Matrix Representation


Introduction


Every linear map \(f \in \mathcal L(\mathcal V, \mathcal W )\) between two vector spaces \(\mathcal V\) and \(\mathcal W\) has a unique matrix representation. In most fields where one uses numerical approaches to solve problems a matrix representation is preferred to do any calculations due to practicality and opportunity of parallelizing computations. This section is a short summary of matrix representation.



Approach to obtain the Matrix Representation


Let \(\mathcal V (\mathbb{F})\) and \(\mathcal W (\mathbb{F})\) be two vector spaces over a field \(\mathbb{F}\) and let \(\mathcal B_{1} = \{b_{1}, ..., b_{n} \}\) and \(\mathcal B_{2} = \{\hat{b_{1}}, ..., \hat{b_{m}} \}\) be two basis for \(\mathcal V\) and \(\mathcal W\) respectively. Further let \(f \in \mathcal L (\mathcal V, \mathcal W )\) be a homomorphism.

Then for \(f(b_{j}) \in \mathcal W\)

/c/there exists uniquely defined coordinates/c/

\(\alpha_{1, j}, ..., \alpha_{n, j} \in \mathbb{F}\) such that:


\[f(b_{j}) = (\hat{b_{1}}, ..., \hat{b_{m}}) \begin{bmatrix} \alpha_{1, j} \\ \alpha_{2, j} \\ \vdots \\ \alpha_{n, j} \end{bmatrix}\]


This is a system of equations that can be extended so that we obtain one expression for all basis elements in $ \mathcal B_{1}$. For this denote $ A = [ \alpha_{i, j} ] \in \mathbb{F}^{m \times n} $ as the matrix of all coordinates from that system of equation:


\[\begin{align} f(b_{1}, ..., b_{n}) = (f(b_{1}), ..., f(b_{n})) = (\hat{b_{1}}, ..., \hat{b_{m}}) \ A \end{align}\]


Then we obtain for any \(v = \sum_{i=1}^{\\n} \lambda_{i} b_{i} \in \mathcal V\) :


\[\begin{align} f(v) &= f \ (\sum_{i=1}^{\\n} \lambda_{i} b_{i}) \\ &= (f(b_{1}), ..., f(b_{n})) \begin{bmatrix} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{bmatrix} \\ &= (\hat{b_{1}}, ..., \hat{b_{m}}) \ A \begin{bmatrix} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{bmatrix} \end{align}\]


The matrix representation of \(f\) is given by \(A\) and is denoted by \([f]_{B_{1}, B_{2}}\). The coordinates of \(f(v) = w \in \mathcal{W}\) is given by \(\Biggl( [f]_{B_{1}, B_{2}} \begin{bmatrix} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{bmatrix} \Biggl)\) where \({\lambda_{1}, ..., \lambda_{n}}\) are the coordinates of \(v\).


Let $f \in \mathcal{L}(\mathbb{R}^{3}, \mathbb{R}^{2})$ and $\mathcal{B}_1 = \left\{ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \right\} $ be a basis of $\mathbb{R}^{3}$ and $\mathcal{B}_2 = \left\{ \begin{pmatrix} 3 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \end{pmatrix} \right\} $ be a basis of $\mathbb{R}^{2}$. Then $f$ is defined by: $$ f \ : \mathbb{R}^{3} \longrightarrow \mathbb{R}^{2} , \quad \begin{pmatrix} v_1 \\ v_2 \\ v_3 \end{pmatrix} \longmapsto \begin{pmatrix} v_1 + 3v_2 \\ v_1 + 2v_3 \end{pmatrix} \ $$
Find the coordinates of $f\Biggl( \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} \Biggl)$ w.r.t. $\mathcal{B}_2$.

SOLUTION:

1) We need to calculate the image of the basis $\mathcal{B}_1$ under f:

$ f \Biggl( \Biggl( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \Biggl) \Biggl) = \Biggl(f \Biggl( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\Biggl), f \Biggl( \begin{pmatrix} 0 \\ 2 \\ 1 \end{pmatrix}\Biggl), f \Biggl( \begin{pmatrix} 0 \\ 0 \\ 2 \end{pmatrix} \Biggl) \Biggl) = \Biggl( \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \begin{pmatrix} 6 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 4 \end{pmatrix}\Biggl)$


2) Now we want to obtain the matrix representation $[f]_{B_1, B_2}$:

$[f]_{B_1, B_2} = \begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix}^{-1} \begin{bmatrix} 1 & 6 & 0 \\ 1 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 1/3 & 0 \\ -1/6 & 1/2 \end{bmatrix} \begin{bmatrix} 1 & 6 & 0 \\ 1 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 1/3 & 2 & 0 \\ 1/3 & 0 & 2 \end{bmatrix} $


3) Next we need to find the coordinates of $\begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix}$ w.r.t. $\mathcal{B}_1$:

$\Phi_{\mathcal{B}_1} \Biggl( \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} \Biggl) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 1 & 2 \end{bmatrix}^{-1} \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1/2 & 0 \\ 0 & -1/4 & 1/2 \end{bmatrix} \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} = \begin{bmatrix} 3 \\ 3 \\ -1/2 \end{bmatrix} $


4) Last but not least the coordinates of $f\Biggl( \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} \Biggl)$ w.r.t. $\mathcal{B}_2$:

$\Phi_{\mathcal{B}_2} \Biggl( f \Biggl( \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} \Biggl) \Biggl) = [f]_{B_1, B_2} \begin{bmatrix} 3 \\ 3 \\ -1/2 \end{bmatrix} = \begin{bmatrix} 1/3 & 2 & 0 \\ 1/3 & 0 & 2 \end{bmatrix} \begin{bmatrix} 3 \\ 3 \\ -1/2 \end{bmatrix} = \begin{bmatrix} 7 \\ 0 \end{bmatrix} $


Let's check our result:

$$ \begin{pmatrix} 21 \\ 7 \end{pmatrix} = f\Biggl( \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} \Biggl) = \Biggl( \begin{pmatrix} 3 \\ 1 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \end{pmatrix} \Biggl) \Phi_{\mathcal{B}_2} \Biggl( f \Biggl( \begin{pmatrix} 3 \\ 6 \\ 2 \end{pmatrix} \Biggl) \Biggl) = \begin{bmatrix} 3 & 0 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 7 \\ 0 \end{bmatrix} = \begin{pmatrix} 21 \\ 7 \end{pmatrix} \qquad \qquad \qquad \square $$



Commutative Diagram


Let \(\mathcal{V}(\mathbb{F})\) be an n-dimensional vector space and \(\mathcal{B}_1, \ \mathcal{B}_2\) are some basis of \(\mathcal{V}\). \(\mathcal{W}(\mathbb{F})\) is an m-dimensional vector space with basis \(\mathcal{\hat{B}}_1, \ \mathcal{\hat{B}}_2\). The following commutative diagram captures the mappings of the (1) homomorphism, (2) coordinates, (3) identity and their matrix representations:


\[\begin{align} &(1) \quad f \ : \mathcal V \longrightarrow \mathcal W , \quad v \longmapsto f(v) = w\\ &(2) \quad \Phi_B \ : \mathcal V \longrightarrow \mathbb{K}^{n, 1} , \quad v = \sum_{i=1}^{\\n} \lambda_{i} b_{i} \longmapsto \Phi(v) := \begin{bmatrix} \lambda_{1} \\ \vdots \\ \lambda_{n} \end{bmatrix} \\ &(3) \quad \mathcal{I}d_{\mathcal V} \ : \mathcal V \longrightarrow \mathcal V , \quad \mathcal v \longmapsto \mathcal{I}d_{\mathcal V}(v) = v \end{align}\]



The dotted arrows are the mappings from the matrix representations.
To obtain $$ [\mathcal{I}d_{\mathcal V}]_{B_1, B_2} $$ use the same method from before for f but consider f to be an endomorphism and then let f be the identity map.


Hint : Notice that the matrix representations only map between the coordinates of vectors.





Properties


Coordinate Mapping


  1.   The coordinate mapping \(\Phi_{\mathcal{B}}\) is an isomorphism.
  2.   \(\Phi_{\mathcal{B}_2}(f(v)) = [f]_{\mathcal{B}_1, \mathcal{B}_2} \ \Phi_{\mathcal{B}_1}(v) ; \quad \forall \ v \in \mathcal{V}\)
  3.   A coordinate transformation is defined by \(\Phi_{\mathcal{B}_2} = [\mathcal{I}d_{\mathcal{V}}]_{\mathcal{B}_1, \mathcal{B}_2} \ \circ \ \Phi_{\mathcal{B}_1}\). This is useful if you consider \(\mathcal{B}_1\) to be the canonical basis.



Basis Transformation


  1.   The matrix representation of the identity mapping \(\mathcal{I}d_{\mathcal{V}}\) is the basis transformation from \(\mathcal{B}_1\) to \(\mathcal{B}_2\).
  2.   The basis transformation \([\mathcal{I}d_{\mathcal{V}}]_{\mathcal{B}_1, \mathcal{B}_2}\) is an isomorphism.
  3.   \(([\mathcal{I}d_{\mathcal{V}}]_{\mathcal{B}_1, \mathcal{B}_2})^{-1} = [\mathcal{I}d_{\mathcal{V}}]_{\mathcal{B}_2, \mathcal{B}_1}\)



Matrix Representation


  1.   \(f = \Phi_{\mathcal{B}_2}^{-1} \ [f]_{\mathcal{B}_1, \mathcal{B}_2} \ \Phi_{\mathcal{B}_1}\)
  2.   For \(f \in \mathcal{L}(\mathcal{V}, \mathcal{W})\) and \(g \in \mathcal{L}(\mathcal{W}, \mathcal{U})\) and \(\mathcal{V}, \mathcal{W}, \mathcal{U}\) being finite dimensional with basis \(\mathcal{B}_1, \mathcal{B}_2, \mathcal{B}_3\), respectively, \([g \circ f]_{\mathcal{B}_1, \mathcal{B}_3} = [g]_{\mathcal{B}_2, \mathcal{B}_3} \ [f]_{\mathcal{B}_1, \mathcal{B}_2}\).
  3.   Let \(v_1, ..., v_n \in \mathcal{V}\) be linearly independent and \(dim(\mathcal{W}) = m \in \mathbb{N}\). Further, let \(f \in \mathcal{L}(\mathcal{V}, \mathcal{W})\). Iff \(\textbf{rank}([f]_{\mathcal{B}_1, \mathcal{B}_2}) = m\) then \((v_1, ..., v_n) [f]_{\mathcal{B}_1, \mathcal{B}_2} = (w_1, ..., w_m) \in \mathcal{W}\) are linearly independent.




Applications


  Diagonal representation

Let \(\mathcal{V}(\mathbb{F})\) be a finite dimensional vector space and \(f\) an endomorphism. If \(\{\hat{v}_1, ..., \hat{v}_n \}\) are the eigenvectors of \(f\) choose \(\mathcal{B} = \{\hat{v}_1, ..., \hat{v}_n \}\) as the basis. Then we get as the matrix representation of \(f\):

\[[f]_{\mathcal{B}} = \begin{bmatrix} \lambda_1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \lambda_n \end{bmatrix}\]

where \(f(\hat{v}_i) = \lambda_i \hat{v}_i\).


Some text.



  Differential operator representation


Let \(p = \sum_{k=0}^{n} \alpha_k t^k \in \mathbb{R}[t]_{\leq n} \ \land \ \mathcal{B} = \{ t^k \}_{k=0}^{n}\) be the canonical basis then the derivative of p(t) is given by:

\[D(p(t)) = D \Biggl( \sum_{k=0}^{n} \alpha_k t^k \Biggl) = \sum_{k=1}^{n} k \alpha_k t^{k - 1} = \begin{bmatrix} 1 & t & \ldots & t^n \end{bmatrix} \begin{bmatrix} \alpha_{1} \\ 2 \alpha_2 \\ \vdots \\ n \alpha_{n} \\ 0 \end{bmatrix}\]


Therefore, the representation matrix of the operator, \(D\), has to fulfill the following equation:

\[[D]_{\mathcal{B}} \begin{bmatrix} \alpha_0 \\ \alpha_1 \\ \vdots \\ \alpha_{n-1} \\ \alpha_n \end{bmatrix} = \begin{bmatrix} \alpha_{1} \\ 2 \alpha_2 \\ \vdots \\ n \alpha_{n} \\ 0 \end{bmatrix}\]


Solving that equation we obtain the following form:

\[[D]_{\mathcal{B}} = \begin{bmatrix} \begin{array}{c|c} 0_{n, 1} & M \\ \hline 0 & 0_{1, n} \end{array} \end{bmatrix} \\\]

where \(M = \textbf{diag}(1, ..., n) \in \mathcal{M}_{n \times n}(\mathbb{R})\).