Elementary Linear Algebra For University of Lethbridge Math 1410

We ended the previous section by stating that invertible matrices are important. Since they are, in this section we study invertible matrices in two ways. First, we look at ways to tell whether or not a matrix is invertible, and second, we study properties of invertible matrices (that is, how they interact with other matrix operations).

In the last section we stated Theorem 4.4.7, in which we listed several properties of a matrix that equivalent to that matrix being invertible, including the fact that an \(n\times n\) matrix must have rank \(n\) in order to be invertible.

We begin this section by collecting additional properties that are equivalent to matrix a matrix being invertible. Some of these results were established in the previous section, but we state them again here for ease of reference.

Note that the theorem uses the phrase “the following statements are equivalent.” Recall that when two or more statements are equivalent, it means that the truth of any one of them implies that the rest are also true; if any one of the statements is false, then they are all false. So, for example, if we determined that the equation \(A\vx = \zero\) had exactly one solution (and \(A\) was an \(n\times n\) matrix) then we would know that \(A\) was invertible, that \(A\vx = \vb\) had only one solution, that the reduced row echelon form of \(A\) was \(I\text{,}\) etc.

So we came up with a list of statements that are all equivalent to the statement “\(A\) is invertible.” Again, if we know that if any one of them is true (or false), then they are all true (or all false).

Theorem 4.5.1 states formally that if \(A\) is invertible, then \(A\vx = \vb\) has exactly one solution, namely \(A^{-1}\vb\text{.}\) What if \(A\) is not invertible? What are the possibilities for solutions to \(A\vx = \vb\text{?}\)

We know that \(A\vx = \vb\) cannot have exactly one solution; if it did, then by our theorem it would be invertible. Recalling that linear equations have either one solution, infinitely many solutions, or no solution, we are left with the latter options when \(A\) is not invertible. This idea is important and so we’ll state it again as a Key Idea.

In Theorem 4.5.1 we’ve come up with a list of ways in which we can tell whether or not a matrix is invertible. At the same time, we have come up with a list of properties of invertible matrices — things we know that are true about them. (For instance, if we know that \(A\) is invertible, then we know that \(A\vx = \vb\) has only one solution.)

We now go on to discover other properties of invertible matrices. Specifically, we want to find out how invertibility interacts with other matrix operations. For instance, if we know that \(A\) and \(B\) are invertible, what is the inverse of \(A+B\text{?}\) What is the inverse of \(AB\text{?}\) What is “the inverse of the inverse?” We’ll explore these questions through an example.

Let’s summarize the results of this example. If \(A\) and \(B\) are both invertible matrices, then so is their product, \(AB\text{.}\) We demonstrated this with our example, and there is more to be said. Let’s suppose that \(A\) and \(B\) are \(n\times n\) matrices, but we don’t yet know if they are invertible. If \(AB\) is invertible, then each of \(A\) and \(B\) are; if \(AB\) is not invertible, then either \(A\) or \(B\) is not invertible.

In short, invertibility “works well” with matrix multiplication. However, we saw that it doesn’t work well with matrix addition. Knowing that \(A\) and \(B\) are invertible does not help us find the inverse of \((A+B)\text{;}\) in fact, the latter matrix may not even be invertible.

Let’s do one more example, then we’ll summarize the results of this section in a theorem.

The matrix \(A\) in the previous example is a diagonal matrix: the only nonzero entries of \(A\) lie on the diagonal. The relationship between \(A\) and \(A^{-1}\) in the above example seems pretty strong, and it holds true in general. We’ll state this and summarize the results of this section with the following theorem.

We end this section with a comment about solving systems of equations “in real life.” Solving a system \(A\vx = \vb\) by computing \(A^{-1}\vb\) seems pretty slick, so it would make sense that this is the way it is normally done. However, in practice, this is rarely done. There are two main reasons why this is the case.

First, computing \(A^{-1}\) and \(A^{-1}\vb\) is “expensive” in the sense that it takes up a lot of computing time. Certainly, our calculators have no trouble dealing with the \(3 \times 3\) cases we often consider in this textbook, but in real life the matrices being considered are very large (as in, hundreds of thousand rows and columns). Computing \(A^{-1}\) alone is rather impractical, and we waste a lot of time if we come to find out that \(A^{-1}\) does not exist. Even if we already know what \(A^{-1}\) is, computing \(A^{-1}\vb\) is computationally expensive — Gaussian elimination is faster.

Secondly, computing \(A^{-1}\) using the method we’ve described often gives rise to numerical roundoff errors. Even though computers often do computations with an accuracy to more than 8 decimal places, after thousands of computations, rounding off can cause big errors. (A “small” \(1,000 \times 1,000\) matrix has \(1,000,000\) entries! That’s a lot of places to have roundoff errors accumulate!) It is not unheard of to have a computer compute \(A^{-1}\) for a large matrix, and then immediately have it compute \(AA^{-1}\) and not get the identity matrix. (The result is usually very close, with the numbers on the diagonal close to 1 and the other entries near 0. But it isn’t exactly the identity matrix.)

Therefore, in real life, solutions to \(A\vx = \vb\) are usually found using the methods we learned in Section 3.6. It turns out that even with all of our advances in mathematics, it is hard to beat the basic method that Gauss introduced a long time ago.