# 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 $$\begin{bmatrix}1&1\\1&-1\\2&0\\0&2\end{bmatrix}\begin{bmatrix}3\\-3\end{bmatrix}?‌​ – t.b. Jul 24 '11 at 20:39 You know: it's probably a lot less work to just type the question you want, instead of posting these images which contain a lot of irrelevant information... – Arturo Magidin Jul 24 '11 at 21:08 ## 3 Answers Suppose you have a linear transformation T\colon\mathbb{R}^n\to\mathbb{R}^m. If you know what happens to the standard basis \mathbf{e}_1,\ldots,\mathbf{e}_n of \mathbb{R}^n, then you know what happens to every vector in \mathbb{R}^n, because given any vector \mathbf{v}=(a_1,\ldots,a_n), we have:$$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,$$so we can compute T\mathbf{v} by multiplying \mathbf{v}^t by the matrix A. The matrix A is the "standard matrix of T" (with respect to the standard bases of \mathbb{R}^n and \mathbb{R}^m). So you have computed A; you know \mathbf{v}. Now you just need to multiply A by \mathbf{v}^t to get the (transpose of the) image of \mathbf{v} under T. - The matrix you've written down is correct. If you have a matrix M and a vector v, the image of v means Mv. Something is a bit funny with the notation in your question. Your matrix is 4x2, so it operates on column vectors of height two (equivalently, 2x1 matrices). But the vector given is a row vector. Still, it seems clear that what you need to calculate is the product Mv that Theo wrote down in the comment. Do you know how to do that? - I would like to do it in a systematic way. It is instructive to see how OP fits in the following steps. Let V (resp. W) be an n (resp. m) dimensional vector space over \mathbb{R}. Let$$\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