0
$\begingroup$

Let $L: \mathbb{R}^2\rightarrow \mathbb{R}^2$ be a linear operator. If $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$, find the value of $L((7,5)^T).$

Is there a way to solve these kinds of problems? I only know if $L(\alpha v_1+\beta v_2)=\alpha L(v_1)+\beta L(v_2)$, then the vector space is said to be a linear transformation.

$\endgroup$
0
$\begingroup$

Let your transformation matrix $L=\begin{bmatrix}a&b\\c&d\end{bmatrix}$. Then $L((1,2)^T)=(-2,3)^T$ and $L((1,-1)^T)=(5,2)^T$ give $$\begin{align}&a+2b=-2 &c+2d=3\\&a-b=5 &c-b=2.\end{align}$$ Solving we get $L=\begin{bmatrix}\dfrac{8}{3}&\dfrac{-7}{3}\\\dfrac{7}{3}&\dfrac{1}{3}\end{bmatrix}$. $\:$So $L((7,5)^T)=\begin{bmatrix}7\\18\end{bmatrix}$.

$\endgroup$
  • $\begingroup$ Thanks! But why let the transformation matrix L=(a b c d). Doesn't R2 mean the row or column vectors whose has 2 dimensions? I thought R2 means R 2×1(2 rows,1 column) $\endgroup$ – Shadow Z Nov 18 '18 at 10:35
  • $\begingroup$ @ShadowZ: If we have a linear transformation from $R^n$ ton $R^m$, the corresponding matrix associated with the transformation will have dimension $m\times n$ $\endgroup$ – Yadati Kiran Nov 18 '18 at 10:37
  • $\begingroup$ Okay!Thank you! $\endgroup$ – Shadow Z Nov 18 '18 at 10:41
  • $\begingroup$ Okay!Thank you! $\endgroup$ – Shadow Z Nov 18 '18 at 10:41
0
$\begingroup$

Right, so you have to find $\alpha$ and $\beta$ with $$ \alpha(1,2)+\beta(1,-1)=(7,5)$$

$\endgroup$
  • 1
    $\begingroup$ I figured α equals 4 and β equals 3, then I write L(7,5)^T=4*(-2,3)^T+3*(5,2)^T and I get the answer is (7,18)^T, is it correct? $\endgroup$ – Shadow Z Nov 18 '18 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.