Understanding range of linear transformation

Question

Here are two different problems in my book with their solutions that confuse me.

1. Define the linear transformation $T$ by $T(x) = Ax$, and find $range(T)$:

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

The solution is to transpose $A$ and reduce it

$$ A^T=\begin{bmatrix}1 & 0 \\ -1 & 1 \\ 2 & 2 \\ \end{bmatrix}= \begin{bmatrix}1 & 0 \\ 0& 1 \\ 0&0 \\ \end{bmatrix} $$

Therefore $range(T)=\{(1,0),(0,1)\}=R^2$

2. Define the linear transformation $T$ by $T(x) = Ax$, and find $range(T)$: $$ A=\begin{bmatrix}5 &-3 \\ 1& 1 \\ 1 & -1 \\ \end{bmatrix} $$

Again, the solution is to transpose $A$ and reduce it

$$ A^T=\begin{bmatrix}5 & 1 & 1 \\ -3 & 1 & -1 \\ \end{bmatrix}=\begin{bmatrix}1 & 0 & 1/4 \\ 0 & 1 & -1/4 \\ \end{bmatrix} $$

Therefore $range(T)=\{(4s,4t,s-t)\ \text {for all real s and t}\}$

There are three things that I do not understand:

(a). My book defines the range of T: V→W as the set of all vectors w in W that are images of vectors in V. What does this have to do with transposing and then row reducing?

(b). Is it possible to arrive at the same result without transposing, but merely reducing the matrix as is?

(c ). The calculation in example 2 does not make sense to me. The way I would proceed after reducing is like this:

$$x_1=-x_3/4 \\ x_2=x_3/4 \\ x_3=t$$ $$ \mathbb x=\begin{bmatrix}x_1 \\ x_2\\ x_3 \\ \end{bmatrix}= \begin{bmatrix}-t/4 \\ t/4\\ t \\ \end{bmatrix} $$

Where does the extra $s$ come from?

Answer 1

Let’s look at what your definition of the range of $T$ means in terms of the columns of $A$. When you right-multiply a matrix by a vector (an $n\times1$ matrix), the result is a linear combination of the columns of the matrix, specifically, $$Ax = \sum_i x_iA_i,$$ where $A_i$ stands for the $i$th column of $A$. Thus if $T:x\mapsto Ax$, the range of $T$ is precisely the span of the columns of $A$—the column space of $A$.

Transposing $A$ changes its column space into the row space of $A^T$, and the nonzero rows of $A^T$ form a basis for its row space, so this is one way to compute a basis for the range of $T$. However, you can certainly row-reduce $A$ instead. The difference is that the RREF of $A$ doesn’t give you a basis for its columns space directly: the columns of $A$ that correspond to pivot columns of its RREF are linearly independent and form a basis for the column space. Finding a basis via the first method generally produces a “nicer” basis than the second method: the first $\operatorname{rank}T$ elements of the resulting basis vectors consist of a single $1$ with the rest of those elements $0$.

As to your last question, you’ve solved the wrong problem: you’ve computed the solution to the equation $A^Tx=0$. This is the null space of $A^T$, which is the orthogonal complement of the space that we’re trying to find. The row space of $A^T$ consists of all linear combinations of the two rows of the reduced matrix, that is all combinations $s(1,0,1/4)+t(0,1,-1/4)=(s,t,(s-t)/4)$, which has been multiplied by $4$ to make it look prettier. (You can do this because the constants $s$ and $t$ are arbitrary.)

Answer 2

Row-reducing of transposed matrix (or equally column-reducing of original matrix) doesn't change image, so we can perform it make matrix simpler (and row reducing of original matrix potentially changes range).

The reason is that $i$-th column of matrix is exactly image of $e_i$. And two matrices $A$ and $B$ s.t. $A e_i = B e_i$ for $i \neq i_0$ and $B_i = \alpha A e_i + \beta Ae_j$ with $\alpha \neq 0$ have the same range, so we can make elementary transformations on columns.

For the last part, reduced $A$ corresponds to a function $f\colon\mathbb R^2 \to \mathbb R^3$ s.t. $f((s, t)) = (s, t, \frac{s - t}{4})$. So the image of $f$ (and of $A$) is $\{(s, t, \frac{s - t}{4})\} = \{(4s, 4t, s - t)\}$. What you found is a kernel of $A^T$ - it's orthogonal complement to image of $A$.