# Second method for finding matrix operator $f$ in base $\beta_2$

There is a linear operator $$f:R^2 \rightarrow R^2$$ with $$f(x,y)=(4x-y, 2x+y)$$ and there is base $$\beta_2=\{(1,3)(2,5)\}$$ in $$R^2$$.

I need to find matrix of operator $$f$$ in base $$\beta_2$$.

I got the result by first getting matrix $$T=\begin{bmatrix}1 & 2\\3 & 5\end{bmatrix}$$.

Then I found the inverse of $$T$$ which is $$T^{-1}=\begin{bmatrix}-5 & 2\\3 & -1\end{bmatrix}$$

And final component that I needed was $$F_{\beta_{canon}}$$ which was $$F_{\beta_{canon}}=\begin{bmatrix}4 & -1\\2 & 1\end{bmatrix}$$

Finally to get $$F_{\beta_{2}}$$ I used equation $$F_{\beta_{2}}=T^{-1}*F_{\beta_{canon}}*T$$ and I got:

$$F_{\beta_{2}}=\begin{bmatrix}5 & 3\\-2 & 0\end{bmatrix}$$

I'm just wondering what is the other method of finding given operator, because this one is relatively slow.

First, compute $$f(1,3)$$ and $$f(2,5)$$ and put them as a linear combination of the same elements of $$\beta_2$$, like this $$f(1,3) = (1,5) = 5(1,3)+(-2)(2,5)$$ $$f(2,5) = (3,9) = 3(1,3)+0(2,5)$$ Then, put the coefficients in the columns of your matrix as they come out $$\begin{pmatrix} 5&3 \\ -2&0 \end{pmatrix}$$
• That's right, to evaluate $f$ in the point $(1,3)$ just substitute $x$ with $1$ and $y$ with $3$ in the expression $(4x-y,2x+y)$. And yes, you had to solve TWO system of linear equations. That's why I say that the other method is the most efficient. – Azif00 Aug 14 at 1:26