Linear Algebra and Its Applications, Exercise 2.4.15

Exercise 2.4.15. For each of the following matrices

A = \begin{bmatrix} 1&1&0 \\ 0&1&1 \end{bmatrix} \qquad M = \begin{bmatrix} 1&0 \\ 1&1 \\ 0&1\end{bmatrix} \qquad T = \begin{bmatrix} a&b \\ 0&a \end{bmatrix}

find a left inverse and right inverse if they exist.

Answer: We begin with the 2 by 3 echelon matrix A. Since A has two pivots its rank r = 2. Since A has two rows we also have r = m so that A has a (3 by 2) right inverse. It does not have a left inverse since r \ne n = 3.

We now attempt to find a right inverse C. We must have AC = I or

\begin{bmatrix} 1&1&0 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} c_{11}&c_{12} \\ c_{21}&c_{22} \\ c_{31}&c_{32} \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}

Multiplying the two rows of A by the first column of C we must have c_{11} + c_{21} = 1 and c_{21} + c_{31} = 0. One way to achieve this is to set c_{11} = 1 and c_{21} = c_{31} = 0.

Multiplying the two rows of A by the second column of C we must have c_{12} + c_{22} = 0 and c_{22} + c_{32} = 1. One way to achieve this is to set c_{12} = c_{22} = 0 and c_{32} = 1. We then have

C = \begin{bmatrix} 1&0 \\ 0&0 \\ 0&1 \end{bmatrix}

so that

AC = \begin{bmatrix} 1&1&0 \\ 0&1&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 0&0 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I

We now consider the matrix M. The first and third rows of M are linearly independent, and the second row of M is equal to the sum of the first and third rows. The rank of M is therefore r = 2. Since M has two columns we also have r = n so that M has a (3 by 2) right inverse. It does not have a right inverse since r \ne m = 3.

We now attempt to find a left inverse B of M such that BM = I. We could proceed as we did above, but there is a shortcut: Note that M = A^T so that we are looking for B such that BA^T = I. But BA^T = AB^T and we already have a matrix C such that AC = I. We can therefore choose B such that B^T = C or B = C^T:

B = \begin{bmatrix} 1&0&0 \\ 0&0&1 \end{bmatrix}

We then have

BM = \begin{bmatrix} 1&0&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 1&0 \\ 1&1 \\ 0&1\end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = I

so that B = C^T is a left inverse of M.

Finally we consider the 2 by 2 triangular matrix T. If a \ne 0 then T has two pivots and rank r = 2. In this case since r = m = n both a left inverse B and right inverse C exist and are the same: T is invertible with T^{-1} = B = C.

From the standard formula for the inverse of a 2 by 2 matrix (see page 42) we have

T^{-1} = 1/(a \cdot a - b \cdot 0) \begin{bmatrix} a&-b \\ -0&a \end{bmatrix}

= 1/a^2 \begin{bmatrix} a&-b \\ 0&a \end{bmatrix} = \begin{bmatrix} 1/a&-b/a^2 \\ 0&1/a \end{bmatrix}

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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