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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?

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    $\begingroup$ Yes, and the matrix is that with columns $L(e_1), L(e_2),..., L(e_n)$. $\endgroup$ – arts Dec 2 '17 at 13:15
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    $\begingroup$ 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. $\endgroup$ – arts Dec 2 '17 at 13:22
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    $\begingroup$ All finite dimensional vector spaces of dimension $n$ are isomorphic to $\mathbb R^n$,... just pick a basis $\endgroup$ – klirk Dec 2 '17 at 13:35
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    $\begingroup$ @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. $\endgroup$ – arts Dec 2 '17 at 13:46
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    $\begingroup$ @user3053216 you can set as solved if you are ok $\endgroup$ – user Dec 3 '17 at 9:58
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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 $$m^j_i=<y_j,\phi(x_i)>.$$

Further, for an arbitrary vector $ v = \sum_{i=1}^n a^i x_i \in V$ (with some coefficients $a_i$), we have that $$\phi(v)=\phi( \sum_{i=1}^n a^i x_i) = \sum_{i=1}^n a^i \phi(x_i) = \sum_{i=1}^n a^i m^j_i y_i.$$ Form 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: What should be obvious, but maybe it still needs to be adressed as pointed out by a comment to the question:
$M$ 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:

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

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  • $\begingroup$ Finite dimensionality is not needed, only existence of basis. $\endgroup$ – arts Dec 2 '17 at 13:29
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    $\begingroup$ @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. $\endgroup$ – Mark S. Dec 2 '17 at 13:42
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    $\begingroup$ 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 $\endgroup$ – klirk Dec 2 '17 at 13:42
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    $\begingroup$ @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 $\endgroup$ – klirk Dec 2 '17 at 13:49
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    $\begingroup$ @arts What is the definition of matrix you are working with? $\endgroup$ – klirk Dec 2 '17 at 13:52
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Yes it's alway possible use matrices for linear maps!

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

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    $\begingroup$ 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. $\endgroup$ – Mark S. Dec 2 '17 at 13:44
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    $\begingroup$ 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" $\endgroup$ – klirk Dec 2 '17 at 15:14

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