1
$\begingroup$

If $\mathbf{A} = \begin{pmatrix} 2&-1&5&8 \\ 2&4&0&5 \\ 1&3&-1&4 \end{pmatrix}$, let $\mathbf{x}$ denote the column vector $(x_1, x_2, x_3, x_4)^T$, and let the linear map $T$ be given by the equation $T(\mathbf{x}) = \mathbf{Ax}$ then (a) find and describe the kernel (null space), (b) find $T(3e_1-2e_2-e_3)$; (c) solve $T(\mathbf{x}) = 3e_1-2e_2-2e_3.$

I row reduced the matrix to $\begin{pmatrix} 1&0&2&0 \\ 0&1&-1&0 \\ 0&0&0&1 \end{pmatrix} $ then found the kernel to be $\mathbf{x} = x_{3}(-2,1,1,0)^{T}$. It's a line through the origin but I don't know anything else about it. Also, I need help with the rest.

Also, I thought $T(3e_1-2e_2-e_3)=\begin{pmatrix} 1&0&2&0 \\ 0&1&-1&0 \\ 0&0&0&1 \end{pmatrix}(3,-2,-1)$ but the answer is a vector.

$\endgroup$
  • 1
    $\begingroup$ Well, for part (b), note that your transformation is from $\Bbb R^4 \to \Bbb R^3$. So you need to think of $3e_1 - 2e_2 - e_3$ as $3e_1 - 2e_2 - e_3 + 0e_4$, with the "missing" basis vector's $0$ coefficient. $\endgroup$ – pjs36 Mar 4 '16 at 19:01
  • $\begingroup$ @pjs36 Thank you! I get $T(3e_1-2e_2-e_3)=\begin{pmatrix} 1&0&2&0 \\ 0&1&-1&0 \\ 0&0&0&1 \end{pmatrix}(3,-2,-1,0)^{T} = (1,-1,0)$ is this the way to do it? Also, regarding (a), what's the geometric interpretation of $\mathbf{x} = x_{3}(-2,1,1,0)^{T}$? $\endgroup$ – studrayght5 Mar 4 '16 at 19:25
1
$\begingroup$

The row reduction looks good, and that vector is indeed a basis for the kernel. I'm not sure I'd write ${\bf x} = x_3(-2, 1, 1, 0)$, it's kind of analogous to writing $x = \Bbb R$ when you mean to say that $x$ can be anything in $\Bbb R$. So perhaps write $\ker T = \{x_3(-2, 1, 1, 0): x_3 \in \Bbb R\}$.

I honestly don't think there's a better way to describe the kernel than you have, just saying it's a one-dimensional subspace, hence a line through the origin in $\Bbb R^4$.


As mentioned in the comment, since the transformation $T$ is a map $\Bbb R^4 \to \Bbb R^3$, we need to rewrite $3e_1-2e_2-2e_3$ as $3e_1-2e_2-2e_3+0e_4$, adding in the "missing" basis vector.

But you can't use the row-reduced ${\bf A}$ to compute $T(3e_1-2e_2-2e_3)$, you need to use the original matrix (you should get $(-2, -2, -1)$). To see why this is, consider the dilation matrix ${\bf D} = \begin{pmatrix}2&0\\0&2\end{pmatrix}$. It should be pretty easy to see that ${\bf D}$ sends a vector $(x, y)$ to $(2x, 2y)$, but if you row-reduce $D$ to the identity, then you'll incorrectly think $D$ sends $(x, y)$ to itself (in fact, you'll think all invertible matrices do the same thing as the identity matrix)!


But for part (c), you can row-reduce the augmented matrix $\left(\begin{array}{cccc|c} 2&-1&5&8&3 \\ 2&4&0&5&-2 \\ 1&3&-1&4&-2 \end{array}\right)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.