Course Notes for Linear Algebra

I’ve now defined linear transformations. It is possible that a linear transformation is invertible, much like a function of numbers. Let me give the definition.

Since I’ve encoded transformations as matrices, I can use matrices to understand the inverses of linear transformations.

Before returning to inverse transformations, I’m going to talk about the algebraic idea of inversion and reciprocals. In number systems, when there is an identity, there usually is a way of getting back to that identity. Zero is the identity for addition. For any number \(a\text{,}\) there exists a number \((-a)\) so that \(a + (-a) = 0\) gets back to the identity (this obviously works in numbers systems that allow negatives, such as \(\ZZ\)). One is the identity in multiplication. For any non-zero number \(a\text{,}\) there is the number \(\frac{1}{a}\) so that \(a \frac{1}{a} = 1\) gets back to the identity (this obviously works in numbers systems that allow reciprocals). For matrices, I have already defined the identity matrix, which is the identity for matrix multiplication. If \(A\) is a \(k \times l\) matrix, then \(AI_l = I_k A = A\text{.}\) For matrices, multiplying by the identity doesn’t change the matrix.

For matrices, I have the same question as I had for numbers systems. For any matrix \(M\text{,}\) is there another matrix \(N\) such that \(MN = I\text{?}\) Is there a way to get back to the identity? Multiplication of numbers already showed that no such number need exist, since I cannot divide by zero. For matrices, I have to be even more cautious. For now, here is the formal definition.

I should note a couple of thing about this definition. First, it only applies to square matrices. I don’t even try to invert non-square matrices. Second, both multiplications are required: \(MM^{-1}\) and \(M^{-1}M\text{.}\) Both are required because matrix multiplication is non-commutative. In general, these two orders could result in different products. In this definition, I insist that both orders produce the identity matrix.

The definition is good, but I am left with the problem of determining which matrices have inverses and calculating those inverses. It turns out there is a convenient algorithm using techniques I have already presented. I will write an extended matrix \((M|I_n)\) with the original matrix and the identity matrix next to each other, separated by a vertical line. Then I row reduce, treating the conglomeration as one matrix with long rows. If the row reduction of \(M\) results in the identity on the left, then the matrix has the form \((I|M^{-1})\text{;}\) the right-hand side will be the inverse.

This algorithm leads to two observations. First, obviously, it is a way to calculate inverses. But it also is a condition: an \(n \times n\) matrix is invertible, in this algorithm, only if it row reduces to the identity. This is part of a general result, which is the first of several conditions for invertible matrices.