Linear Algebra and Its Applications, Exercise 1.6.11

Exercise 1.6.11. For each of the following criteria, give examples of matrices A and B that satisfy the criteria:

(i) Both A and B are invertible, but their sum A+B is not invertible.

(ii) Neither A nor B is invertible, but their sum A+B is invertible.

(iii) A, B, and their sum A+B are all invertible.

For (iii), in addition show that B^{-1} + A^{-1} is also invertible, and find an expression for its inverse. Hint: use the fact that

A^{-1}(A + B)B^{-1} = B^{-1} + A^{-1}

Answer: (i) I is the simplest example of an invertible matrix, being its own inverse. If I is flipped horizontally to form a matrix like the first example in exercise 1.6.10, that matrix is also invertible, and like I is its own inverse. So we can choose

A = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = A^{-1}, B = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} = B^{-1}


A+B = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}

which is not invertible.

(ii) If we modify the matrix A in (i) by removing the top left entry of 1 then we obtain a matrix that is not invertible, ditto if we remove the bottom right entry. We can therefore choose A and B as follows:

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

These are not invertible, but their sum is I which is invertible:

A + B = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} = (A + B)^{-1}

(iii) A 2 by 2 matrix is invertible as long as (ad – bc) is nonzero. We can therefore again use I as our first matrix, and modify it slightly to obtain a second invertible matrix:

A = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}, B = \begin{bmatrix} 1&0 \\ 0&2 \end{bmatrix}

We then have

A +B = \begin{bmatrix} 1&0 \\ 0&3 \end{bmatrix}

which is also invertible. A is of course its own inverse, and we also have

B^{-1} = \begin{bmatrix} 1&0 \\ 0&\frac{1}{2} \end{bmatrix}, (A + B)^{-1} = \begin{bmatrix} 1&0 \\ 0&\frac{1}{3} \end{bmatrix}

Continuing, let’s first prove the formula that was given as a hint:

A^{-1}(A + B)B^{-1} = (A^{-1}A + A^{-1}B)B^{-1} = (I + A^{-1}B)B^{-1}

= B^{-1} + A^{-1}BB^{-1} = B^{-1} + A^{-1}

If A, B, and A+B are all invertible then we have the following:

(B^{-1}+A^{-1}) B(A+B)^{-1}A = A^{-1}(A + B)B^{-1}B(A+B)^{-1}A

= A^{-1}(A + B)(A+B)^{-1}A = A^{-1}A = I


B(A+B)^{-1}A(B^{-1}+A^{-1}) = B(A+B)^{-1}AA^{-1}(A + B)B^{-1}

= B(A+B)^{-1}(A + B)B^{-1} = BB^{-1} = I

so that

(B^{-1}+A^{-1})^{-1} = B(A+B)^{-1}A

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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