# How to find out the linear transformation from its matrix with respect to non-standard bases?

Here is one example:

Find the functional form of the linear operator $f$ on $\mathbb{R}^2$ whose matrix with respect to the basis $\mathcal{B}=\{(1,1),(1,-1)\}$ is $$\begin{bmatrix} 2 & -1\\3&5\end{bmatrix}.$$

Hint: From the first column, we can deduce that $$f(1,1) = 2 (1,1) + 3(1,-1) = (5,-1)$$ From the second column, we can deduce that $$f(1,-1) = -(1,1) + 5(1,-1) = (4,-6)$$ In the standard matrix of $f$, the first column will be $f(1,0)$, and the second column will be $f(0,1)$.

Alternatively: let $$M = \pmatrix{2&-1\\3&5}$$ We know that the matrix $$P = \pmatrix{1&1\\1&-1}$$ implements a change of basis from the standard basis to $\mathcal B$. If we take $A$ to denote the matrix that we're looking for, we have $$PM = AP \implies A = PMP^{-1}$$ which can be directly computed.

Being $f$ a linear map we have to remember that:

$$f(ax) = af(x) \\f(a+b) = f(a)+f(b)$$

The matrix representation of a linear map, after choosing a base for the domain and for the image (in our case $\mathcal{B} = \{(1,1),(1,-1)\})$ is done by evaluating the map on the basis and then writing down the resultant vector in the basis of the image: the column of the matrix are the coefficients of that vector.

## Long solution

In our case then we have that: $$f((1,1)) = 2(1,1)+3(1,-1)\\ f((1,-1)) = -1(1,1)+5(1,-1))$$ All we have to do now is to write down our basis vectors in the standard base $\mathcal{C} = \{(1,0), (0,1)\}$. This can be easily done: $$\color{red}{(1,1)} = 1(1,0)+1(0,1)\\ \color{blue}{(1,-1)} = 1(1,0)-1(0,1)$$ We can then utilise the fact that $f$ is a linear map, the first becomes $$f(\color{red}{(1,0)+(0,1)}) = 2((1,0)+(0,1))+3((1,0)-(0,1)) \\ f((1,0))+f((0,1)) = 5(1,0)-1(0,1)$$ and the second $$f(\color{blue}{(1,0)-(0,1)}) = -((1,0)+(0,1))+5((1,0)-(0,1)) \\ f((1,0))-f((0,1)) = 4(1,0)-6(0,1)$$ All we need to find now is the values of $f((0,1))$ and $f((1,0))$ which is easily done by solving the system of the fist and second equalities. Doing the calculations we get: $$2f((1,0)) = 9(1,0)-7(0,1)\\-2f((0,1)) = -(1,0)-5(0,1)$$ So the matrix in the standard base is $$[f]_{\mathcal{C}}^{\mathcal{C}} = \left(\begin{matrix} {9\over 2} & {1\over 2}\\ -{7\over 2}& {5\over 2}\end{matrix}\right)$$ The explicit representation is easily found: take a vector $\mathbf{v} = (x,y)$ then $$f(\mathbf{v}) = [f]_{\mathcal{C}}^{\mathcal{C}}\mathbf{v} = \left({9\over 2}x+{1\over 2}y, -{7\over 2}x+{5\over 2}y\right)$$

## Fast, easy, better solution


Let consider the matrix

$$M=\begin{bmatrix} 1 & 1\\1&-1\end{bmatrix}$$

which represents the change of basis form the basis $\mathcal{B}$ to the standard basis, therefore from

$$w_\mathcal{B}= A v_\mathcal{B}\quad w_\mathcal{B}= \begin{bmatrix} 2 & -1\\3&5\end{bmatrix}v_\mathcal{B}$$

and

• $w=M w_\mathcal{B}\implies w_\mathcal{B}=M^{-1} w$

• $v=M v_\mathcal{B}\implies v_\mathcal{B}=M^{-1} v$

we have

$$w_\mathcal{B}= A v_\mathcal{B}$$

$$M^{-1} w= A M^{-1} v$$

$$w= MA M^{-1} v$$