 0A+0B-C+D$

therefore change of basis matrix from $C \rightarrow B$ can be written as \begin{bmatrix} 1 & -1 & 0 &0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

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B)to find \begin{bmatrix} 1 & -2 \\ 3 & 4\\ \end{bmatrix} with repect to C $[v]_C=[I^C_B]^{-1}[v]_B = [I^B_C] \cdot [v]_B$

Subsituting 2nd matrix found above $C \rightarrow B$ in above expression

Here $[v]_B = 1E_{11}+2E_{12}+3E_{21}+4E_{22} = \begin {bmatrix} 1 \\2 \\3\\ 4 \end{bmatrix}$

we will get $\begin {bmatrix} -1 \\-1 \\-1\\ 4 \end{bmatrix}$

Answer 2

For part (A) you have to express each element of the basis of $\mathcal{C}$ as a linear combination of the elements of the basis $\mathcal{B}$.

For instance $$A= 1\cdot E_{11} + 0 \cdot E_{21} + 0 \cdot E_{12} + 0\cdot E_{22}$$ $$B = 1\cdot E_{11} + 0 \cdot E_{21} + 1 \cdot E_{12} + 0\cdot E_{22}$$ $$C= 1\cdot E_{11} + 1 \cdot E_{21} + 1 \cdot E_{12} + 0\cdot E_{22}$$ $$D = 1\cdot E_{11} + 1 \cdot E_{21} + 1 \cdot E_{12} + 1\cdot E_{22}$$

Now, I'm going to assume $P_{\mathcal{B} \leftarrow \mathcal{C}}$ means changing coordinates from $\mathcal{C}$-coordinates into $\mathcal{B}$-coordinates. So the matrix $P_{\mathcal{B} \leftarrow \mathcal{C}}$is given by writing the the coefficients of each linear combination as a column, so

$$P_{\mathcal{B} \leftarrow \mathcal{C}}= \begin{pmatrix} 1 & 1 & 1 & 1 \\ 0 & 0 &1 & 1\\ 0 & 1 & 1&1 \\ 0 & 0& 0 & 1\end{pmatrix}$$

Likewise, using the same method you should get that the matrix that changes $\mathcal{B}$-coordinates into $\mathcal{C}$-coordinates should be

$$P_{\mathcal{C} \leftarrow \mathcal{B}}= \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & -1 &1 & 0\\ 0 & 1 & 0&-1 \\ 0 & 0& 0 & 1\end{pmatrix}$$

Try to do part (B) with this.