# Find the spanning set of the range of the linear transformation $T(x)=Ax$.

Let $$A= \begin{bmatrix} -4 & -4 & 12 & 0 \\ -4 & -4 & 12 & 0 \\ 4 & -2 & 0 &-6 \\ 1 &-4 &7 &-5 \\ \end{bmatrix}$$

Find the spanning set of the range of the linear transformation $T(x)=Ax$.

I have found the row reduced echelon form of A.

$$RREF(A)= \begin{bmatrix} 1 & 0 & -1 & -1 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$

I don't know what to do with it after.

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Have you tried of finding the Reduced Row Echelon Form of $A$? – S. Snape Feb 4 '13 at 8:16
@BabakSorouh I found the RREF(A). The first part of the question was to find the spanning set of Ax=0 and I got that part. But I don't know what to do with this part. – user60912 Feb 4 '13 at 8:24

It $T(v)=Av,~~~v\in\mathbb K^4$, then $$T(v)= \begin{pmatrix} 1 & 0 & -1 & -1 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix}=\begin{pmatrix} x-z-t \\ y-2z+t \\ 0 \\ 0 \\ \end{pmatrix}$$ but $$\begin{pmatrix} 1 & 0 & -1 & -1 \\ 0 & 1 & -2 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}\approx\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}$$ so $$T(v)= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix}\begin{pmatrix} x \\ y \\ z \\ t \\ \end{pmatrix}=\begin{pmatrix} x \\ y \\ 0 \\ 0 \\ \end{pmatrix}=x\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \\ \end{pmatrix}+y\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix}$$

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+1 for all the typing and formatting involved! – amWhy Feb 6 '13 at 3:05

The range of $T$ is the column space of $A$. So the columns of $A$ already form a spanning set. If you want to find a linearly independent spanning set, you should find a column echelon form of $A$ instead of a row echelon form. I found that $\{(-4-4,4,1)^T,\,(0,0,-6,-5)^T\}$ is an answer, but depending on the column operations you perform, you may get a different answer.

Edit: for a starter, $$\begin{bmatrix} -4 & -4 & 12 & 0 \\ -4 & -4 & 12 & 0 \\ 4 & -2 & 0 &-6 \\ 1 &-4 &7 &-5 \\ \end{bmatrix} \stackrel{C_2-C_1,\, C_3+3C_1}{\longrightarrow} \begin{bmatrix} -4 & 0 & 0 & 0 \\ -4 & 0 & 0 & 0 \\ 4 & -6 & 12 &-6 \\ 1 &-5 &10 &-5 \\ \end{bmatrix} \,\longrightarrow\cdots \begin{bmatrix} -4 & 0 & 0 & 0 \\ -4 & 0 & 0 & 0 \\ 4 & -6 & 0 & 0 \\ 1 &-5 & 0 & 0 \\ \end{bmatrix}$$

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Did you mean an answer like what I did? Thanks – S. Snape Feb 4 '13 at 8:56
Hmm, no, row echelon form is irrelevant here. – user1551 Feb 4 '13 at 9:42

There is no need to use row echelon form here...the question is only asking for a spanning set.

Consider a general vector $v = (x,y,z,w)^T$, then

$Av = (-4x-4y+12z,-4x-4y+12z,4x-2y-6w,x-4y+7z-5w)$

$= x(-4,-4,4,1) + y(-4,-4,-2,-4) + z(12,12,0,7) + w(0,0,-6,-5)$

so those four vectors form a spanning set for the range. Notice that they are the columns of $A$.

Of course this isn't the BEST choice for a spanning set since they aren't linearly independent...but that would be your aim if you were finding a basis.

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