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Suppose $B= \left\{v_1,v_2\right\}$ and $B'= \left\{w_1,w_2 \right\}$ given by $w_1=v_1+v_2$ , $w_2=2v_1+3v_2$ are bases of $V$ and $\mathbf{T}: V \rightarrow V$ A linear transformation defined by $\mathbf{T}(v_1)= 3v_1-2v_2$, $\mathbf{T}(v_2)= v_1+4v_2$.

Calculate $[A_T]_{B,B'}$

I can't seem to find the correct matrix that is $$ \left[ {\begin{array}{cc} 8 & 11 \\ -2 & -1 \\ \end{array} } \right] $$

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$T(w_1) = T(v_1+ v_2) = T(v_1) + T(v_2) = 4v_1+ 2v_2 =8 w_1 - 2w_2$ $T(w_2) = T(2v_1+ 3v_2) = 2T(v_1) + 3T(v_2) = (2\cdot3+3\cdot1) v_1+ (2\cdot-2 + 3\cdot 4)v_2 =9 v_1 + 8w_2 = 11w_1 -w_2$

$T\left(\begin{bmatrix} w_1\\w_2\end{bmatrix}\right)=\begin{bmatrix} 8&-2\\11&-1\end{bmatrix}\begin{bmatrix} w_1\\w_2\end{bmatrix}$

Which is the traspose of what you show....

different approach:

$\begin{bmatrix} w_1\\w_2\end{bmatrix} = \begin{bmatrix} 1&1\\2&3\end{bmatrix}\begin{bmatrix} v_1\\v_2\end{bmatrix}$

$\begin{bmatrix} v_1\\v_2\end{bmatrix} = \begin{bmatrix} 3&-1\\-2&1\end{bmatrix}\begin{bmatrix} w_1\\w_2\end{bmatrix}$

$T\left(\begin{bmatrix} v_1\\v_2\end{bmatrix}\right)= \begin{bmatrix} 3&-2\\1&4\end{bmatrix}\begin{bmatrix} v_1\\v_2\end{bmatrix}$

$T\left(\begin{bmatrix} w_1\\w_2\end{bmatrix}\right) = T\left(\begin{bmatrix} 1&1\\2&3\end{bmatrix}\begin{bmatrix} v_1\\v_2\end{bmatrix}\right)=\begin{bmatrix} 1&1\\2&3\end{bmatrix}\begin{bmatrix} 3&-2\\1&4\end{bmatrix}\begin{bmatrix} v_1\\v_2\end{bmatrix} = \begin{bmatrix} 1&1\\2&3\end{bmatrix}\begin{bmatrix} 3&-2\\1&4\end{bmatrix}\begin{bmatrix} 3&-1\\-2&1\end{bmatrix}\begin{bmatrix} w_1\\w_2\end{bmatrix}$

$T\left(\begin{bmatrix} w_1\\w_2\end{bmatrix}\right)=\begin{bmatrix} 8&-2\\11&-1\end{bmatrix}\begin{bmatrix} w_1\\w_2\end{bmatrix}$

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