# Find the image of a vector by using the standard matrix (for the linear transformation T)

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 $$\begin{bmatrix}1&1\\1&-1\\2&0\\0&2\end{bmatrix}\begin{bmatrix}3\\-3\end{bmatrix}?‌​ – t.b. Jul 24 '11 at 20:39 You know: it's probably a lot less work to just type the question you want, instead of posting these images which contain a lot of irrelevant information... – Arturo Magidin Jul 24 '11 at 21:08 ## 2 Answers Suppose you have a linear transformation T\colon\mathbb{R}^n\to\mathbb{R}^m. If you know what happens to the standard basis \mathbf{e}_1,\ldots,\mathbf{e}_n of \mathbb{R}^n, then you know what happens to every vector in \mathbb{R}^n, because given any vector \mathbf{v}=(a_1,\ldots,a_n), we have:$$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.$$So if you know T\mathbf{e}_i for each i, we can get T\mathbf{v} for every \mathbf{v}. The standard matrix of T is a way of keeping track of precisely this information, and making it easy to perform the computation above. What we are using is the fact that if A is a matrix, and we let \mathbf{a}_i be the ith column of A, that is,$$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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