We are asked to find the standard matrix $A$ for $T$:
Consider the transformation $T : \mathbb{R^3} \rightarrow \mathbb{R^4}$ given by
$$T(x_1, x_2, x_3) = (x_1 + x_2 + x_3, x_2 + x_3, 3x_1 + x_2, 2x_2 + x_3)$$
for every $$(x_1,x_2,x_3) \in \mathbb{R^3}$$
The standard matrix is as follows:
$$ A = \left[\begin{array}{rrr}
1 & 1 & 1\\
0 & 1 & 1\\
3 & 1 & 0\\
0 & 2 & 1\\
\end{array}\right]
$$
From $T(e_1)=(1,0,3,0)^T, T(e_2)=(1,1,1,2)^T, T(e_3)=(1,1,0,1)^T$
Its row echelon form is as follows:
$$ ref(A) = \left[\begin{array}{rrr}
1 & 1 & 1\\
0 & 1 & 1\\
0 & 0 & 1\\
0 & 0 & 0\\
\end{array}\right]
$$
The question is asking:
By reducing $A$ to row echelon form, determine the dimension of the kernel of $T$ and the dimension of the range of $T$.
We need to juxtapose (augment) the original matrix with the zero vector?
$$ A = \left[\begin{array}{rrr}
1 & 1 & 1 &| 0\\
0 & 1 & 1 &| 0\\
3 & 1 & 0 &| 0\\
0 & 2 & 1 &| 0\\
\end{array}\right]
$$
Then let $x_3=t, x_1=-t, …$? The dimension is the number of vectors that span so I would imagine it to be 1? How would I find the range?
Best Answer
If $\ ref(A)= \begin{bmatrix} 1 & 1 & 1 \\ 0&1&1 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}$, then the only solution to $\begin{bmatrix} 1 & 1 & 1 \\ 0&1&1 \\ 3&1&0 \\ 0&2&1 \end{bmatrix} \begin{bmatrix} x_1\\x_2 \\x_3 \end{bmatrix}= \begin{bmatrix} 0\\0 \\0\\0 \end{bmatrix}$ is the zero
vector, which is never a member of any spanning set. Thus, the kernel has dimension zero. By the rank-nullity theorem, $dim(im(A))+dim(ker(A))=3.$ Thus, the image has dimension three. Notice that each column of the row reduced matrix has a pivot. Then the range is the span of $\Big\{\begin{bmatrix} 1\\0 \\3\\0 \end{bmatrix},\begin{bmatrix} 1\\1 \\1\\2 \end{bmatrix},\begin{bmatrix} 1\\1 \\0\\1 \end{bmatrix} \Big\}$