Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My task is to find a matrix of linear transformation $\varphi$ in basis $A,B$

$\varphi:\mathbb{R}^{2}\to\mathbb{R}^{4} \varphi((x_{1},x_{2}))=(3x_{1}+x_{2},x_{1}+5x_{2},-x_{1}+4x_{2},2x_{1}+x_{2})$

$\mathcal{A}=\{(3,1),(4,2)\} \mathcal{B}=\{(1,0,1,0),(0,1,1,1),(0,1,2,3),(0,0,0,1)\}$

How I've started:

$M_{st}^{st}(\varphi)=\left[\begin{matrix}3 & 1\\ 1 & 5\\ -1 & 4\\ 2 & 1 \end{matrix}\ \right] $

$M_{B}^{st}(id)=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 1 & 1 & 2 & 0\\ 0 & 1 & 3 & 1 \end{array}\right] $

$M_{A}^{st}(id)=\left[\begin{matrix}3 & 4\\ 1 & 2 \end{matrix}\right] $

$M_{st}^{A}(id)=(M_{A}^{st}(id))^{-1} $

$\left[\begin{matrix}3 & 4 & 1\\ 1 & 2 & & 1 \end{matrix}\right]\sim\left[\begin{matrix}1 & 0 & 1 & -2\\ 1 & 2 & & 1 \end{matrix}\right]\sim\left[\begin{matrix}1 & 0 & 1 & -2\\ 0 & 2 & -1 & 3 \end{matrix}\right]\sim\left[\begin{matrix}1 & 0 & 1 & -2\\ 0 & 1 & -\frac{1}{2} & \frac{3}{2} \end{matrix}\right] $

$M_{st}^{A}(id)=\left[\begin{matrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2} \end{matrix}\right] $

$M(id)_{st}^{A}\cdot M_{st}^{st}(\varphi)=M_{st}^{A}(\varphi)$

$M_{st}^{A}(\varphi)=\left[\begin{matrix}1 & -2\\ -\frac{1}{2} & \frac{3}{2} \end{matrix}\right]\cdot\left[\begin{matrix}3 & 1\\ 1 & 5\\ -1 & 4\\ 2 & 1 \end{matrix}\right] = ???$

I was doing everything with my algorithm. But I did something wrong. Could someone point me where and how to fix it?

Thanks in advance!

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Assuming that vectors in $\mathbb{R}^2$ and $\mathbb{R}^4$ are represented by column vectors, you should find $M_B^{st}(id)^{-1}M_{st}^{st}(\varphi)M_A^{st}(id)$ instead. If you adopt a row vector convention, just transpose the resulting matrix.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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