# Is a linear map (transformation) always a matrix multiplication

I am studying linear maps. It is defined as a linear map $L$ which transforms a vector from dimension $n$ to dimension $k$

$L:\mathbb{R}^n \rightarrow \mathbb{R}^k$

This seems to me as a matrix multiplication (from $x$ to $y$):

$y = Ax$

My question is, is this correct, and further, can a linear map always be written as a matrix multiplication?

• Yes, and the matrix is that with columns $L(e_1), L(e_2),..., L(e_n)$.
– arts
Dec 2, 2017 at 13:15
• By the way, this is only true for linear maps from $\mathbb{R}^n$ to $\mathbb{R}^k$. For vector spaces of a different nature, it is clearly not true since the vectors don't need to be column matrices to begin with.
– arts
Dec 2, 2017 at 13:22
• All finite dimensional vector spaces of dimension $n$ are isomorphic to $\mathbb R^n$,... just pick a basis Dec 2, 2017 at 13:35
• @klirk Which gives you a matrix multiplication between the $\mathbb{R}^n$'s, not the original spaces. Exactly the subtle distinction I wanted him to understand.
– arts
Dec 2, 2017 at 13:46
• @user3053216 you can set as solved if you are ok
– user
Dec 3, 2017 at 9:58

The answer is yes. If you have a linear map $$\phi: V \to W$$, between finite dimensional vector spaces of dimension $$n$$ resp $$k$$, then this gives rise to a matrix in the following way:

Choose a basis $$\{x_i\}$$ of $$V$$ and $$\{y_1\}$$ of $$W$$.
Then the matrix corresponds to how $$\phi$$ acts on the $$x_i$$ in terms of $$y_i$$.

As $$\phi(x_i)\in W$$, we can find coefficients $$m^j_i$$ such that $$\phi(x_i)=\sum_{j=1}^k m^j_i y_j.$$The coefficients $$m^j_i$$ correspond to the entries of the matrix $$M$$ representing $$\phi$$.

In particular, if $$\{y_i\}$$ are an orthogonal basis, we can calculate $$m^j_i$$ by $$m^j_i=.$$

Further, for an arbitrary vector $$v = \sum_{i=1}^n a^i x_i \in V$$ (with some coefficients $$a_i$$), we have that $$\phi(v)=\phi( \sum_{i=1}^n a^i x_i) = \sum_{i=1}^n a^i \phi(x_i) = \sum_{i=1}^n a^i m^j_i y_i.$$ From the formula for multiplicating a vector with a matrix, we see that in this basis, the components of $$\phi(v)$$ correspond to the entries of $$Mv$$

Edit: A short remark as a reply to a comment:
$$M$$ is not $$\phi$$. $$\phi$$ is a linear map between $$V$$ and $$W$$, whereas $$M$$ is a matrix and thus induces a linear map between $$\mathbb R^n$$ and $$\mathbb R^k$$ by $$x \mapsto Mx$$. $$M$$ only represents $$\phi$$, that is the following diagram commutes:

$$\newcommand{\ra}{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} V & \ra{\phi} & W \\ \da{} && \da {} \\ \mathbb R^n & \ra{M \cdot} & \mathbb{R}^k \\ \end{array} ,$$ where the vertical maps are the isomorphisms given by choosing a basis.

• Finite dimensionality is not needed, only existence of basis.
– arts
Dec 2, 2017 at 13:29
• @arts without finite dimensionality, the information needed to specify the linear transformation isn't a finite length grid of coefficients, and would not be called a matrix. Dec 2, 2017 at 13:42
• Assuming the axiom of choice, every vector space has a basis. If the dimension is infinite, the basis is uncountable. I wouldn't call the resulting object a Matrix. Sometimes (for example in quantum physics), people work with orthogonal systems (countable) and call the $m^j_i$ as defined in my answer matrixelements Dec 2, 2017 at 13:42
• @arts You are right, the linear combinations are finite. However, there are still uncuntably many basis vectors and the matrix would need to include information about how $\phi$ acts on all of them Dec 2, 2017 at 13:49
• @arts What is the definition of matrix you are working with? Dec 2, 2017 at 13:52

Yes it's alway possible use matrices for linear maps!

https://en.wikipedia.org/wiki/Linear_map

• As klirk said, this works for the maps between finite dimensional spaces, but especially if the domain is not spanned by a finite set, you wouldn't have a matrix. Dec 2, 2017 at 13:44
• I just wanted to add, that in that book, the coefficients in the Fourier expansion with respect to an orthogonal "basis" are used to define the infinite matrix. But such an orthogonal system is not a basis in the usual sense. As @arts wrote: "You need to study what it means to be a basis. Bases generate all vectors of the space by finite linear combinations." But an orthogonal system is not able to achieve this. In fact, the definition in this book agrees with that I wrote previously: "Sometimes, people work with orthogonal systems and call the m_ji as defined in my answer matrixelements" Dec 2, 2017 at 15:14
• @user Is (AX)B where A, B, and X are matrices also a linear transformation? Sep 11, 2020 at 15:29
• @Media Yes of course, if $(AX)B$ are compatible for the multiplication they represent a linear transformation.
– user
Sep 11, 2020 at 15:37