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

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

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

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

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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