Picking up the thread of Remark 2.2.10, we observe that the cancellation property enjoyed in real number algebra is a consequence of the fact that every nonzero real number \(a\ne 0\) has a multiplicative inverse, denoted \(a^{-1}\) or \(1/a\text{,}\) that satisfies \(aa^{-1}=1\text{.}\) Indeed, “canceling” the \(a\) in the equation \(ab=ac\) (assuming \(a\ne 0\)) is really the act of multiplying both sides of this equation by the multiplicative inverse \(a^{-1}\text{.}\)
Ever on the lookout for connections between real number and matrix algebra, we ask whether there is a sensible analogue of multiplicative inverses for matrices. We have seen already that identity matrices \(I_n\) play the role of multiplicative identities for \(n\times n\) matrices, just as the number \(1\) does for real numbers. This suggests we should restrict our attention to \(n\times n\) matrices. The following definition is then the desired analogue of the multiplicative inverse of a nonzero real number.
The phrase “an inverse” in Definition 2.3.1 is somewhat jarring. Shouldn’t we speak of the inverse of a matrix? Not surprisingly, if a matrix is invertible, then it has one and only one inverse. As intuitive as this fact may seem, however, it still requires proof.
Without any additional theory at our disposal, to show a matrix \(A\) is invertible we must exhibit an inverse. The onus is on us to find a matrix \(B\) satisfying both \(AB=I\) and \(BA=I\text{.}\) (Remember: since we cannot assume \(BA=AB\text{,}\) we really need to show both equations hold.)
By the same token, to show \(A\) is not invertible we must show that an inverse does not exist: that is, we must prove that there is no \(B\) satisfying \(AB=BA=I\text{.}\) The next example illustrates this technique for a variety of matrices.
The inverse of the matrix \(\begin{amatrix}[rr]2\amp 1\\ 3\amp 2
\end{amatrix}\) is \(\begin{amatrix}
2\amp -1\\ -3\amp 2
\end{amatrix}
\text{.}\) Indeed, we have
The matrix \(A= \begin{amatrix}[rrr]
1\amp 1 \amp 1\\
1\amp 1 \amp 1\\
1\amp 1\amp 1
\end{amatrix}\) is not invertible. Indeed, using the row method of matrix multiplication, we see that given any matrix \(B\text{,}\) each row of \(AB\) is given by
As the preceding example illustrates, deciding whether a matrix is invertible is not so straightforward, especially if the matrix is large. For the \(2\times 2\) case, however, we have a relatively simple test for invertibility. (We will generalize this to the \(n\times n\) case in Section 2.5.)
Assume \(ad-bc=0\text{.}\) If \(A=\boldzero\text{,}\) then \(A\) is not invertible, as we saw in the example above. Thus we can assume \(A\) is nonzero, in which case \(B=\begin{amatrix}[rr]d\amp -b\\ -c\amp a
\end{amatrix}\) is also nonzero. An easy computation shows
Sage has a number of useful tools related to invertibility. The boolean function is_invertible() tests for invertibility, and the method inverse() computes the inverse of an invertible matrix. Below we generate a random matrix with rational coefficients, test whether it is invertible, and compute its inverse if it is invertible. The density=0.5 ensures that roughly half of the matrix entries are zero; and this in turn increases the likelihood that the matrix is singular, for reasons that will become somewhat clearer later.
When the matrix is invertible, verify that \(AA^{-1}=A^{-1}A=I\text{.}\) If you like, use the blank Sage cell to compute \(AA^{-1}\) and \(A^{-1}A\text{.}\)
Try increasing the density setting in random_element() (e.g., density=0.75, density=.875) and see if the matrix is more or less likely to be invertible.
The next theorem tells us that invertibility is preserved by matrix multiplication: that is, if \(A\) and \(B\) are invertible \(n\times n\) matrices, then so is \(C=AB\text{.}\)
Assume \(A\) and \(B\) are invertible. The statement of the theorem proposes a candidate for the inverse of \(AB\text{:}\) namely, \(C=B^{-1}A^{-1}\text{.}\) We need only show that \(C\) satisfies \(C(AB)=(AB)C=I\text{.}\) Here goes:
More generally, if \(A_1,A_2,\dots A_r\) are invertible \(n\times n\) matrices, then their product \(A_1A_2\cdots A_r\) is invertible. Furthermore, we have in this case
We prove by induction on the number \(r\) of matrices, \(r\geq 1\text{,}\) that if the \(A_i\) are invertible, then the proposed inverse formula is valid.
Base step: \(r=1\).
For \(r=1\text{,}\) the inverse formula reads \(A_1^{-1}=A_1^{-1}\text{,}\) which is clearly true.
For the induction step we assume that the inverse formula is valid for any collection of \(r-1\) invertible matrices, and then show it is valid for any collection of \(r\) invertible matrices. Let \(A_1,A_2,\dots, A_r\) be invertible \(n\times n\) matrices. Define \(A=A_1A_2\cdots A_{r-1}\text{.}\) Then
Whenever confronted with a logical implication of the form \(\mathcal{P}\implies\mathcal{Q}\text{,}\) where \(\mathcal{P}\) and \(\mathcal{Q}\) denote arbitrary propositions, you should always ask whether the implication “goes the other way”. In other words, does the converse implication \(\mathcal{Q}\implies \mathcal{P} \) also hold?
Where is the flaw in our logic here? The second statement only allows us to conclude that there is some mystery matrix \(C\) satisfying \((AB)C=C(AB)=I\text{.}\) We cannot yet say that \(C=B^{-1}A^{-1}\text{,}\) as this formula from Theorem 2.3.7 only applies when we already know that \(A\) and \(B\) are both invertible. But this is exactly what we are trying to prove! As such we are guilty here of “begging the question”, or petitio principii in Latin.
We end this section by exploring how the matrix inverse operation fits into our matrix algebra. First, we can now use the inverse operation to define matrix powers of the form \(A^r\text{,}\) where \(A\) is a square matrix and \(r\) is an arbitrary integer.
Let \(f(x)=\anpoly\) be a polynomial with real coefficients. For any square matrix \(A\) of size \(n\times n\text{,}\) we define the matrix \(f(A)\) as
It is both easy and perilous to forget the identity matrix in the term \(a_0I_n\) appearing in (2.11). Take caution not to make this mistake; without an identity matrix of appropriate size, the expression \(f(A)\) simply does not make sense.
We took care to heed the warning in Remark 2.3.12, making sure to include \(I_3\) for \(f(A)\) (identity_matrix(3)) and \(I_2\) for \(f(B)\) (identity_matrix(2)). Interestingly, Sage is smart enough to figure out what we mean even if we are sloppy in this regard.
The following properties hold for all matrices \(A, B\text{,}\) all scalars \(c\in \R\text{,}\) and all integers \(r,s\in\Z\) for which the given expression makes sense.
The proofs of the first three statements are elementary, and closely resemble proofs of similar results in real number algebra. We leave these as an (unassigned) exercise.
For the fourth statement to make sense, we must assume that \(A\) is invertible. The claim here is that \(A^{-1}\) is invertible, and that its inverse is \(A\) itself. To prove this we need only show \(A^{-1}A=AA^{-1}=I\text{,}\) which follows from the definition of the inverse.
Suppose \(A\) is invertible with inverse \(A^{-1}\text{.}\) To see that \(A^T\) is invertible, with inverse as specified in (2.12), we need only show that
For the other direction, assume \(A^T\) is invertible. Setting \(B=A^T\text{,}\) we see that \(A=(A^T)^T=B^T\text{.}\) By the first implication, we know that if \(B\) is invertible, then so is \(B^T=A\text{.}\)
SOLUTION: The matrix \(\displaystyle A = \begin{bmatrix} 1 \amp 1 \\ c \amp c^2 \end{bmatrix}\) is invertible provided the columns of \(A\) are linearly independent, which will be the case if \(c \neq c^2\text{.}\) Thus, we require that \(c\neq 0\) and \(c \neq 1\text{.}\)
Each \(A\) below is invertible. Find \(A^{-1}\) by guess and check. You may want to use the row or column method of matrix multiplication to justify your answer.
Let \(A=\begin{bmatrix}
1\amp 1\\ 0\amp 1
\end{bmatrix}\text{.}\) Find a formula for \(A^r\text{,}\) where \(r\geq 1\) is an integer. Justify your answer using a proof by induction.
Let \(A=[1]_{n\times n}\text{,}\) the \(n\times n\) matrix consisting of all ones. Find a formula for \(A^r\text{,}\) where \(r\geq 1\) is an integer. Justify your answer using a proof by induction.
Let \(p(x)=x^2-5x+c\text{,}\) where \(c\in \R\) is some fixed scalar. Suppose \(A\) is an \(n\times n\) matrix satisfying \(p(A)=\boldzero_{n\times n}\text{.}\)
Note that each step (base and induction) of your induction on \(r\) will require an argument that uses induction on \(s\text{!}\) This is sometimes called double induction. For example, in the base step \(r=1\) you must show that
Let \(p(x)=\anpoly\) and \(q(x)=\bmpoly\) be polynomials with real coefficients. For any square matrix \(A\text{,}\) show that the matrices \(p(A)\) and \(q(A)\) commute: i.e.,
