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.


Believe it or not, this method is the fastest and most efficient way to compute the matrices referred to certain basis. This is seen in larger dimensions and saves you solving many systems of equations. Alternatively, you could use the definition :

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 $$\begin{pmatrix} 5&3 \\ -2&0 \end{pmatrix}$$

  • $\begingroup$ I think I got it but just to make sure. To get (1,5) we just incuded (1,3) into (4x-y, 2x+y) and to get the last part we had to solve equation with 2 unknown factors: x(1,3)+y(2,5)=(1,5). We need to do the same thing for the lower row. There isn't any shortcut that I'am missing? $\endgroup$ – LukaMrko Aug 14 at 1:14
  • $\begingroup$ 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. $\endgroup$ – Azif00 Aug 14 at 1:26

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