# Kernel/image of a linear transformation

My question is regarding how to find the image point of the following linear transformation when the given point is in $\mathbb{R}^3$ and the transformation matrix is $3 \times 4$? I'm not sure if it makes sense to find the image. Thanks.

Consider the matrix $\textbf{A} = \begin{pmatrix} 2 & -1 & 5& 8\\ 2 & 4 & 0 & 5\\ 1 & 3 & -1 & 4 \end{pmatrix}$.

Let $\overline{x}$ denote the column vector $(x_1,x_2,x_3,x_4)^t$. What's the kernel of the linear transformation given by $T(\overline{x}) = \textbf{A} \overline{x}$? What's the image point of $T(3\mathbf{e}_1-2\mathbf{e}_2-\mathbf{e}_3)$?

I've done the kernel bit.

$T(\overline{x} ) = \overline{0} \implies \text{Reff}(\mathbf{A}) \cdot \overline{x} = \begin{pmatrix} 1 & 0 & 2 & 0\\ 0 & 1 & -1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \overline{0} \implies$ $x_1+2x_3 = 0 ,~ x_2-x_3 = 0, ~ x_4 = 0$ so the solution is $x_3(-2,1,1,0)$, which is a line in through the origin in $\mathbb{R}^4$, right?

• Do you mean T(3e1−2e2−e3)? – user9750060 Jul 23 '18 at 14:21
• @user9750060 Yes, thanks. – Ryan Jul 23 '18 at 14:27
• Might be a silly question, but what is the exponent t? Are you denoting the transpose? – user9750060 Jul 23 '18 at 14:37
• @user9750060 Transpose. I think they are saving space to write the column vector as a transpose of a row vector. – Ryan Jul 23 '18 at 14:42

$T$ is a linear map from $\mathbf{R}^4 \to \mathbf{R}^3$ with transformation matrix $\mathbf{A}$.

Your kernel is correct. Multiplying your matrix $\textbf{A}$ with the vector $v:=(-2,1,1,0)^t$ gives $\vec 0$, so you easily see that your vector actually lies in the kernel. Linear algebra tells you, that the kernel of a linear map is a subspace of the domain-vector space. So from the above calculation, you immediately see that the line spanned by $v$ lies inside the kernel. Possibly, it could be a plane a higher dimensional subspace. But since the matrix $\mathbf A$ and, equivalently, $\text{Reff}(\mathbf A)$ have $\text{rank}=3$, the kernel is $1$-dimensional (see rank-nullity). So it is exactly your line.

You get the image of $3\mathrm e_1 - 2 \mathrm e_2 - \mathrm 3_3$ by (left-)multiplication with $\textbf A$ ($3\mathrm e_1-2\mathrm e_2-\mathrm e_3$ lies in $\mathbf{R}^4$, the image point lies in $\mathbf{R}^3$). I guess you know how to multiply a matrix and a vector?

Abstractly, you can use that your transformation is linear to break up the term $T(3\mathrm e_1-2\mathrm e_2-\mathrm e_3)$. Then you only need to know what $\textbf{A}$ does with the standard-base-vectors. And $T$ maps the vector $\mathrm e_1=(1,0,0,0)^t$ to $(2,2,1)^t$, the first column of $\textbf A$.

• Thank you very much. When you left multiplication, what does that mean $\overline{x} \mathbf{A}$ instead of $Ax$? – Ryan Jul 23 '18 at 14:57
• Ah, sorry, I mean leftmultiplication of the vector $\overline{x}$ by $\textbf A$. This is $\textbf A \overline{x}$. – Babelfish Jul 23 '18 at 14:59
• Okay, this is confusing me because the number of columns of A have to match the number of rows of x, otherwise multiplication is not defined. I feel like I'm missing something obvious about all of this. – Ryan Jul 23 '18 at 15:02
• Yeah, but this is the case. $\mathbf{A}$ has four columns, $3\mathrm e_1-2\mathrm e_2-\mathrm e_3$ has 4 rows... – Babelfish Jul 23 '18 at 15:04
• Oooh, $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3 \in \mathbb{R}^4$. I was going crazy thinking nothing makes sense. I was thinking $x \in \mathbb{R}^3$ for some reason. Many thanks again. – Ryan Jul 23 '18 at 15:05