# Linear transformations

• Dec 4th 2009, 10:56 AM
SirOJ
Linear transformations

(b) Let
L : R3 ->R2 be the linear transformation defined by

L
(x, y, z) = (x y, y z)

(i) Find the standard (2
× 3) matrix representation of L and compute L(1, 1, 1).

I know how to show something is a linear transformation but unsure how to start this question. Just the starting point is all i'm looking for...
• Dec 4th 2009, 10:58 AM
Bruno J.
Find the image of $(1,0,0),(0,1,0),(0,0,1)$ under this transformation. The vectors you will obtain will be the columns of the matrix.
• Dec 4th 2009, 12:03 PM
SirOJ
Quote:

Originally Posted by Bruno J.
Find the image of $(1,0,0),(0,1,0),(0,0,1)$ under this transformation. The vectors you will obtain will be the columns of the matrix.

I'm a little confused as to what you mean above.. Is there any way you could elaborate a small bit?
• Dec 4th 2009, 12:18 PM
Bruno J.
Sure!

You have :

$L(1,0,0)=(1,0)$

$L(0,1,0)=(-1,1)$

$L(0,0,1)=(0,-1)$

and the three vectors on the right are the columns of the matrix which represents $L$ with respect to the standard basis $\beta$. So we have

$[L]_\beta=\left(\begin{array}{ccc}1& -1& 0 \\ 0 &1 &-1 \\ \end{array}\right)$.
• Dec 4th 2009, 12:25 PM
Krizalid
it's worth to say that this procedure only applies when finding the associated matrix in the standard basis.
• Dec 5th 2009, 06:00 AM
HallsofIvy
But to generalize: If L is a linear transformation from vector space U, with dimension n, to vector space V, with dimension m, then it can be written in matrix form for given basis $\{u_1, u_2, ..., u_n\}$ of U and given basis $\{v_1, v_2, ..., v_m\}$ of V (changing bases will result in different but "equivalent" matrices representing the same transformation).

Take $L(u_1)$ and write it as a linear combination of the basis $\{v_1, v_2, ...v_n\}$, which you can do since $L(u_1)$ is in V. Say, $L(u_1)= a_1v_1+ a_2v_2+ ...+ a_mv_m$. Then $\begin{bmatrix}a_1 \\ a_2 \\ \cdot \\\cdot \\\cdot \\ a_m \end{bmatrix}$ is the first column in the matrix. Similarly, writing $L{u_2}$ as a linear combination of $v_1, v_2, ..., v_m$ gives the second column, etc.
• Dec 6th 2009, 09:08 AM
SirOJ
Quote:

Originally Posted by HallsofIvy
But to generalize: If L is a linear transformation from vector space U, with dimension n, to vector space V, with dimension m, then it can be written in matrix form for given basis $\{u_1, u_2, ..., u_n\}$ of U and given basis $\{v_1, v_2, ..., v_m\}$ of V (changing bases will result in different but "equivalent" matrices representing the same transformation).

Take $L(u_1)$ and write it as a linear combination of the basis $\{v_1, v_2, ...v_n\}$, which you can do since $L(u_1)$ is in V. Say, $L(u_1)= a_1v_1+ a_2v_2+ ...+ a_mv_m$. Then $\begin{bmatrix}a_1 \\ a_2 \\ \cdot \\\cdot \\\cdot \\ a_m \end{bmatrix}$ is the first column in the matrix. Similarly, writing $L{u_2}$ as a linear combination of $v_1, v_2, ..., v_m$ gives the second column, etc.

I'm new to a lot of this stuff so some of that definition didn't really make sense to me... You are saying that i would have to approach a problem like the one below differently, correct?

(b) Let
L : R3 ->R2 be the linear transformation defined by

L
(x, y, z) = (x + y + z, y + z)

(i) Find the standard (2
× 3) matrix representation of L and compute L(0,1, 1).