DLA Theory

Section 10.5 Theory

In this section.

We have discussed the reasoning behind many of the below facts in Section 10.3, so we will omit some of the formal proofs.

Subsection 10.5.1 Adjoints and inverses

First, we record the adjoint inversion formula we have discovered.

Theorem 10.5.1. Inversion by adjoint.

If \(\det A \neq 0\) then \(A\) is invertible, with \(\inv{A} = \frac{1}{\det A} \, \adj A\text{.}\)

Remark 10.5.2.

Based on our computations for the \(2\times 2\) case in Subsection 10.4.1, if \(A\) is \(2\times 2\) then the statement of the theorem above is exactly the same as Proposition 5.5.4.

Subsection 10.5.2 Determinants determine invertibility

As we saw in Subsection 10.3.2, there is a stronger connection between the determinant and invertibility, which we now state here more formally by adding a new statement to Theorem 6.5.2.

Theorem 10.5.3. Characterizations of invertibility.

For a square matrix \(A\text{,}\) the following are equivalent.

Matrix \(A\) is invertible.
Every linear system that has \(A\) as a coefficient matrix has one unique solution.
The homogeneous system \(A\uvec{x} = \zerovec\) has only the trivial solution.
There is some linear system that has \(A\) as a coefficient matrix and has one unique solution.
The rank of \(A\) is equal to the size of \(A\text{.}\)
The RREF of \(A\) is the identity.
Matrix \(A\) can be expressed as a product of some number of elementary matrices.
The determinant of \(A\) is nonzero.

In particular, a square matrix is invertible if and only if its determinant is nonzero.

Remark 10.5.4.

In the last sentence of the theorem, the connecting phrase “if and only if” between the two conditions is just a different way to say that the two conditions are equivalent. And recall that conditions are equivalent when they have to be either all true or all false at the same time. Rephrasing in terms of the “all false” scenario, we could also say that a square matrix is singular if and only if its determinant is zero.

Subsection 10.5.3 Determinant formulas

Here we collect the determinant formulas from Subsections 10.3.3–10.3.7. First we look at a special case, previously considered in Discovery 10.4 and Subsection 10.3.3, of the multiplicative formula for determinants.

Lemma 10.5.5. Determinant is multiplicative: elementary case.

If \(E\) is an elementary matrix and \(A\) is a square matrix of the same size, then

\begin{gather} \det (E A) = (\det E) (\det A)\text{.}\label{equation-more-det-theory-multiplicative-elem-version}\tag{\(\star\)} \end{gather}

Proof.

There are three cases to consider here, based on the type of elementary matrix we are dealing with.

Case \(E\) represents swapping rows.

The product \(E A\) represents the result of swapping two rows in \(A\text{,}\) so

\begin{equation*} \det (E A) = -\det A \end{equation*}

(Part 2 of Proposition 9.4.2).

But also \(\det E = -1\) (Part 1 of Proposition 9.4.5), so

\begin{equation*} (\det E) (\det A) = (-1) (\det A) = -\det A \end{equation*}

as well.

This establishes (\(\star\)) in this case.

Case \(E\) represents multiplying a row by \(k\).

The product \(E A\) represents the result of multiplying that row of \(A\) by \(k\text{,}\) so

\begin{equation*} \det (E A) = k \det A \end{equation*}

(Part 4 of Proposition 9.4.2).

But also \(\det E = k\) (Part 2 of Proposition 9.4.5), so

\begin{equation*} (\det E) (\det A) = k \det A \end{equation*}

as well.

This establishes (\(\star\)) in this case.

Case \(E\) represents adding a multiple of one row to another.

The product \(E A\) represents the result of adding a multiple of a row to another in \(A\text{,}\) so \(\det (E A)\) is equal to \(\det A\text{.}\) But also \(\det E = 1\) (Part 3 of Proposition 9.4.5), so

\begin{equation*} \det (E A) = \det A = (1) (\det A) = (\det E) (\det A), \end{equation*}

establishing (\(\star\)) in this case.

With the above lemma established, we can consider the general multiplicative formula for determinants.

Proposition 10.5.6. Determinant is multiplicative: general case.

A determinant of a product of square matrices is the product of the determinants of those matrices. In particular, the following hold.

If \(M\) and \(N\) are square matrices of the same size, then
\begin{equation*} \det (M N) = (\det M) (\det N). \end{equation*}
If \(M_1, M_2, \dotsc, M_{\ell-1}, M_\ell \) are square matrices of the same size, then
\begin{equation*} \det (M_1 M_2 \dotsm M_{\ell-1} M_\ell) = (\det M_1) (\det M_2) \dotsm (\det M_{\ell-1}) (\det M_\ell)\text{.} \end{equation*}

Proof outlines.

There are two cases to consider.

Case \(M\) is invertible.

In this case, \(M\) can be expressed as a product of elementary matrices (Theorem 10.5.3), and so Lemma 10.5.5 can be repeatedly applied to obtain the desired equality.

In Discovery 10.5 and Subsection 10.3.4, we worked under the assumption that \(M\) could be expressed as a product of three elementary matrices, but the calculations and logic used there would work no matter how many elementary matrices were required in a product expression for \(M\text{.}\)
Comment.

A more formal proof would require using the principal of mathematical induction on the number of elementary matrices required in a product expansion for \(M\text{,}\) but that is beyond the scope of this book.
Case \(M\) is singular.

In this case, \(\det M = 0\) by our newly added statement in the list of Theorem 10.5.3, so we have

\begin{equation*} \text{RHS} = (\det M) (\det N) = 0 \cdot \det N = 0 \end{equation*}

as well. But we also know that if \(M\) is singular, then the product \(M N\) must also be singular (Statement 1 of Proposition 6.5.8). So again we can apply the equivalence of Statement 1 and Statement 8 of Theorem 10.5.3 to obtain

\begin{equation*} \text{LHS} = \det (M N) = 0. \end{equation*}

Since both LHS and RHS are equal to \(0\text{,}\) they are equal to each other.
This result can be obtained by repeated applications of the formula in Statement 1, one \(M_i\) at a time.

Comment.

Again, a more formal proof would require mathematical induction on the number of matrices in the product.

Remark 10.5.7.

We can now understand the formula \(\det (k A) = k^n \det A\) as a special case of Proposition 10.5.6. Using \(M = k I\) and \(N = A\text{,}\) we have

\begin{equation*} \det (k A) = \det \bbrac{(k I) A} = \bbrac{\det (k I)} (\det A) = k^n \det A \text{.} \end{equation*}

(See Statement 2 of Proposition 8.5.2.)

Lemma 10.5.5 and the proof of Proposition 10.5.6 connect to Proposition 9.4.2 (which includes the formula \(\det (k A) = k^n \det A\) as one of its statements) by the fact that an \(n \times n\) scalar matrix \(k I\) is the product of \(n\) elementary matrices, one for each of the \(n\) operations multiply row \(R_j\) by \(k\).

Proposition 10.5.8. Determinant of an inverse.

The determinant of an inverse is the inverse of the determinant. That is, if \(N\) is an invertible matrix then \(\det (\inv{A}) = \inv{(\det A)}\text{.}\)

Subsection 10.5.4 Cramer's rule

Finally, we formally record Cramer's rule (discussed in Subsection 10.3.8).

Theorem 10.5.9. Cramer's rule.

If system \(A\uvec{x} = \uvec{b}\) has invertible square coefficient matrix \(A\text{,}\) then the value of variable \(x_j\) in the one unique solution to the system is

\begin{equation*} x_j = \frac{\det A_j}{\det A} \text{,} \end{equation*}

where \(A_j\) is the matrix obtained by replacing the \(\nth[j]\) column of \(A\) by \(\uvec{b}\text{.}\)