Suppose $A = \{x_1, x_2\}$ is a basis for some vector space $V$, and that $T: V \rightarrow V$ is a linear transformation. Let $B = \{y_1, y_2\}$ be another basis for $V$ with $y_1 = x_1 + x_2$ and $y_2 = 2x_1 + 3x_2$. If $$T_A = \begin{bmatrix} 3 & 1 \\ -2 & 4\end{bmatrix}$$ I wish to find $T_B$. How would I go about doing this? Thanks.
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$\begingroup$ Did you know what is the change-of-coordinate matrix from $A$ to $B$? $\endgroup$– mactonJan 25, 2021 at 9:19
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$\begingroup$ @macton This is all I've been given $\endgroup$– CBBAMJan 25, 2021 at 9:22
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$\begingroup$ sorry for my bad formulation. Can you write down the change-of-coordinate matrix from $A$ to $B$ given that $y_1 = x_1 + x_2$ and $y_2 = 2x_1 + 3x_2$? $\endgroup$– mactonJan 25, 2021 at 9:24
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$\begingroup$ @macton Wouldn't it be $\begin{bmatrix} 1 & 2\\ 1 & 3 \end{bmatrix}$? $\endgroup$– CBBAMJan 25, 2021 at 9:30
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1$\begingroup$ Correct. Now did you know that $$T_B = Q^{-1}T_AQ$$ where $Q$ is the matrix you give? $\endgroup$– mactonJan 25, 2021 at 9:33
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1 Answer
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We only have to "read" the images of the vectors $x_1$ and $x_2$ and then write them in function of $y_1$ and $y_2$.
$T(x_1)=3x_1-2x_2$ and $T(x_2)=x_1+4x_2$. Since the map is linear we can now observe that
$$T(y_1)=T(x_1+x_2)=T(x_1)+T(x_2)=4x_1+2x_2$$
$$T(y_2)=T(2x_1+3x_2)=2T(x_1)+3T(x_2)=6x_1-4x_2+3x_1+12x_2=9x_1+8x_2.$$
Now we have $T(y_1)=8y_1-2y_2$ and $T(y_2)=11y_1-y_2$, therefore
$$\mathcal M_B^B(T)=T_B=\begin{pmatrix} 8 && 11\\-2 && -1\end{pmatrix}.$$
In order to find $T(y_i)$ I solved the systems $$T(y_1)=\alpha y_1+\beta y_2\text{ and }T(y_2)=\alpha y_1+\beta y_2$$
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$\begingroup$ How did you conclude from $T(y_1) = 4x_1 + 2x_2$ that $T(y_1) = 8y_1 - 2y_2$? $\endgroup$– CBBAMJan 25, 2021 at 9:41
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1$\begingroup$ I edited my answer:-) I find the coefficients $\alpha$ and $\beta$ to write the images of the vectors $y_1,y_2$ as a linear combination of $B$. $\endgroup$– VajraJan 25, 2021 at 9:47