Matrices and Change of Basis

Section 13.4 Matrices and Change of Basis

Subsection Matrices of Linear Transformations

“No one can be told what the matrix is. You have to see it for yourself.”
―Morpheus

Definition 13.59. Matrix of Free Module Homomorphism.

Let \(R\) be a non-zero commutative ring and let \(V,W\text{,}\) be \(R\)-modules of finite rank \(n\) and \(m\text{,}\) respectively. Let \(B=\{b_1, \dots, b_n\}\) and \(C=\{c_1, \dots, c_m\}\) be ordered bases of \(V\) and \(W\text{.}\) Given an \(R\)-module homomorphism \(f: V \rightarrow W\text{,}\) we define elements \(a_{i j} \in R\) for \(1 \leqslant i \leqslant m\) and \(1 \leqslant j \leqslant n\) by the formulas

\begin{equation*} f\left(b_{i}\right)=\sum_{j=1}^{m} a_{j, i} c_{j}. \end{equation*}

The matrix

\begin{equation*} [f]_{B}^{C}=\left[\begin{array}{cccc} a_{1,1} & a_{1,2} & \cdots & a_{1, n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2, n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m, 1} & a_{m, 2} & \cdots & a_{m, n} \end{array}\right] \end{equation*}

is said to represent the homomorphism \(f\) with respect to the bases \(B\) and \(C\text{.}\) In the case that \(V\) and \(W\) are \(F\)-vector spaces, the matrix \([f]_{B}^{C}\) is called the matrix of the linear transformation \(f\) with respect to the bases \(B\) and \(C\text{.}\)

Remark 13.60.

The notion that a homomorphism between finitely generated free modules can be represented as a matrix is extremely important! It allows us to translate information we know about matrices into information we know about homomorophisms and linear transformations, and vice versa.

Example 13.61. Identity Automorphism of Free \(R\)-Module.

If \(\id_V: V \to V\) is the identity automorphism of an \(n\)-dimensional free \(R\)-module \(V\text{,}\) then for any basis \(B\) of \(V\) we have \(\id_V(b_i) = b_i\) for all \(i\) and hence

\begin{equation*} [\id_V]^B_B = I_n. \end{equation*}

Example 13.62. Matrix Representing the Derivative Operator.

Let \(P_{3}\) denote the the \(F\)-vector space of polynomials of degree at most \(3\) (including the zero polynomial) and consider the linear transformation \(d: P_{3} \rightarrow P_{3}\) given by taking the derivative \(d(f)=f^{\prime}\text{.}\) Let \(B=\left\{1, x, x^{2}, x^{3}\right\}\text{.}\) Then

\begin{equation*} [f]_{B}^{B}=\left[\begin{array}{llll} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right]. \end{equation*}

Lemma 13.63. Linear Transformation between Vector Spaces is Matrix Multiplication.

Let \(F\) be a field and consider a linear transformation \(f: V \rightarrow W\text{,}\) where \(V=F^{n}\) and \(W=F^{m}\text{.}\) Consider also the standard ordered bases \(B\) and \(C\text{,}\) i.e. \(b_{i}=e_{i} \in V\) and \(c_{i}=e_{i} \in W\text{.}\) Then for any

\begin{equation*} v=\left[\begin{array}{c} l_{1} \\ \vdots \\ l_{n} \end{array}\right]=\sum_{i} l_{i} b_{i} \end{equation*}

in \(V\) we have

\begin{equation*} f\left(\sum l_{i} b_{i}\right)=\sum_{i} l_{i} f\left(b_{i}\right) \end{equation*}

Each \(f\left(b_{i}\right)\) can be written uniquely as a linear combination of the \(c_{j}\)’s as in (2.2.1):

\begin{equation*} f\left(b_{i}\right)=\sum_{j} a_{j, i} c_{j} \end{equation*}

Then we get

\begin{equation*} f(v)=\sum_{i} l_{i}\left(\sum_{j} a_{j, i} c_{j}\right)=\sum_{j}\left(\sum_{i} a_{j, i} l_{i}\right) c_{j}. \end{equation*}

In other words, we have

\begin{equation*} f(v)=\left[\begin{array}{c} \sum_{i} a_{1, i} l_{i} \\ \vdots \\ \sum_{i} a_{m, i} l_{i} \end{array}\right] =\left[\begin{array}{cccc} a_{1,1} & a_{1,2} & \cdots & a_{1, n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2, n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m, 1} & a_{m, 2} & \cdots & a_{m, n} \end{array}\right] \cdot\left[\begin{array}{c} l_{1} \\ \vdots \\ l_{n} \end{array}\right]=[f]_{B}^{C} \cdot v \end{equation*}

Then for any

\begin{equation*} v=\sum_{i} l_{i} b_{i} \end{equation*}

in \(V\) we have

\begin{equation*} f\left(\sum l_{i} b_{i}\right)=\sum_{i} l_{i} f\left(b_{i}\right) \end{equation*}

Each \(f\left(b_{i}\right)\) is uniquely expressible as a linear combination of the \(c_{j}\) ’s, say

\begin{equation*} f\left(b_{i}\right)=\sum_{j} a_{j, i} c_{j}. \end{equation*}

Then we get

\begin{equation*} f(v)=\sum_{i} l_{i}\left(\sum_{j} a_{j, i} c_{j}\right)=\sum_{j}\left(\sum_{i} a_{j, i} l_{i}\right) c_{j}. \end{equation*}

In other words, we have

\begin{equation*} f(v)=[f]_{B}^{C} \cdot v \end{equation*}

where

\begin{equation*} [f]_{B}^{C}= \left[\begin{array}{cccc} a_{1,1} & a_{1,2} & \cdots & a_{1, n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2, n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m, 1} & a_{m, 2} & \cdots & a_{m, n} \end{array}\right] \end{equation*}

and \([f]_{B}^{C} \cdot v\) denote the usual rule for matrix multiplication.

This says that any linear transformation \(f: F^{n} \rightarrow F^{m}\) is given by multiplication by a matrix, since we noticed above that \(f(v)=[f]_{B}^{C} \cdot v\text{.}\) The same type of statement holds for free modules over commutative rings, and we will show it below in Theorem 2.32.

Theorem 13.64. \(R\)-map Bewteen Free Modules is Matrix Multiplication.

Let \(R\) be a commutative ring with \(1 \neq 0\text{.}\) Let \(V\) and \(W\) be finitely generated free \(R\)-modules of ranks \(n\) and \(m\) respectively. Fixing ordered bases \(B\) for \(V\) and \(C\) for \(W\) gives an isomorphism of \(R\)-modules

\begin{equation*} \operatorname{Hom}_{R}(V, W) \cong \mathrm{M}_{m, n}(R) \quad f \mapsto[f]_{B}^{C} \end{equation*}

If \(V=W\text{,}\) so that in particular \(m=n\text{,}\) and \(B=C\text{,}\) then the above map is an \(R\)-algebra isomorphism \(\operatorname{End}_{R}(V) \cong \mathrm{M}_{n}(R)\text{.}\)

Proof.

Let \(\varphi: \operatorname{Hom}_{R}(V, W) \rightarrow \mathrm{M}_{m, n}(R)\) be defined by \(\varphi(f)=[f]_{B}^{C}\text{.}\) We need to check that \(\varphi\) is a homomorphism of \(R\)-modules, which translates into \([f+g]_{B}^{C}=[f]_{B}^{C}+[g]_{B}^{C}\) and \([\lambda f]_{B}^{C}=\lambda[f]_{B}^{C}\) for any \(f, g \in \operatorname{Hom}_{R}(V, W)\) and \(\lambda \in R\text{.}\) Let \(A=[f]_{B}^{C}\) and \(A^{\prime}=[g]_{B}^{C}\text{.}\) Then

\begin{equation*} (f+g)\left(b_{i}\right)=f\left(b_{i}\right)+g\left(b_{i}\right)=\sum_{j} a_{j, i} c_{j}+\sum_{j} a_{j, i}^{\prime} c_{j}=\sum_{j}\left(a_{j, i}+a_{i, j}^{\prime}\right) c_{j} \end{equation*}

gives \([f+g]_{B}^{C}=A+A^{\prime}\) and

\begin{equation*} (\lambda f)\left(b_{i}\right)=\lambda\left(\sum_{j} a_{j, i} c_{j}\right)=\sum_{j}\left(\lambda a_{j, i}\right) c_{j} \end{equation*}

gives \([\lambda f]_{B}^{C}=\lambda A\text{.}\)

Finally, the argument described in Example 2.31 also works for any ring \(R\text{,}\) and it can be adapted for any two chosen basis \(B\) and \(C\text{,}\) showing that \(\varphi\) is a bijection.

Subsection Change of Basis

“Change brings opportunity.”
―Nido R. Qubein

As we noted in Section 13.2, bases of free modules are rarely unique. Each basis provides its own individual “langauge” of constucting elements of the module. Different bases may lend themselves to different situations and mathematicians, and thus being able to “translate” between these langauges is extremely important.

Definition 13.65. Change of Basis Matrix.

Let \(V\) be a finitely generated free module over a commutative ring \(R\text{,}\) and let \(B\) and \(C\) be bases of \(V\text{.}\) Let \(\operatorname{id}_{V}\) be the identity map on \(V\text{.}\) Then \(\left[\operatorname{id}_{V}\right]_{B}^{C}\) is a matrix called the change of basis matrix from \(B\) to \(C\text{.}\)

Remark 13.66.

In Theorem 13.71 we will show that \(\left[\mathrm{id}_{V}\right]_{B}^{C}\) is invertible with inverse \(\left(\left[\mathrm{id}_{V}\right]_{B}^{C}\right)^{-1}=\left[\mathrm{id}_{V}\right]_{C}^{B}\text{.}\)

Example 13.67. CoB and \(P_n\).

Consider \(V = P_2\text{,}\)

Where P_2 is as defined in Example 13.62.

let \(B = \{1, x, x^2\}\) and \(B' = \{1,x-2,(x-2)^2\}\) be bases of \(V\text{.}\) We calculate the change of basis matrix. We have

\begin{equation*} \begin{aligned}\id_V(1) &=1 ,\\ \id_V(x) &=2\cdot1+1\cdot(x-2), \\ \id_V(x^2) &=4\cdot1 +4\cdot(x-2)+1\cdot(x-2)^2. \end{aligned} \end{equation*}

Thus, the change of basis matrix is given by

\begin{equation*} [\id_V]_B^{B'} = \begin{bmatrix} 1 & 2 & 4\\ 0 & 1 & 4\\ 0 & 0 & 1 \end{bmatrix}. \end{equation*}

Lemma 13.68. CoB and Identity.

If \(V, W, U\) are finitely generated free \(R\)-modules spaces with ordered bases \(B\text{,}\) \(C\text{,}\) and \(D\text{,}\) and if \(f: V \rightarrow W\) and \(g: W \rightarrow U\) are \(R\)-module homomorphisms, then

\begin{equation*} [g \circ f]_{D}^{B}=[g]_{D}^{C} \cdot[f]_{C}^{B} \end{equation*}

Proof.

Given \(v \in V\text{,}\) we have

\begin{equation*} (f \circ g)(v)=f(g(v))=f\left([g]_{B}^{C} v\right)=[f]_{C}^{D}\left([g]_{B}^{C} v\right)=\left([f]_{C}^{D}[g]_{B}^{C}\right)v \end{equation*}

so \([f \circ g]_{B}^{B}=[f]_{B}^{B}[g]_{B}^{B}\text{.}\)

Definition 13.69. Similar Matrices.

Let \(F\) be a finitely generated free module over a commutative ring \(R\text{.}\) Two \(R\)-module homomorphisms \(f, g: F \rightarrow F\) are similar if there is an \(R\)-module isomorphism \(h: V \rightarrow V\) such that \(g=h \circ f \circ h^{-1}\text{.}\)

Two \(n \times n\) matrices \(A\) and \(B\) with entries in \(R\) are similar if there is an invertible \(n \times n\) matrix \(P\) such that \(B=P A P^{-1}\text{.}\)

The notion of similar matrices aligns with the idea of moving between bases. Translate your data into the language using your new basis (using \(P\)), perform the desired transformations (using \(A\)), and then translate this new data back into the original basis (using \(P\inv\))

Remark 13.70.

For elements \(A, B \in \mathrm{GL}_{n}(R)\text{,}\) the notions of similar and conjugate are the same.

Theorem 13.71. Similar CoB Matrices.

Let \(V, W\) be finitely generated free modules over a commutative ring \(R\text{,}\) let \(B\) and \(B^{\prime}\) be bases of \(V\text{,}\) let \(C\) and \(C^{\prime}\) be bases of \(W\text{,}\) and let \(f: V \rightarrow W\) be a homomorphism. Then

\begin{equation*} [f]_{B^{\prime}}^{C^{\prime}}=\left[\mathrm{id}_{W}\right]_{C}^{C^{\prime}}[f]_{B}^{C}\left[\mathrm{id}_{V}\right]_{B^{\prime}}^{B} \end{equation*}

In particular, if \(g: V \rightarrow V\) is an \(R\)-module homomorphism, then \([g]_{B}^{B}\) and \([g]_{B^{\prime}}^{B^{\prime}}\) are similar.

Proof.

Since \(f=\mathrm{id}_{W} \circ f \circ \mathrm{id}_{V}\text{,}\) by Lemma 13.68 we have

\begin{equation*} [f]_{B^{\prime}}^{C^{\prime}}=\left[\mathrm{id}_{W}\right]_{C}^{C^{\prime}}[f]_{B}^{C}\left[\mathrm{id}_{V}\right]_{B^{\prime}}^{B} \end{equation*}

Setting \(V=W, B=C, B^{\prime}=C^{\prime}\text{,}\) and \(f=\mathrm{id}_{V}\) we have \(\left[\mathrm{id}_{V}\right]_{B^{\prime}}^{B^{\prime}}=\left[\mathrm{id}_{V}\right]_{B}^{B^{\prime}}\left[\operatorname{id}_{V}\right]_{B}^{B}\left[\operatorname{id}_{V}\right]_{B^{\prime}}^{B}\text{.}\) Notice that \(\left[\operatorname{id}_{V}\right]_{B}^{B}=\left[\operatorname{id}_{V}\right]_{B^{\prime}}^{B^{\prime}}=I\) is the identity matrix, so the previous formula says that

\begin{equation*} I=\left[\mathrm{id}_{V}\right]_{B}^{B^{\prime}} I\left[\mathrm{id}_{V}\right]_{B^{\prime}}^{B} \end{equation*}

Setting \(P=\left[\mathrm{id}_{V}\right]_{B}^{B^{\prime}}\text{,}\) we notice that the previous identity gives \(P^{-1}=\left[\mathrm{id}_{V}\right]_{B^{\prime}}^{B}\text{.}\)

Now set \(V=W, B=C, B^{\prime}=C^{\prime}\) and \(f=g\) to obtain

\begin{equation*} [g]_{B^{\prime}}^{B^{\prime}}=\left[\mathrm{id}_{V}\right]_{B}^{B^{\prime}}[g]_{B}^{B}\left[\mathrm{id}_{V}\right]_{B^{\prime}}^{B}=P[g]_{B}^{B} P^{-1}. \end{equation*}

Subsection Elementary Basis Operations

We now come to certain special changes of basis and their matrices:

Definition 13.72. Elementary Basis Change Operations.

Let \(R\) be a commutative ring with \(1 \neq 0\text{,}\) let \(M\) be a free \(R\)-module of finite rank \(n\text{,}\) and let \(B = \{b_1,\dots ,b_n\}\) be an ordered basis for \(M\text{.}\) An elementary basis change operation on the basis \(B\) is one of the following three types of operations:

(Type I) Replacing \(b_i\) by \(b_i + rb_j\) for some \(i \neq j\) and some \(r\in R\text{.}\)
(Type II) Replacing \(b_i\) by \(ub_i\) for some \(i\) and some unit \(u\) of \(R\text{,}\)
(Type III) Swapping the positions of \(b_i\) and \(b_j\) for some \(i \neq j\text{.}\)

Definition 13.73. Elementary Row Operations.

Let \(R\) be a commutative ring with \(1 \neq 0\text{.}\) An elementary row (column) operation on a matrix \(A \in \Mat_{m,n}(R)\) is one of the following three types of operations:

Type I.
Adding an element of \(R\) times a row (column) of \(A\) to a different row column of \(A\text{.}\)
Type II.
Multiplying a row (column) of \(A\) by a unit of \(R\text{.}\)
Type III.
Interchanging two rows (columns) of \(A\text{.}\)

Definition 13.74. Elementary Matrix.

Let \(R\) be a commutative ring with \(1 \neq 0\text{.}\) An elementary matrix over \(R\) is an \(n \times n\) matrix obtained from \(I_n\) by applying a single elementary column operation (or, equivalently, a single elementary column operation). In more detail:

Type I.
For \(r \in R\) and \(1 \leq i,j \leq n\) with \(i \neq j\text{,}\) let \(E_{i,j}(r)\) be the type I elementary matrix with \(1\)’s on the diagonal, \(r\) in the \((i,j)\) position, and \(0\) everywhere else.
Type II.
For \(u \in R^\times\) and \(1\leq i \leq n\) let \(E_i(u)\) be the type II elementary matrix with \((i,i)\) entry \(u\text{,}\) \((j,j)\) entry \(1\) for all \(j \neq i\text{,}\) and \(0\) everywhere else.
Type III.
For \(1 \leq i,j \leq n\) with \(i \neq j\text{,}\) let \(E_{(i,j)}\) be the type III elementary matrix with \(1\) in the \((i,j)\) and \((j,i)\) positions and in the \((l,l)\) positions for all \(l\not \in \{i,j\}\text{,}\) and 0 in all other entries.

Summary

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