# Homework Help: Linear transformation ranges

1. Apr 26, 2010

### newtomath

L: R^3 -> R^3 is a linear transformation defined by L(v) =A(v)
A is given as -1 2 0 and w= 1
1 1 1 2
2 -1 1 -1

is w in the range of L?

My understanding is that if a vector exists such that the product of A and this vector = W then it is in the range. I then have the following equations:

-X +2Y = 1
X+Y+Z= 2
2X -Y+Z= -1

no solutions exists, so w is not in the range. Can someone confirm this or offer any other insight?

thanks

2. Apr 27, 2010

### lanedance

bit hard to read is A given as:
$$A = \begin{bmatrix} -1 & 2 & 0\\ 1 & 1 & 1 \\ 2 & -1 & 1 \end{bmatrix}$$

Last edited: Apr 27, 2010
3. Apr 27, 2010

### lanedance

what you've done sounds reasonable, though i haven't checked the math,

note so what you are trying to solve is:
$$A \textbf{x} = \textbf{x}$$

If A was invertible a solution must exist given by:
$$\textbf{x} = A^{-1} \textbf{x}$$
so you could check that $det(A) = 0$

Now, consider the action of A on a unit vector:
$$\textbf{e}_1 = (1,0,0)^T$$
$$A\textbf{e}_1 = \textbf{a}_1$$
where $\textbf{a}_1$ is the first column vector of $A$

Now every vector can be wirtten in terms of the basis of unit vectors, so the range of A is in fact the column space of A. So in effect you are checking w is not in the column space of A.

4. Apr 27, 2010

### HallsofIvy

Internet readers do not respect spaces so better to use LaTex.

If
$$A= \begin{bmatrix}-1 & 2 & 0 \\ 1 & 1 & 1 \\ 2 & - 1 & 1\end{bmatrix}$$
and
$$w= \begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix}$$

is w in the range of A?
That will be true if and only if there exist x, y, z such that
[tex]\begin{bmatrix}-1 & 2 & 0 \\ 1 & 1 & 1 \\ 2 & - 1 & 1\end{bmatrix}\begin{bmatrix}x \\ y \\ z\end{bmatrix}= \begin{bmatrix} 1 \\ 2 \\ -1\end{bmatrix}[/itex]
(Click on those to see the code.)

That can be written as the three equations, -x+ 2y= 1, x+ y+ z= 2, 2x- y+ z= -1.

IF those three equations have a solution (not necessarily unique) then w is in the range of A.

It might be simpler to check the determinant of A. If the determinant is not 0, the range of A is all of $R^3$ and so any w is in it.

If the determinant of a is 0, then you would have to check if w is in the "columns space". In that case, it is probaly simpler to solve the three equations above.