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Was wondering if anyone can help out with the following problem:

Use the standard matrix for the linear transformation $T$ to find the image of the vector $\mathbf{v}$, where $$T(x,y) = (x+y,x-y, 2x,2y),\qquad \mathbf{v}=(3,-3).$$

I found out the standard matrix for $T$ to be: $$\begin{bmatrix}1&1\\1&-1\\2&0\\0&2\end{bmatrix}$$

From here I honestly don't know how to find the "image of the vector $\mathbf{v}$". Does anyone have any suggestions?

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Your matrix is correct. Did you try multiplying $$\begin{bmatrix}1&1\\1&-1\\2&0\\0&2\end{bmatrix}\begin{bmatrix}3\\-3\end{bmatrix}?$‌​$ – t.b. Jul 24 '11 at 20:39
You know: it's probably a lot less work to just type the question you want, instead of posting these images which contain a lot of irrelevant information... – Arturo Magidin Jul 24 '11 at 21:08
up vote 1 down vote accepted

Suppose you have a linear transformation $T\colon\mathbb{R}^n\to\mathbb{R}^m$. If you know what happens to the standard basis $\mathbf{e}_1,\ldots,\mathbf{e}_n$ of $\mathbb{R}^n$, then you know what happens to every vector in $\mathbb{R}^n$, because given any vector $\mathbf{v}=(a_1,\ldots,a_n)$, we have: $$T\mathbf{v} = T\Bigl(a_1\mathbf{e}_1+\cdots+a_n\mathbf{e}_n\Bigr) = a_1T\mathbf{e}_1+\cdots a_nT\mathbf{e}_n.$$ So if you know $T\mathbf{e}_i$ for each $i$, we can get $T\mathbf{v}$ for every $\mathbf{v}$.

The standard matrix of $T$ is a way of keeping track of precisely this information, and making it easy to perform the computation above. What we are using is the fact that if $A$ is a matrix, and we let $\mathbf{a}_i$ be the $i$th column of $A$, that is, $$A = (\mathbf{a}_1\;|\;\cdots\;|\;\mathbf{a}_n),$$ and you multiply $A$ by an $n\times 1$ column vector, then the result of the product is the same as taking an appropriate linear combination of the columns, to wit, if $\mathbf{v} = (a_1,\ldots,a_n)$, then: $$A\mathbf{v}^t = A\left(\begin{array}{c}a_1\\\vdots\\a_n\end{array}\right) = a_1\mathbf{a}_1 + \cdots + a_n\mathbf{a}_n$$ (where $\mathbf{v}^t$ is the transpose of $\mathbf{v}$).

That means that if we make a matrix $A$ wuch that $\mathbf{a}_i$ is $(T\mathbf{e}_i)^t$, then we have $$A(\mathbf{v})^t = \left((a_1T\mathbf{e}_1)^t + \cdots + (a_nT\mathbf{e}_n)^t\right)^t = (T\mathbf{v})^t,$$ so we can compute $T\mathbf{v}$ by multiplying $\mathbf{v}^t$ by the matrix $A$. The matrix $A$ is the "standard matrix of $T$" (with respect to the standard bases of $\mathbb{R}^n$ and $\mathbb{R}^m$).

So you have computed $A$; you know $\mathbf{v}$. Now you just need to multiply $A$ by $\mathbf{v}^t$ to get the (transpose of the) image of $\mathbf{v}$ under $T$.

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The matrix you've written down is correct. If you have a matrix $M$ and a vector $v$, the image of $v$ means $Mv$.

Something is a bit funny with the notation in your question. Your matrix is 4x2, so it operates on column vectors of height two (equivalently, 2x1 matrices). But the vector given is a row vector. Still, it seems clear that what you need to calculate is the product $Mv$ that Theo wrote down in the comment. Do you know how to do that?

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I would like to do it in a systematic way. It is instructive to see how OP fits in the following steps.

Let $V$ (resp. $W$) be an $n$ (resp. $m$) dimensional vector space over $\mathbb{R}$. Let $$\alpha=(v_1,\cdots,v_n)$$ be an ordered basis in $V$ and $$\beta=(w_1,\cdots,w_m)$$ an ordered basis in $W$. For any vector $x\in V$, denote its coordinate w.r.t. the basis $\alpha$ as $$ [x]_\alpha=(x_1,\cdots,x_n)^T $$ and for any vector $y\in W$, denote its coordinate w.r.t. the basis $\beta$ as $$ [y]_\beta=(y_1,\cdots,y_m)^T. $$ Let $T:V\to W$ be a linear transformation. Let $[T]_\alpha^\beta$ denotes the matrix for $T$ w.r.t. the bases $\alpha$ and $\beta$, i.e., $$ [T]^\alpha_\beta=[[Tv_1]_\beta,\cdots,[Tv_n]_\beta]. $$ Note in particular that $[T]^\alpha_\beta$ is an $m\times n$ matrix.

Given $x\in V$, we have $$ [Tx]_\beta=[T(x_1v_1+\cdots+x_nv_n )]_\beta\\ =x_1[T(v_1)]_\beta+\cdots x_n[T(v_n)]_\beta\\ =[T]_\beta^\alpha[x]_\alpha $$

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This is an old question in almost five years ago... Hope this would be useful for someone. – Jack May 26 at 3:09

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