1
$\begingroup$

Let $A = \begin{bmatrix} -1 & -4 & -8 \\ 8 & -7 & 4 \\ \end{bmatrix}_.$

Define the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ by $T(x) = Ax$.

Let $u = \begin{bmatrix} -2\\ 1\\ 2\\ \end{bmatrix}$ and $v = \begin{bmatrix} a\\ b\\ c\\ \end{bmatrix}_.$

Find the images of $u$ and $v$ under $T$.

I'm not sure exactly how to do this. What do they mean by find the images? Can someone help me out?

$\endgroup$
7
  • 1
    $\begingroup$ Just apply the operator $T$ to the vectors $u$ and $v$. The image of a vector under an operator is the vector one obtains by applying the operator to the starting vector. $\endgroup$
    – Dave
    Jun 20, 2016 at 19:50
  • $\begingroup$ What do you mean by "Just apply the operator $T$ to the vectors $u$ and $v$" .. Can you elaborate? $\endgroup$
    – Yusha
    Jun 20, 2016 at 19:52
  • $\begingroup$ If you have an example, I'm sure I could mimic it. $\endgroup$
    – Yusha
    Jun 20, 2016 at 19:52
  • 1
    $\begingroup$ It is nothing but matrix-vector multiplication. Just google it and open the wikipedia page $\endgroup$
    – b00n heT
    Jun 20, 2016 at 19:52
  • 1
    $\begingroup$ So in the case of the image for $u$ under $T$ you would "plug in" $u$ into the equation for $T$, which is $T(u)=Au $. Then you just have to multiply the vector $u$ by the matrix $A$ to obtain its image. Repeat for $v$. $\endgroup$
    – Dave
    Jun 20, 2016 at 19:56

3 Answers 3

Reset to default
3
$\begingroup$

The term "the image of $u$ under $T$" refers to $T(u) = Au$. All that you have to do is multiply the matrix by the vectors.

$\endgroup$
1
$\begingroup$

Turned out this was simple matrix multiplication. $T(u) = \begin{bmatrix} -18\\ -15\\ \end{bmatrix}$ and $T(v) = \begin{bmatrix} -a-4b-8c\\8a-7b+4c\end{bmatrix}$

$\endgroup$
0
$\begingroup$

Consider projecting a $3D$ object on a $2D$ surface. (Just like a shadow of an object). Image means the projection of the object (here a set of vectors) from $\mathbb{R}^3$ on $\mathbb{R}^2$.. So the projection matrix(also known as the operator , since it operates on vectors) being given, therefore multiply each of the vectors$(\in \mathbb{R}^3)$ with $T$ to get their image (projection) on $\mathbb{R}^2$

So if $v_1$ and $v_2$ are the vectors, their images will be $Tv_1$ and $Tv_2$ respectively.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .