# Linear transformation between bases

1. Sep 6, 2009

### gothlev

Hi !

I am a little bit confused with notation in the following:

Let A=

$$\begin{bmatrix} 2 & 3 & 4 \\ 8 & 5 & 1 \\ \end{bmatrix}$$

and consider A as a linear transformation mapping $$\mathbb{R}^3$$ to $$\mathbb{R}^2$$. Find the matix representation of A with respect to the bases

$$\begin{bmatrix} 1\\ 1\\ 0\\ \end{bmatrix} ,$$ $$\begin{bmatrix} 0\\ 1\\ 1\\ \end{bmatrix} ,$$ $$\begin{bmatrix} 1\\ 0\\ 1\\ \end{bmatrix}$$ of $$\mathbb{R}^3$$ and

$$\begin{bmatrix} 3\\ 1\\ \end{bmatrix} ,$$ $$\begin{bmatrix} 2\\ 1\\ \end{bmatrix}$$ of $$\mathbb{R}^2$$

It seems to be a lot of A´s in here with different meanings, and I suppose it is what confuses me :(. Anyway I solved it as follows:

$$\begin{bmatrix} 3 & 2\\ 1 & 1\\ \end{bmatrix}^{-1} *$$ $$\begin{bmatrix} 2 & 3 & 4\\ 8 & 5 & 1\\ \end{bmatrix} *$$ $$\begin{bmatrix} 1 & 0 & 1\\ 1 & 1 & 0\\ 0 & 1 & 1\\ \end{bmatrix} =$$ $$\begin{bmatrix} -21 & -5 & -12\\ 34 & 11 & 21\\ \end{bmatrix}$$
I am still not sure that I have not confused myself with all the different A´s :( Am I on the right track or completely lost ?

2. Sep 6, 2009

### loveequation

I think you are right but no guarantees. Just as a vector has different representations (components) in different bases, a matrix has different representations in different bases.

3. Sep 6, 2009

### HallsofIvy

In general, to find the matrix representation of A, from U to V, with $\{u_1, u_2, ..., u_n\}$ a basis for U and $\{v_1, v_2, ..., v_m\}$ a basis for V:

Apply A to each of $\{u_1, u_2, ..., u_n\}$ in turn and write the result as a linear combination of the $\{v_1, v_2, ..., v_m\}$. The coefficients form the columns of the matrix.

for example, here
$$u_1= \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$$
so
$$Au_1= \begin{bmatrix}2 & 3 & 4 \\8 & 5 & 1 \\\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}5 \\ 13\end{bmatrix}$$

Now, [5, 13]= a[3, 1]+ b[2, 1] gives 3a+ 2b= 5 and a+ b= 13. Multiplying the second equation by 2, 2a+ 2b= 26 and, subtracting that from the first equation a= -21. -21+ b= 13 gives b= 34. The first column of the matrix you want is
$$\begin{bmatrix}-21 \\ 34\end{bmatrix}$$

Last edited by a moderator: Sep 6, 2009
4. Sep 6, 2009

### gothlev

Thx for the replies. Thank you for a very clear and good explanation (HallsofIvy), the book I am reading is very compact and does not give very good explanations. There was a typo in the end of your reply:

$$\begin{bmatrix}-21 \\ 13\end{bmatrix}$$

should be

$$\begin{bmatrix}-21 \\ 34\end{bmatrix}$$

5. Sep 6, 2009

### HallsofIvy

Right! Thanks.

No, I'll go back and edit my post so I can claim I never made that silly mistake!