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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?

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  • $\begingroup$ The matrix should be 4x4 after the transformation, since it goes from $\mathbb{R^3} \rightarrow \mathbb{R^4}$, the transformation brings 4 vectors with 3 components into a space of 4 vectors of 4 components, i.e. a square matrix; you can find the kernel of a square matrix by simply solving the associated linear system. $\endgroup$ – james42 May 30 '15 at 23:57
  • $\begingroup$ It seems to me your "ref" is wrong, you should eliminate all the ones in the upper diagonal. Then you'll obtain $$\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$ This says that $\dim\rm im\,=3$, and that the map is (in informal terms) "essentially" $(x_1,x_2,x_3)\mapsto (x_1,x_2,x_3,0)$ so that it is injective. $\endgroup$ – Pedro Tamaroff May 31 '15 at 22:26
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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 $\Big\{\begin{bmatrix} 1\\0 \\3\\0 \end{bmatrix},\begin{bmatrix} 1\\1 \\1\\2 \end{bmatrix},\begin{bmatrix} 1\\1 \\0\\1 \end{bmatrix} \Big\}$

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