# How to find the linear transformation associated with a given matrix?

Good day, I have a little doubt: It is well known that given two bases (or even one if we consider the canonical basis) of a vector space, every linear transformation $T:V \rightarrow W$ can be represented as a matrix, but since this is an isomorphism between $L(V,W)$ and $\mathbb{M}_{m\times n}$ where the latter represents the space $m\times n$ matrices on the same field in which are defined respectively vector spaces. That's where my question comes up, I know find the matrix associated with the linear transformation, but not know how to move from the matrix transformation, ie given any matrix, find the linear transformation that defines it. I wish you could please explain the theoretical process and to see a practical example. Thank you very much, I know it's definitely something silly, but I'm still a student.

The columns of the matrix tell us where the basis vectors of the domain are mapped, in terms of the basis vectors of the codomain. Since every vector in the domainis a linear combination of the basis vectors (in a unique way), we can extrapolate, in a sense, the image of any given vector. Let $A$ be an $m\times n$ matrix (with coefficients in a field $F$) with columns $A_1, ..., A_n$. Let $V$ be an $n-$dimensional $F-$vector space, and $W$ an $m-$dimensional $F-$vector space, with ordered bases $(v_1, ..., v_n)$ and $(w_1, ..., w_m)$, respectively. Finally, let $T$ be the linear transformation associated with $A$, and let $v\in V$ with $v = c_1v_1 + ... + c_n v_n$ (remember, this expression for $v$ as a linear combination of basis vectors is unique). Then $$T(v) = T(c_1v_1+...+c_nv_n) = c_1T(v_1)+...+c_nT(v_n) = c_1A_1 + ... + c_nA_n$$ So, this is how the matrix lets us calculate the image of any vector. Notice that the expression on the right is just the matrix $A$ multiplied by the vector $[c_1, c_2, ..., c_n]^T$.

For example, if we let $V = W = \Bbb{R}^2$ (considered as $\Bbb{R}-$vector spaces) with the standard basis, let $$A = \left[ \begin{array}{ccc} 1 & 2 \\\ 2 & 3 \end{array} \right]$$ And let $T$ be the linear transformation associated with $A$, and let $v = [1,5]^T$. Then $$T(v) = 1\cdot T(e_1) + 5\cdot T(e_2) = [1,2]^T+5[2,3]^T = [11,17]^T = A\cdot v$$

• Thank you for your contribution! But I do not understand something, the vector $(1,5)$ where it came from? assuming we are working with the canonical basis. And in the last expression $Av$, this seems to be a vector not the algebraic expression of $T$ ?. Please, you can better explain your example? – Hendrik Matamoros Aug 25 '16 at 13:52
• I just chose an arbitrary vector to show how you can take any vector, and use the matrix to find its image under the associated linear transformation. – florence Aug 25 '16 at 14:07
• Sure, I understand, but that's not my question, my question is given the matrix $A$ I want to find is the linear transformation $T$ associated with $A$, I did not mean vector image. Can you explain this please? – Hendrik Matamoros Aug 25 '16 at 14:11
• I don't quite get what you mean by "find the linear transformation"; a linear transformation is just a special function which assigns an image to each element of the domain. If you know how to find the image of a given vector, then you know everything about the linear transformation. – florence Aug 25 '16 at 14:46

I am not sure what you expect, say $T$ is your linear transform, and $A$ represents it in the basis for $V = \{a_1,\dots,a_n\}$ and $W = \{b_1,\dots,b_m\}$.

Then we say $Ta_j = \sum_i A_{ij}b_i.$

This should completely define what $T$ "does". Typically, one defines the action of $T$ on the basis vectors $a_i$ in order to determine what $T$ "looks like". Otherwise, one can not write down much about $T$. Please clarify what type of answer you are expecting if this is not sufficient.

• If we are dealing with a linear operator $T:V \to V,$ then we can determine some more properties that hold for $T$ independent of basis, such as tr($T$), det($T$), rank($T$), etc. but these things do not fully characterize $T$. As stated, the action of $T$ between two bases is enough to fully define the transformation. – Merkh Aug 25 '16 at 12:59

Suppose that you have an $m,n$ matrix $A$. Choose a basis $B$ of $V$ and another one $B'$ of $W$. The linear transformation associated with $A$ relative to the bases $B$ and $B'$ is $T(v) = Av$, where $v$ is to be written as a column whose entries are the coefficients of $v$ in the basis $B$ and the resulting column $T(v)$ has entries which are the coefficients of $T(v)$ in the basis $B'$. If you choose other bases, you get different linear transformations

• Thank you so much for your contribution!, but can you explain me with this matrix A=\begin{bmatrix}{3}&{2}\\{-8}&{1}\end{bmatrix}.? Please. – Hendrik Matamoros Aug 25 '16 at 13:01
• Assuming the canonical basis of $\mathbb R^2$, the associated linear transformation is for $v=(v_1,v_2)$, $T(v) = Av^T = (3v_1 +2v_2, -8v_1+v_2)^T$ (where $^T$ is the transpose) – user290300 Aug 25 '16 at 13:05
• By the way, it was not nice of you to switch the accepted answer. Make up your mind then choose (please don't re-accept my answer) – user290300 Aug 25 '16 at 13:06
• Oh, thank you very much for the explanation, I ask you a million apologize for the inconvenience, I'm really new to the forum, I figured I could choose several answers, all seems good contribution and I am grateful for it. It has never been my intention to hurt anyone's feelings – Hendrik Matamoros Aug 25 '16 at 13:11
• You are very welcome, there's no need to apologize. Welcome to MSE! Almost all answers over here are good contributions and people carefully typeset equations with Mathjax and organize their answers. So yes, it is difficult to choose among answers – user290300 Aug 25 '16 at 13:14