# Vectors inside of a linear transformation.

so I am working on an algebra question and I need some help understanding what the question is asking.

I found that the standard matrix of the linear transformation: $$T: \mathbb{R}^2 \to \mathbb{R}^3 = \begin{bmatrix}1 & 0\\0 & 3\\5 & 8\end{bmatrix}$$

And now I am being asked to find$$T\left(\begin{bmatrix}1\\2\end{bmatrix}\right)$$

What exactly am I being asked to find? Am I supposed to find what matrix transforms T to \begin{bmatrix}1\\2\end{bmatrix} or am I supposed to multiply that into T? Any help would be appreciated. Thanks.

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Is it actually $\mathbb{R}^3\rightarrow\mathbb{R}^2$ or was that supposed to be $\mathbb{R}^2\rightarrow\mathbb{R}^3$? – Peter Woolfitt Jul 13 '14 at 3:07
Oops I wrote it wrong.. – Sofia June Jul 13 '14 at 3:10

At the end of the day, since $T$ is a linear transformation from $\mathbb{R}^{2}$ to $\mathbb{R}^{3}$, that means you plug in vectors from $\mathbb{R}^{2}$ (these are vectors with only 2 components) and the transformation outputs a vector in $\mathbb{R}^{3}$. The output will be a vector with three components because it lives in $\mathbb{R}^{3}$. So you want to know where the vector $\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$ is being sent to under the transformation $T$. It is being sent to a vector in $\mathbb{R}^{3}$ (a vector with 3 components), but which one?? I show you how to figure that out below using the fact that $T$ is a linear transformation.

Note that $T$ is a linear transformation, so we know that $T(cv) = cT(v)$ for any $v$ in $\mathbb{R}^{2}$ and any real number $c$.

Also, we have $T(v_{1} + v_{2}) = T(v_{1}) + T(v_{2})$ for any $v_{1}$ and $v_{2}$ in $\mathbb{R}^{2}$.

Both of these together give the general property that if $c$ and $d$ are real numbers, and $v_{1}$ and $v_{2}$ are vectors in $\mathbb{R}^{2}$, then $T(c v_{1} + d v_{2}) = cT(v_{1}) + dT(v_{2})$.

Why am I telling you this? Because this is the key to figuring out $T \left (\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}\right )$. You have to realize that we can write $\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$ as the sum $\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + 2*\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$. So, we get:

$T \left ( \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} \right ) = T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} + 2*\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right ) = T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right ) + 2*T \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right )$.

Now, since every linear transformation $T$ follows the rule that $T(x) = Ax$ for an associated matrix $A$, which you found, we have $T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right ) = A*\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$, and this product is just the first column of the matrix $A$, so $T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right ) = \begin{bmatrix} 1 \\ 0 \\ 5\\ \end{bmatrix}$.

Similarly, $T \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right ) = A*\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$, and this product is just the second column of the matrix $A$, so $T \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right ) = \begin{bmatrix} 0 \\ 3 \\ 8\\ \end{bmatrix}$.

So, we have $T \left ( \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} \right ) = T \left ( \begin{bmatrix} 1 \\ 0 \\ \end{bmatrix} \right ) + 2*T \left ( \begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right ) = \begin{bmatrix} 1 \\ 0 \\ 5\\ \end{bmatrix} + 2*\begin{bmatrix} 0 \\ 3 \\ 8\\ \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \\ 5\\ \end{bmatrix} + \begin{bmatrix} 0 \\ 6 \\ 16\\ \end{bmatrix} = \begin{bmatrix} 1 \\ 6 \\ 21\\ \end{bmatrix}$.

Therefore, the final answer is $T \left ( \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} \right ) = \begin{bmatrix} 1 \\ 6 \\ 21\\ \end{bmatrix}$, that is, under this specific transformation $T$ with associated matrix $A$, the vector $\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$ (which lives in $\mathbb{R}^{2}$, since it has 2 components) is being sent to the vector $\begin{bmatrix} 1 \\ 6 \\ 21\\ \end{bmatrix}$ (which lives in $\mathbb{R}^{3}$, since it has 3 components).

Hope that helps!

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I just realized I solved this problem the "hard way". I'm not sure why this was the first way to solve the problem that popped into my head. As I said in my post, because $T$ is a linear transformation, it has an associated matrix $A$ such that $T(x) = Ax$ for all vectors $x$ in $\mathbb{R}^{2}$. – user46944 Jul 13 '14 at 3:49
So, that means $T( \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} ) = A*\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$. You have $A$ in your problem description, so your final answer is whatever product you get when you multiply $A$ and $\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}$ together. Your answer should agree with my answer above, but this way is much faster and more direct. Sorry for showing you the long way first! – user46944 Jul 13 '14 at 3:51
actually I should thank you for taking the time and showing me the long way, it helped me understand the whole concept a lot better :) – Sofia June Jul 13 '14 at 4:30
Good, I'm glad! :) – user46944 Jul 13 '14 at 4:37

Yes, you just multiply the matrix with this vector. You should understand that the matrix is a representation of $T$ in a particular basis. $T$ is a linear transformation--a linear function--and we use matrices and matrix multiplication to carry out these calculations because matrix-vector multiplication is also linear.

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