5. Matrix Representation of Linear Maps#

We now know what are linear maps, now let’s look at how to represent those linear maps as matrix multiplication.

Definition#

First, let’s look at a statement:

Important

Any linear map between finite dimensional vector spaces can be represented as matrix multiplication.

There are two important pieces of information within this statement, that is:

  1. We are considering linear map only

  2. The vector spaces that we are operating need to be finite dimensional

Now, let’s see how can we represent a linear map as a matrix multiplication.

Given a linear map \(\mathcal{A}: U \rightarrow V\), this means that \(\mathcal{A}(u) = v\) (\(\mathcal{A}\) operates on an element \(u\) of domain \(U\) gives an element \(v\) of co-domain \(V\)). We want to write this as \(Au = v\) with \(A\) now is a matrix. This means that \(A\) is a matrix representation of the linear map \(\mathcal{A}\).

Warning

The \(u, v\) in \(\mathcal{A}(u) = v\) and the \(u, v\) in matrix multiplication \(Au = v\) are not necessarily the same. We will see more why this is the case later.

Bases#

Let’s first construct bases for both our domain and co-domain. This is an important step because the matrix representation depends on the bases of both the domain and co-domain. In another word, if you have the bases for both the domain and co-domain, and the definition of the linear map \(\mathcal{A}\), you can then construct the matrix \(A\). If you change the bases, then matrix \(A\) will change.

Assume that \(U\) is an n-dimensional vector space, and \(V\) is an m-dimensional vector space, given the basis \(\{u_j\}^n_{j=1}\) and \(\{v_j\}^m_{j=1}\) for each vector space respectively. Recall the meaning of basis: \(\{u_j\}^n_{j=1}\) is the set of basis for vector space \(U\) iff any vectors in \(U\) can be represented as a linear combination of \(u_j\). In addition, the coordinate of a vector \(x \in U\) w.r.t the choice of basis is unique, that is:

\[ \forall x \in U \exists! \xi = \{\xi_i\}^n_{i=1} \ni x = \Sigma_{j=1}^{n}\xi_j u_j \]

With \(\xi_i\) are the unique coordinate of \(x\) wrt choice of basis in \(U\). From here, we have the following derivation:

\[\begin{split} \begin{align*} \mathcal{A}(x) &= \mathcal{A}(\Sigma_{j=1}^{n}\xi_j u_j) \\ &= \Sigma_{j=1}^{n}\xi_j\mathcal{A}(u_j) && \text{superposition of linear map} \tag 1 \\ \end{align*} \end{split}\]

Given that each \(\mathcal{A}(u_j)\) is a linear map of element from \(U \rightarrow V\), i.e. \(\mathcal{A}(u_j) \in V\), we then have:

\[ \begin{align*} \mathcal{A}(u_j) = \Sigma_{i=1}^{m} a_{ij} v_i \tag 2 \end{align*} \]

Think of it as the operation of linear map \(\mathcal{A}\) on the j-th basis of \(U\) to get the representation in term of basis elements in \(V\).

From (1) and (2), we have:

\[\begin{split} \begin{align*} \mathcal{A}(x) &= \Sigma_{j=1}^{n}\xi_j\mathcal{A}(u_j) \\ &= \Sigma_{j=1}^{n}\xi_j\Sigma_{i=1}^{m} a_{ij} v_i \\ &= \Sigma_{i=1}^{m}(\Sigma_{j=1}^{n} a_{ij} \xi_j) v_i \\ &= \Sigma_{i=1}^{m} \eta_i v_i \tag 3 \end{align*} \end{split}\]

But as stated above, the coordinate of a vector in a vector space w.r.t the choice of basis is unique, hence, what we end up is a linear equation describing how the coordinates of vector \(x\) change under the mapping \(\mathcal{A}\).

Linear Equation#

From (3), we then have:

\[ \eta_i = \Sigma_{j=1}^{n} a_{ij} \xi_j \forall i = \{1\dots m\}, j = \{1 \dots n\} \tag 4 \]

Rewriting this in matrix form, we have:

\[ \eta = A \xi \tag 5 \]

Where \(A\) is a matrix whose elements are the \(a_{ij}\) that we have just constructed. \(A\) takes the coordinates representation of \(x\) in the domain and maps to the coordinate representation of the vector in the co-domain w.r.t choices of bases \(\{u_i\}_{i=1}^n, \{v_j\}_{j=1}^m\).

We can now see that \(\mathcal{A}(u) = v \not \leftrightarrow Au = v\), but rather \(\mathcal{A}(u) = v \leftrightarrow A \xi = \eta\).

Note

\(\mathcal{A}(u) = v \not \leftrightarrow Au = v\) is correct if we use standard basis elements in \(\mathbb{R}^n\), then the vectors themselves are the coordinates. But in general, \(A\) operates on the coordinate representation of the vector rather than the vector.

Overall, the key takeaway is:

Important

The \(j^\text{th}\) column of the matrix \(A\) is \(\mathcal{A}(u_j)\) expressed w.r.t the basis elements \(\{v_i\}\)