MHB Solving the Matrix Transformation: $B \to C$

[evinda]

Hello! (Wave)

Let $B=(b_1, b_2)$, $C=(c_1, c_2)$ basis of $\mathbb{R}^2$ and $L$ operator of $\mathbb{R}^2$, the matrix as for $B$ of which is $\begin{pmatrix}
2 & 2\\
1 & 0
\end{pmatrix}$. If $b_1=c_1+2c_2+b_2=c_1+3c_2$ and $A=\begin{pmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{pmatrix}$ the matrix of $L$ as for the basis $C$, what does the number $2a_{11}+3a_{12}+a_{21}+3a_{22}$ equal to? (Thinking)

In order to find the desired quantity, do we use the fact that the composition of C with $\begin{pmatrix}
2 & 2\\
1 & 0
\end{pmatrix}$ is equal to $A$ ? But how do we express the composition mathematically? We cannot multiply $C$ by the matrix, since the dimensions do not agree... (Worried)

Or am I somewhere wrong? (Thinking)

 

[HOI]

Hello! (Wave)

Let $B=(b_1, b_2)$, $C=(c_1, c_2)$ basis of $\mathbb{R}^2$ and $L$ operator of $\mathbb{R}^2$, the matrix as for $B$ of which is $\begin{pmatrix}
2 & 2\\
1 & 0
\end{pmatrix}$. If $b_1=c_1+2c_2+b_2=c_1+3c_2$

Do you mean $b_1= c_1+ 3c_2$ and $b_2= c_1+ 3c_2$?

and $A=\begin{pmatrix}
a_{11} & a_{12}\\
a_{21} & a_{22}
\end{pmatrix}$ the matrix of $L$ as for the basis $C$, what does the number $2a_{11}+3a_{12}+a_{21}+3a_{22}$ equal to? (Thinking)

In order to find the desired quantity, do we use the fact that the composition of C with $\begin{pmatrix}
2 & 2\\
1 & 0
\end{pmatrix}$ is equal to $A$ ? But how do we express the composition mathematically? We cannot multiply $C$ by the matrix, since the dimensions do not agree... (Worried)

Or am I somewhere wrong? (Thinking)

Perhaps you are confused as to what $C$ is. C is the pair of 2-vectors $c_1$ and $c_2$ so we certainly can multiply each by the 2 by 2 matrix A. Or we can represent C itself as a 2 by 2 matrix with $c_1$ and $c_2$ as columns and multiply that matrix by A.

To write a linear operator in a given ordered basis, apply the linear operator to each base vector in turn and use the result (again written as a linear combination of the basis vectors) as columns.

So how do we apply A to $c_1$? From $b_1= c_1+ 2c_2$ and $b_2= c_1+ 3c_2$, subtracting eliminates $c_1$ and gives $c_2= b_2- b_1$. Then $b_1= c_1+ 2c_2= c_1+ 2(b_2- b_1)= c_1+ 2b_2- 2b_1$ so $c_1= b_1- 2b_2+ 2b_1= 3b_1- 2b_2$.

So $Ac_1= A(3b_1- 2b_2)= 3Ab_1- 2Ab_2$. But $Ab_1$ is just the first column of the matrix form for A in basis B: $Ab_1= \begin{pmatrix}2 \\ 1 \end{pmatrix}$. Similarly, $Ab_2$ is the second column: $Ab_2= \begin{pmatrix}2 \\ 0 \end{pmatrix}$. Then $Ac_1= 3\begin{pmatrix}2 \\ 1\end{pmatrix}- 2\begin{pmatrix}2 \\ 0 \end{pmatrix}= \begin{pmatrix}2 \\ 3\end{pmatrix}$.

And $Ac_2= A(b_2- b_1)= Ab_2- Ab_1= \begin{pmatrix}2 \\ 0 \end{pmatrix}- \begin{pmatrix}2 \\ 1 \end{pmatrix}= \begin{pmatrix}0 \\ -1\end{pmatrix}$.

Finally, linear operator A, represented as a matrix in basis C, is the matrix having those vectors as columns:
$\begin{pmatrix}2 & 0 \\ 3 & -1 \end{pmatrix}$.