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A linear transformation $T: \Bbb R^2\to \Bbb R^2$ is given by

$T(i) = i+j; T(j) = 2i -j$.

With respect to the basis $\{i,j\}$, The matrix of $T = \left(\begin{array}{cc}1 & 2 \\1 & -1\end{array}\right).$

What is the matrix of $T$ if the basis $\{i,j\}$ is replaced by $e_1 = (i-j)$
and $e_2=3i+j$?

The solution is: $\left(\begin{array}{cc}-7/4 & -1/4 \\1/4 & 7/4\end{array}\right)$.

I am not able to come up with this solution. Here's my attempt:

$T(e_1) = T(i) - T(j) = i+j -(2i -j) = -i + 2j$

$T(e_2) = 3T(i) + T(j) = 3(i+j) +2i-j = 5i+2j$.

Thus, $T$ would be $\left(\begin{array}{cc}-1 & 5 \\2 & 2\end{array}\right)$.


Second Question:

If you let $T = \left(\begin{array}{cc}1 & 2 \\1 & -1\end{array}\right).$ Then what is $T$ relative to $e_1,e_2$?

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  • $\begingroup$ You have to express $T(e_1)$ and $T(e_2)$ in the basis $\{e_1,e_2\}$. $\endgroup$ – Guest Mar 6 '15 at 20:46
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If $T(i)=(1,1)$ and $T(j)=(2,-1)$ and $e_1=i-j=(1,-1)$ and $e_2=3i+j=(3,1)$, then $$T(e_1)=T(i-j)=T(i)-T(j)=(-1,2)=a_1e_1+b_1e_2 \,\text{(say)}$$ and $$T(e_2)=T(3i+j)=3(1,1)+(2,-1)=(5,2)=a_2e_1+b_2e_2\,\text{(say)}.$$ Thus the matrix of $T$ w.r.t. the new basis $\{e_1,e_2\}$ is $\begin{pmatrix} a_1 & a_2\\ b_1 & b_2 \end{pmatrix}$ and you need to find the values of $a_1,a_2,b_1$ and $b_2$. The above systems of equations reduces to $$a_1+3b_1=-1\\ -a_1+b_1=2$$ and

$$a_2+3b_2=5\\ -a_2+b_2=2.$$ Solve these equations to obtain $a_1=-\frac{7}{4},b_1=\frac{1}{4},a_2=-\frac{1}{4}$ and $b_2=\frac{7}{4}$.

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  • $\begingroup$ My confusion is that I don't see why I shouldn't do this: $T(i) = (1,1) = a_1e_1 + b_1e_2$.$T(j) = (2,1) = a_2e_1+b_2e_2$. Solve for the $a$s and $b$'s. $\endgroup$ – larry Mar 7 '15 at 1:41
  • $\begingroup$ $T = \left(\begin{array}{cc}a_1 & a_2 \\ b_1 & b_2\end{array}\right)$? Why? $T*e_1$ does not give you the correct answer. $\endgroup$ – larry Mar 11 '15 at 0:41
  • $\begingroup$ @larry: Sorry I did not get you. Can explain more what are you asking about? $\endgroup$ – user149418 Mar 11 '15 at 13:28
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$e_1+e_2=4i$;$i=\frac{e_1+e_2}{4} = P(i |j)^T$;$e_2-3e_1=4j$;$j=\frac{e_2-3e_1}{4} = P(i| j)^T$.

Now compute the Matrix $P^{-1}$ and your $T$ in the new Basis is given by $P^{-1}TP$.

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  • $\begingroup$ If you defined $T=\left(\begin{array}{cc}1 & 2 \\1 & -1\end{array}\right)$, and you wanted to find $T$ relative to new bases $e_1,e_2$ would T now be $\left(\begin{array}{cc} -.5 & 1.25 \\.5 & .25\end{array}\right)$? $\endgroup$ – larry Mar 6 '15 at 21:49
  • $\begingroup$ T(i) = (1,1)' = a(e1) + b(e2), where a,b = -.5,.5. And you do something similar for T(j). Doesn't this give you T in a new bases? $\endgroup$ – larry Mar 6 '15 at 21:59
  • $\begingroup$ I don't want to compute inverses for this problem. Second, the notation has changed and it's more confusing. $\endgroup$ – larry Mar 6 '15 at 22:01
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The matrix of the transformation of basis is: $$ P= \begin{bmatrix} 1&3\\ -1&1 \end{bmatrix} $$ and $$ P^{-1}= \dfrac{1}{4} \begin{bmatrix} 1&-3\\ 1&1 \end{bmatrix} $$ Your transformed matrix is: $$ P^{-1}TP= \dfrac{1}{4} \begin{bmatrix} 1&-3\\ 1&1 \end{bmatrix} \begin{bmatrix} 1&2\\ 1&-1 \end{bmatrix} \begin{bmatrix} 1&3\\ -1&1 \end{bmatrix}= \dfrac{1}{4} \begin{bmatrix} -7&-1\\ 1&7 \end{bmatrix} $$


Added after comments.

enter image description here

The figure illustrate how operate the given transformation in the two basis.

Indices $\{i,j\}$ refer to the canonical basis, indices $\{1,2\}$ to the new basis $\mathbf{e_1},\mathbf{e_2}$.

In the figure we have the vector $\mathbf{v}=\mathbf{v_{i,j}}=[1,2]^T$ (in canonical basis), that is transformed in $\mathbf{v'}=\mathbf{v'_{i,j}}=[5,-1]^T$.

In the new basis we have: $$ \mathbf{v_{1,2}}= P^{-1}\mathbf{v_{i,j}}= \dfrac{1}{4}\left[ \begin{array}{cccc} -5\\ 3 \end {array} \right] $$ and, for the tranformed vector: $$ \mathbf{v'_{1,2}}= P^{-1}\mathbf{v'_{ij}}= \left[ \begin{array}{cccc} 2\\ 1 \end {array} \right]= P^{-1}T_{i,j}\mathbf{v_{i,j}}=P^{-1}T_{i,j}P\,\mathbf{v_{1,2}}=T_{1,2}\mathbf{v_{1,2}} $$ so we have: $$ T_{1,2}=P^{-1}T_{i,j}P= \dfrac{1}{4} \left[ \begin{array}{cccc} -7&-1\\ 1&7 \end {array} \right] $$

I hope this can be helpful.

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  • $\begingroup$ What if you come across a problem where $P$ is not invertible? How would you get around this? $\endgroup$ – larry Mar 6 '15 at 22:15
  • $\begingroup$ I'm also trying to figure out what the $T$ is relative to $e_1,e_2$. $\endgroup$ – larry Mar 6 '15 at 22:16
  • $\begingroup$ If $P$ is not invertible we hawe not a transformation of basis. $\endgroup$ – Emilio Novati Mar 6 '15 at 22:24
  • $\begingroup$ math.stackexchange.com/questions/1178772/… seems to be changing the basis of a transformation matrix, and it's not invertible. $\endgroup$ – larry Mar 6 '15 at 22:28
  • $\begingroup$ What is $T$ relative to $e_1,e_2$? $\endgroup$ – larry Mar 6 '15 at 22:32

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