DLA Theory

Section 22.5 Theory

In this section.

Subsection 22.5.1 Linearity properties of coordinate vectors

Here we finally record the linearity properties of the coordinate vector process.

Theorem 22.5.1.

Suppose \(\basisfont{B}\) is a basis of a finite-dimensional vector space \(V\text{.}\) Then we have the following patterns concerning coordinate vectors relative to \(\basisfont{B}\text{.}\)

The coordinate vector of a sum is the sum of the coordinate vectors. That is,
\begin{equation*} \rmatrixOf{\uvec{v} + \uvec{w}}{B} = \rmatrixOf{\uvec{v}}{B} + \rmatrixOf{\uvec{w}}{B} \end{equation*}
for each pair of vectors \(\uvec{v},\uvec{w}\) in \(V\text{.}\)
The coordinate vector of a scalar multiple is the scalar multiple of the coordinate vector. That is,
\begin{equation*} \rmatrixOf{k \uvec{v}}{B} = k \rmatrixOf{\uvec{v}}{B} \end{equation*}
for each scalar \(k\) and each vector \(\uvec{v}\) in \(V\text{.}\)
The coordinate vector of a linear combination is the corresponding linear combination of the coordinate vectors. That is,
\begin{equation*} \rmatrixOf{k_1 \uvec{v}_1 + k_2 \uvec{v}_2 + \dotsc + k_m \uvec{v}_m}{B} = k_1 \rmatrixOf{\uvec{v}_1}{B} + k_2 \rmatrixOf{\uvec{v}_2}{B} + \dotsc + k_m \rmatrixOf{\uvec{v}_m}{B} \end{equation*}
for each collection of scalars \(k_1,k_2,\dotsc,k_m\) and each collection of vectors \(\uvec{v}_1,\uvec{v}_2,\dotsc,\uvec{v}_m\) in \(V\text{.}\)

Proof.

Write \(\basisfont{B} = \{ \uvec{u}_1, \uvec{u}_2, \dotsc, \uvec{u}_n \}\text{,}\) and suppose \(\uvec{v}\) and \(\uvec{w}\) are vectors in \(V\text{.}\) Then each of \(\uvec{v}\) and \(\uvec{w}\) have a unique expression as a linear combination of the vectors in \(\basisfont{B}\) (Theorem 19.5.3). So we can write

\begin{align*} \uvec{v} \amp = a_1 \uvec{u}_1 + a_2 \uvec{u}_2 + \dotsb + a_n \uvec{u}_n \text{,} \\ \uvec{w} \amp = b_1 \uvec{u}_1 + b_2 \uvec{u}_2 + \dotsb + b_n \uvec{u}_n \text{,} \end{align*}

for some unique collections of scalars \(a_1,a_2,\dotsc,a_n\) and \(b_1,b_2,\dotsc,b_n\text{.}\) These linear combinations tell us the coordinate vectors of \(\uvec{v}\) and \(\uvec{w}\text{:}\)

\begin{align*} \rmatrixOf{\uvec{v}}{B} \amp = (a_1,a_2,\dotsc,a_n) \text{,} \\ \rmatrixOf{\uvec{w}}{B} \amp = (b_1,b_2,\dotsc,b_n) \text{.} \end{align*}

We add linear combinations by collecting like terms, so we have
\begin{equation*} \uvec{v} + \uvec{w} = (a_1 + b_1) \uvec{u}_1 + (a_2 + b_2) \uvec{u}_2 + \dotsb + (a_n + b_n) \uvec{u}_n \text{.} \end{equation*}
The coordinate vector of this sum is then
\begin{align*} \rmatrixOf{\uvec{v} + \uvec{w}}{B} \amp = (a_1 + b_1, a_2 + b_2, \dotsc, a_n + b_n) \\ \amp = (a_1, a_2, \dotsc, a_n) + (b_1, b_2, \dotsc, b_n) \\ \amp = \rmatrixOf{\uvec{v}}{B} + \rmatrixOf{\uvec{w}}{B} \text{,} \end{align*}
as desired.
We scalar multiply a linear combination by distributing the scalar across the vector additions, so for every scalar \(k\) we have
\begin{equation*} k \uvec{v} = k a_1 \uvec{u}_1 + k a_2 \uvec{u}_2 + \dotsb + k a_n \uvec{u}_n \text{.} \end{equation*}
The coordinate vector of this scalar multiple is then
\begin{align*} \rmatrixOf{k \uvec{v}}{B} \amp = (k a_1, k a_2, \dotsc, k a_n) \\ \amp = k (a_1, a_2, \dotsc, a_n) \\ \amp = k \rmatrixOf{\uvec{v}}{B} \text{,} \end{align*}
as desired.
This statement is just the combination of the first two.

Remark 22.5.2.

Note that in each of the statements of Theorem 22.5.1, there are different vector operations in use on either side of the equals sign — on the left the vector operation(s) within the coordinate-vector brackets are those used in the vector space \(V\text{,}\) while on the right the vector operation(s) used between the coordinate-vector brackets are those used in \(\R^n\text{.}\) For example, if \(V\) is a space of polynomials, then the plus sign in the notation \(\rmatrixOf{\uvec{v} + \uvec{w}}{B}\) means addition of polynomials (because that's what \(\uvec{v}\) and \(\uvec{w}\) are), while the plus sign in the expression \(\rmatrixOf{\uvec{v}}{B} + \rmatrixOf{\uvec{w}}{B}\) means addition of vectors in \(\R^n\) (because thats what \(\rmatrixOf{\uvec{v}}{B}\) and \(\rmatrixOf{\uvec{w}}{B}\) are).

Subsection 22.5.2 Properties of transition matrices

Here we record and prove those properties explored in Discovery 22.7 and further discussed in Subsection 22.3.4, as well as this initial uniqueness statement.

Proposition 22.5.3. Uniqueness of transition matrices.

For each pair of bases \(\basisfont{B},\basisfont{B}'\) of an \(n\)-dimensional vector space \(V\text{,}\) there exists one unique \(n \times n\) matrix \(P\) such that \(P \matrixOf{\uvec{v}}{B} = \matrixOf{\uvec{v}}{B'} \) holds for every vector \(\uvec{v}\) in \(V\text{.}\)

Proof.

We already know from our discussion in Subsection 22.3.3 that such a matrix exists, so we need to prove that only \(\ucobmtrx{B}{B'}\) is capable of converting between coordinate vectors relative to these two bases.

So suppose that \(P\) is an \(n \times n\) matrix such that

\begin{gather} P \matrixOf{\uvec{v}}{B} = \matrixOf{\uvec{v}}{B'}\label{equation-change-of-basis-theory-convert-pattern}\tag{\(\star\)} \end{gather}

holds for every vector \(\uvec{v}\) in \(V\text{.}\) In particular, it must then hold for every vector in \(\basisfont{B}\text{.}\) Write

\begin{equation*} \basisfont{B} = \{ \uvec{u}_1, \uvec{u}_2, \dotsc, \uvec{u}_n \} \text{.} \end{equation*}

Now, it easy to write one of the vectors in \(\basisfont{B}\) as a linear combination of these basis vectors. For example,

\begin{equation*} \uvec{u}_1 = 1 \uvec{u}_1 + 0 \uvec{u}_2 + \dotsb + 0 \uvec{u}_n \text{,} \end{equation*}

which says that \(\rmatrixOf{\uvec{u}_1}{B} = \uvec{e}_1\text{,}\) the first standard basis vector of \(\R^n\text{.}\) So if we take \(\uvec{v} = \uvec{u}_1\) in (\(\star\)) above, we get

\begin{equation*} P \uvec{e}_1 = \matrixOf{\uvec{u}_1}{B'} \text{.} \end{equation*}

Now, we have seen a couple of times (e.g. Discovery 21.1) that a matrix times a standard basis vector produces the corresponding column of the matrix. In the above, \(P\) times the first standard basis vector of \(\R^n\) produces the first column of \(P\text{.}\) From this we conclude that the first column of \(P\) must be \(\matrixOf{\uvec{u}_1}{B'} \text{,}\) which coincidentally is the same as the first column of \(\ucobmtrx{B}{B'}\text{.}\)

If we repeat this argument with \(\uvec{v} = \uvec{u}_2\) in (\(\star\)), we arrive at the conclusion that the second column of \(P\) must be \(\matrixOf{\uvec{u}_2}{B'} \text{,}\) which coincidentally is the same as the second column of \(\ucobmtrx{B}{B'}\text{.}\) And so on, running through all the vectors of \(\basisfont{B}\text{,}\) until we have concluded that every column of \(P\) is the same as the corresponding column in \(\cobmtrx{B}{B'}\text{,}\) at which point we can also conclude that \(P = \ucobmtrx{B}{B'} \text{.}\)

Proposition 22.5.4. Properties of transition matrices.

In the following, assume \(\basisfont{B}\text{,}\) \(\basisfont{B}'\text{,}\) and \(\basisfont{B}''\) are bases of a finite dimensional vector space \(V\text{.}\)

The transition \(\basisfont{B} \to \basisfont{B}\) is trivial. That is, \(\ucobmtrx{B}{B} = I \text{,}\) the identity matrix.
A chain of transitions \(\basisfont{B} \to \basisfont{B}'\text{,}\) \(\basisfont{B}' \to \basisfont{B}''\) combine to a single transition \(\basisfont{B} \to \basisfont{B}''\text{.}\) That is, \(\ucobmtrx{B'}{B''} \ucobmtrx{B}{B'} = \ucobmtrx{B}{B''} \text{.}\)
The transition \(\basisfont{B}' \to \basisfont{B}\) is the reverse of the transition \(\basisfont{B} \to \basisfont{B}'\text{.}\) That is, \(\ucobmtrx{B'}{B} = \uinvcobmtrx{B}{B'} \text{.}\)

Proof.

Proposition 22.5.3 says that \(\ucobmtrx{B}{B} \) is the one unique matrix so that \(\ucobmtrx{B}{B} \matrixOf{\uvec{v}}{B} = \matrixOf{\uvec{v}}{B} \) holds for every vector \(\uvec{v}\) in \(V\text{.}\) But clearly also \(I \matrixOf{\uvec{v}}{B} = \matrixOf{\uvec{v}}{B} \) holds for every vector \(\uvec{v}\) in \(V\text{,}\) so we must have \(I = \ucobmtrx{B}{B}\text{,}\) as desired.
Proposition 22.5.3 says that \(\ucobmtrx{B}{B''} \) is the one unique matrix so that \(\ucobmtrx{B}{B''} \matrixOf{\uvec{v}}{B} = \matrixOf{\uvec{v}}{B''} \) holds for every vector \(\uvec{v}\) in \(V\text{.}\)

Set \(P = \ucobmtrx{B'}{B''} \ucobmtrx{B}{B'} \text{.}\) Then for every vector \(\uvec{v}\) in \(V\) we have

\begin{align*} P \matrixOf{\uvec{v}}{B} \amp = (\ucobmtrx{B'}{B''} \ucobmtrx{B}{B'}) \matrixOf{\uvec{v}}{B} \\ \amp = \ucobmtrx{B'}{B''} (\ucobmtrx{B}{B'} \matrixOf{\uvec{v}}{B}) \\ \amp = \ucobmtrx{B'}{B''} \matrixOf{\uvec{v}}{B'} \\ \amp = \matrixOf{\uvec{v}}{B''} \text{.} \end{align*}

So \(P\) also satisfies \(P \matrixOf{\uvec{v}}{B} = \matrixOf{\uvec{v}}{B''} \) for every vector \(\uvec{v}\) in \(V\text{.}\) The uniqueness of \(\ucobmtrx{B}{B''} \) then tells us that \(P = \ucobmtrx{B}{B''}\text{,}\) or \(\ucobmtrx{B'}{B''} \ucobmtrx{B}{B'} = \ucobmtrx{B}{B''} \) as desired.
To prove that a matrix is the inverse of another, we only need to show that they multiply to the identity in one order (Proposition 6.5.4 and Proposition 6.5.6). And the first two statements of this proposition tell us that
\begin{equation*} \ucobmtrx{B'}{B} \ucobmtrx{B}{B'} = \ucobmtrx{B}{B} = I \text{,} \end{equation*}
so that we indeed have \(\ucobmtrx{B'}{B} = \uinvcobmtrx{B}{B'} \text{,}\) as desired.

Corollary 22.5.5. Invertibility of transition matrices.

Every transition matrix is invertible.

Proof.

This fact follows immediately from Statement 3 of Proposition 22.5.4.

Subsection 22.5.3 Change of basis in \(\R^n\)

Finally, we record an observation from Subsection 22.3.5.

Proposition 22.5.6. Invertible matrices are transition matrices.

Every invertible matrix is somehow a transition matrix between bases of \(\R^n\text{.}\)

Proof.

If \(P\) is an invertible matrix, then its columns form a basis of \(\R^n\) (Statement 11 of Theorem 21.5.5). Let \(\basisfont{B}\) represent that basis. Then the columns of the transition matrix \(\ucobmtrx{B}{S}\) (where \(\basisfont{S}\) is the standard basis of \(\R^n\) as usual) are just the vectors in \(\basisfont{B}\text{.}\) Therefore, \(\ucobmtrx{B}{S} = P\text{,}\) as desired.