DLA Theory

Section 39.5 Theory

In this section.

As much as possible, we will work relative to the standard inner products on \(\R^n\) and \(\C^n\) instead of just appealing to properties of matrices, to make the future study of adjoint linear operators on an abstract inner product space a generalization of the current theory.

Subsection 39.5.1 Properties of adjoints

Again, we will simultaneously treat the real and complex contexts.

Theorem 39.5.1. Uniqueness of adjoints.

For \(n \times n\) matrix \(A\) there exists one unique adjoint matrix \(\adjoint{A}\) so that

\begin{gather} \inprod{\uvec{u}}{A \uvec{v}} = \inprod{\adjoint{A} \uvec{u}}{\uvec{v}}\label{equation-matrix-adjoints-theory-adjoint-def}\tag{\(\star\)} \end{gather}

holds for every pair of \(n\)-dimensional column vectors \(\uvec{u},\uvec{v}\text{.}\)

Proof idea.

As in the proof of Theorem 36.6.10, an expression

\begin{equation*} \utrans{\uvec{e}}_i A \uvec{e}_j \text{,} \end{equation*}

picks off the \((i,j)\) entry of \(A\text{,}\) where \(\uvec{e}_i,\uvec{e}_j\) are standard basis vectors.

So we can take \(\uvec{u},\uvec{v}\) to be various combinations of standard basis vectors \(\uvec{e}_i,\uvec{e}_j\) in (\(\star\)), and comparing results on either side will tell us about the entries of \(A\) versus the possible entries for an adjoint matrix \(\adjoint{A}\text{.}\)

Proposition 39.5.2. Adjoint of an adjoint.

The adjoint of an adjoint is the original matrix. That is, \(\adjoint{(\adjoint{A})} = A\text{.}\)

Proof.

The validity of the formula \(\adjoint{(\adjoint{A})} = A\) should be obvious in both the real and complex cases, since both transpose and complex conjugate undo each other. Nevertheless, we will give a proof based on the inner product.

Applying Theorem 39.5.1 to \(\adjoint{A}\text{,}\) there is a unique matrix \(\adjoint{(\adjoint{A})}\) so that

\begin{equation*} \inprod{\uvec{u}}{\adjoint{A} \uvec{v}} = \inprod{\adjoint{(\adjoint{A})} \uvec{u}}{\uvec{v}} \end{equation*}

is true for all column vectors \(\uvec{u},\uvec{v}\text{.}\) But then

\begin{align*} \inprod{\adjoint{(\adjoint{A})} \uvec{u}}{\uvec{v}} \amp = \inprod{\uvec{u}}{\adjoint{A} \uvec{v}} \amp \amp\text{(i)}\\ \amp = \lcconj{\inprod{\adjoint{A} \uvec{v}}{\uvec{u}}} \amp \amp\text{(ii)}\\ \amp = \lcconj{\inprod{\uvec{v}}{A \uvec{u}}} \amp \amp\text{(iii)}\\ \amp = \inprod{A \uvec{u}}{\uvec{v}} \amp \amp\text{(iv)}\text{,} \end{align*}

with justifications

definition of the adjoint \(\adjoint{(\adjoint{A})}\text{;}\)
Axiom RIP 1 (real case) or Axiom CIP 1 (complex case), where in the real case the complex conjugate has no effect;
definition of the adjoint \(\adjoint{A}\text{;}\)
again Axiom RIP 1 (real case) or Axiom CIP 1 (complex case).

The equality

\begin{equation*} \inprod{\adjoint{(\adjoint{A})} \uvec{u}}{\uvec{v}} = \inprod{A \uvec{u}}{\uvec{v}} \end{equation*}

holds for all column vectors \(\uvec{u},\uvec{v}\text{,}\) so if we apply it with those vectors replaced by various combinations of standard basis vectors \(\uvec{e}_i,\uvec{e}_j\text{,}\) we will find that the matrices \(A\) and \(\adjoint{(\adjoint{A})}\) have all the same entries.

Proposition 39.5.3. Orthogonal complement is adjoint invariant.

If \(W\) is an \(A\)-invariant subspace of \(\R^n\) or \(\C^n\text{,}\) as appropriate, then \(\orthogcmp{W}\) is \(\adjoint{A}\)-invariant.

Proof idea.

It necessary to show that if \(\uvec{v}\) is a column vector so that

\begin{equation*} \uvecinprod{v}{w} = 0 \end{equation*}

for every column vector \(\uvec{w}\) in \(W\text{,}\) then also

\begin{equation*} \uvecinprod{\adjoint{A} v}{w} = 0 \end{equation*}

for every column vector \(\uvec{w}\) in \(W\text{.}\) This follows from the adjoint condition (\(\star\)) and the fact that \(W\) is \(A\)-invariant, and we leave the details to you, the reader.

Subsection 39.5.2 Properties of product-preserving matrices

As explored in Discovery 39.4 and Discovery 39.5, and discussed in Subsection 39.3.3, product-preserving matrices preserve geometric aspects of vectors. We leave the formal verification of these properties to you, the reader.

Proposition 39.5.4. Product-preserving preserves geometry.

Suppose \(A\) is a \(n \times n\) product-preserving matrix.

For every \(n\)-dimensional column vector \(\uvec{v}\text{,}\) we have \(\norm{A \uvec{v}} = \norm{\uvec{v}}\text{.}\) That is, multiplication by \(A\) preserves lengths.
For every pair of \(n\)-dimensional column vectors \(\uvec{v},\uvec{w}\text{,}\) we have \(\norm{A \uvec{v} - A \uvec{w}} = \norm{\uvec{v} - \uvec{w}}\text{.}\) That is, multiplication by \(A\) preserves distances.
If \(\{\uvec{v}_1, \uvec{v}_2, \dotsc, \uvec{v}_m\}\) is an orthogonal set of \(n\)-dimensional column vectors, then so is \(\{A \uvec{v}_1, A \uvec{v}_2, \dotsc, A \uvec{v}_m\}\text{.}\) That is, multiplication by \(A\) preserves orthogonality.
If \(\{\uvec{v}_1, \uvec{v}_2, \dotsc, \uvec{v}_m\}\) is an orthonormal set of \(n\)-dimensional column vectors, then so is \(\{A \uvec{v}_1, A \uvec{v}_2, \dotsc, A \uvec{v}_m\}\text{.}\) That is, multiplication by \(A\) preserves orthonormality.

The following lemma will help us relate the rows and columns of a product-preserving matrix.

Lemma 39.5.5.

If \(A\) is a product-preserving matrix, then so is \(\utrans{A}\text{.}\)

Proof.

For this proof, we will need to rely on the matrix characterization of product-preserving matrices:

\begin{equation*} \adjoint{A} A = I \text{.} \end{equation*}

In the real case, the above equality says that \(\inv{A} = \utrans{A}\text{,}\) so

\begin{equation*} \utrans{(\utrans{A})} \utrans{A} = A \utrans{A} = I \end{equation*}

as well.

In the complex case, we can turn

\begin{equation*} \adjoint{A} A = I \end{equation*}

into

\begin{equation*} \utrans{A} \cconj{A} = I \end{equation*}

by taking the transpose of both sides. So we have \(\inv{(\utrans{A})} = \cconj{A}\text{.}\) We can use this in our check that \(\utrans{A}\) is unitary, by

\begin{equation*} \adjoint{(\utrans{A})} \utrans{A} = \cconj{A} \utrans{A} = I \text{.} \end{equation*}

Now we characterize product-preserving matrices in different ways relative to the inner product.

Theorem 39.5.6. Characterizations of product-preserving matrices.

The following are equivalent for an \(n \times n\) product-preserving matrix \(A\text{.}\)

Matrix \(A\) is product-preserving.
Matrix \(A\) is invertible, and for every pair of \(n\)-dimensional column vectors \(\uvec{u},\uvec{v}\text{,}\) we have
\begin{equation*} \inprod{\uvec{u}}{A \uvec{v}} = \inprod{\inv{A} \uvec{u}}{\uvec{v}} \text{.} \end{equation*}
In other words, the adjoint of \(A\) is its inverse.
Matrix \(A\) is invertible, and for every pair of \(n\)-dimensional column vectors \(\uvec{u},\uvec{v}\text{,}\) we have
\begin{equation*} \inprod{A \uvec{u}}{\uvec{v}} = \inprod{\uvec{u}}{\inv{A} \uvec{v}} \text{.} \end{equation*}
The columns of \(A\) form an orthonormal basis of \(\R^n\) or \(\C^n\text{,}\) as appropriate.
The rows of \(A\) form an orthonormal basis of \(\R^n\) or \(\C^n\text{,}\) as appropriate.

Proof.

We will prove that each subsequent statement is equivalent to the first.

Statement 1 \(\implies\) Statement 2.

Suppose \(A\) is product-preserving. To verify that \(A\) is invertible, we may verify that the homogeneous system \(A\uvec{x} = \zerovec\) has no nontrivial solutions (Theorem 6.5.2). But if \(\uvec{x} \neq \zerovec \text{,}\) then using Statement 1 of Proposition 39.5.4 we have

\begin{equation*} \norm{A \uvec{x}} = \unorm{x} \neq 0 \text{,} \end{equation*}

so \(A \uvec{x} \neq 0\text{.}\)

For column vectors \(\uvec{u},\uvec{w}\text{,}\) we have

\begin{align*} \inprod{\uvec{u}}{A \uvec{v}} \amp = \inprod{I \uvec{u}}{A \uvec{v}} \\ \amp = \inprod{A \inv{A} \uvec{u}}{A \uvec{v}} \\ \amp = \inprod{\inv{A} \uvec{u}}{\uvec{v}} \text{,} \end{align*}

where in the last step of each calculation we have applied the product-preserving property of \(A\) to the pair \(\inv{A} \uvec{u}, \uvec{v}\) of column vectors.

Statement 2 \(\implies\) Statement 1.

Suppose \(A\) is invertible, and for every pair of column vectors \(\uvec{u},\uvec{v}\text{,}\) we have

\begin{equation*} \inprod{\uvec{u}}{A \uvec{v}} = \inprod{\inv{A} \uvec{u}}{\uvec{v}} \text{.} \end{equation*}

Applying this property to the pair \(A \uvec{u},\uvec{v}\text{,}\) we have

\begin{align*} \inprod{A \uvec{u}}{A \uvec{v}} \amp = \inprod{\inv{A} A \uvec{u}}{\uvec{v}} \\ \amp = \inprod{I \uvec{u}}{\uvec{v}} \\ \amp = \inprod{\uvec{u}}{\uvec{v}} \text{,} \end{align*}

which verifies that \(A\) is product preserving.

Statement 1 \(\iff\) Statement 3.

This can be verified similarly to the verification of the equivalence of Statement 1 and Statement 2 above.

For the next two arguments, write \(\uvec{a}_1,\uvec{a}_2,\dotsc,\uvec{a}_n\) for the columns of \(A\text{,}\) and recall that for each index \(j\) we have

\begin{gather} \uvec{a}_j = A \uvec{e}_j\text{,}\label{equation-matrix-adjoints-theory-cols-by-std-vecs}\tag{\(\star\star\)} \end{gather}

where \(\uvec{e}_j\) is the \(\nth[j]\) standard basis vector.

Statement 1 \(\implies\) Statement 4.

The standard basis is an orthonormal set, and so taking (\(\star\star\)) into consideration, applying Statement 4 of Proposition 39.5.4 allows us to conclude that the columns of \(A\) must also be an orthonormal set. They are therefore also linearly independent (Proposition 37.5.1), and since there are \(n\) of them they must be a basis (of \(\R^n\) or \(\C^n\text{,}\) as appropriate).

Statement 4 \(\implies\) Statement 1.

Suppose the columns of \(A\) are an orthonormal basis of \(\R^n\) or \(\C^n\text{,}\) as appropriate. We must verify that \(A\) is product-preserving. So consider arbitrary \(n\)-dimensional column vectors \(\uvec{u},\uvec{v}\) and their expansions

\begin{align*} \uvec{u} \amp = k_1 \uvec{e}_1 + k_2 \uvec{e}_2 + \dotsb + k_n \uvec{e}_n \text{,} \amp \uvec{v} \amp = m_1 \uvec{e}_1 + m_2 \uvec{e}_2 + \dotsb + m_n \uvec{e}_n \end{align*}

in terms of the standard basis vectors. Then

\begin{equation*} \uvecinprod{u}{v} = k_1 \lcconj{m}_1 + k_2 \lcconj{m}_2 + \dotsb + k_n \lcconj{m}_n \text{,} \end{equation*}

where the complex conjugate is irrelevant in the real case. Using (\(\star\star\)), we obtain expansions

\begin{align*} A \uvec{u} \amp = k_1 \uvec{a}_1 + k_2 \uvec{a}_2 + \dotsb + k_n \uvec{a}_n \text{,} \amp A \uvec{v} \amp = m_1 \uvec{a}_1 + m_2 \uvec{a}_2 + \dotsb + m_n \uvec{a}_n \end{align*}

in terms of the orthonormal basis formed by the columns of \(A\text{.}\) Applying Proposition 37.5.6 for this orthonormal basis, we obtain

\begin{equation*} \inprod{A \uvec{u}}{A \uvec{v}} = k_1 \lcconj{m}_1 + k_2 \lcconj{m}_2 + \dotsb + k_n \lcconj{m}_n \text{,} \end{equation*}

where again the complex conjugate is irrelevant in the real case. Comparing expressions for \(\uvecinprod{u}{v}\) and \(\inprod{A \uvec{u}}{A \uvec{v}}\) above, we see that \(A\) is product-preserving.

Statement 1 \(\iff\) Statement 5.

This equivalence now follows from our already proven equivalence of Statement 1 and Statement 4, combined with Lemma 39.5.5.

The possible values of the determinant of a product-preserving matrix are constrained by the condition \(\adjoint{A} A = I\text{.}\) As this was already discussed in Subsection 39.3.3, we will state the result without proof.

Proposition 39.5.7. Determinant values of product-preserving.

An \(n \times n\) product-preserving matrix \(A\) satisfies \(\abs{\det A} = 1\text{,}\) indicating normal absolute value in the real case and complex modulus in the complex case.

So for a real orthogonal matrix, we have \(\det A = \pm 1\text{,}\) and for a complex unitary matrix, the complex number \(\det A\) must lie on the unit circle in the complex plane.

Finally, we record our observation about transition matrices between orthonormal bases from Discovery 39.7 and Subsection 39.3.3.

Proposition 39.5.8. Product-preserving matrices are inner product space transition matrices.

Every product-preserving matrix is somehow a transition matrix between orthonormal bases of \(\R^n\) or \(\C^n\text{,}\) as appropriate.

Proof.

If \(P\) is a product-preserving matrix, then its columns form an orthonormal basis of \(\R^n\) (Statement 4 of Theorem 39.5.6). Let \(\basisfont{B}\) represent that basis. Then the columns of the transition matrix \(\ucobmtrx{B}{S}\) (where \(\basisfont{S}\) is the standard basis as usual) are just the vectors in \(\basisfont{B}\text{.}\) Furthermore, the standard basis is orthonormal as well, whether considered as the standard basis of \(\R^n\) or of \(\C^n\text{.}\) Therefore, \(P = \ucobmtrx{B}{S}\text{,}\) a transition matrix between orthonormal bases, as desired.