Mathematical Immaturity

2.21 Miscellaneous review exercises on matrices

2. $\quad$ For each of the following statements about $n \times n$ matrices, give a proof or exhibit a counter example.
(a) $\quad$ If $AB + BA = O,$ then $A^2B^3 = B^3A^2.$
(b) $\quad$ If $A$ and $B$ are nonsingular, then $A + B$ is nonsingular.
(c) $\quad$ If $A$ and $B$ are nonsingular, then $AB$ is nonsingular.
(d) $\quad$ If $A,$ $B,$ and $A + B$ are nonsingular, then $A - B$ is nonsingular.
(e) $\quad$ If $A^3 = O,$ then $A - I$ is nonsingular.
(f) $\quad$ If the product of $k$ matrices $A_1 \cdots A_k$ is nonsingular, then each matrix $A_i$ is nonsingular.

Solution.$\quad$
(a) $\quad$ If $AB + BA = O,$ then $A^2B^3 = B^3A^2.$

$\quad$ Proof. $\quad$ If $AB + BA = O,$ then by the distributive laws of matrix multiplication, we have \begin{align*} A^2B + ABA &= A(AB + BA) \\ &= A(O) = O \\ \\ A^2B^3 + ABAB^2 &= (A^2B + ABA)B^2 \\ &= (O)B^2 = O \end{align*} From this, we can see that $A^2B^3 = -ABAB^2.$ But if $AB + BA = O,$ then $BA = -AB$ and $AB = -BA.$ Then, by the associative law of matrix multiplication and the definition of scalar multiplication for matrices, we have \begin{align*} A^2B^3 &= -ABAB^2 \\ &= (-AB)AB^2 \\ &= (BA)AB^2 \\ &= BA(AB)B \\ &= BA(-BA)B \\ &= B(-AB)AB \\ &= B^2A(-BA) \\ &= B^2(-AB)A \\ &= B^2(BA)A \\ &= B^3A^2. \quad \blacksquare \end{align*}

(b) $\quad$ If $A$ and $B$ are nonsingular, then $A + B$ is nonsingular.

$\quad$ Counterexample. $\quad$ We first note that $O$ is singular. Then, set $A = I$ and $B = -I.$ We can easily verify that $A$ and $B$ are nonsingular, but $A + B = O. \quad \blacksquare$

(c) $\quad$ If $A$ and $B$ are nonsingular, then $AB$ is nonsingular.

$\quad$ Proof. $\quad$ (This proof uses a similar argument to that used in Section 8, Exercise 23.)
Denote $A^{-1}$ and $B^{-1}$ as the inverses of $A$ and $B,$ respectively. Then, we can use the associative law to show that $B^{-1}A^{-1} = (AB)^{-1}:$ \begin{align*} (B^{-1}A^{-1})AB &= B^{-1}(A^{-1}A)B \\ &= B^{-1}(I)B \\ &= B^{-1}B \\ &= I. \quad \blacksquare \end{align*}

(d) $\quad$ If $A,$ $B,$ and $A + B$ are nonsingular, then $A - B$ is nonsingular.

$\quad$ Counterexample. $\quad$ Let $A = I$ and let $B$ be any diagonal matrix whose first diagonal element is $1,$ and the remaining $n - 1$ are arbitrary scalars $\neq -1.$ We can easily verify that $A,$ $B,$ and $A + B$ are nonsingular. But because $a_{11} = b_{11},$ the first diagonal element of $A - B$ is zero. And since $A$ and $B$ are diagonal, it follows that the first row and column of $A - B$ are zero. And as we proved in Exercise 1, any matrix with a row or column of zeros is singular. $\quad \blacksquare$

(e) $\quad$ If $A^3 = O,$ then $A - I$ is nonsingular.

$\quad$ Proof. $\quad$ If $A^3 = O,$ then $AA^2 = A^2A = O,$ giving us two cases.
(1) $\quad$ $A = O,$ in which case $A - I = -I,$ which is nonsingular.
(2) $\quad$ $A \neq O$ but $A^2 = O.$ Then, we have $A^2 = (A+I)(A-I) + I = O, $ or \begin{align*} -(A + I)(A - I) &= I \end{align*} which means that $A - I$ is nonsingular. $\quad \blacksquare$

(f) $\quad$ If the product of $k$ matrices $A_1 \cdots A_k$ is nonsingular, then each matrix $A_i$ is nonsingular.

$\quad$ Proof. $\quad$ Let $(A_1 \cdots A_k)^{-1}$ be the inverse of $A_1 \cdots A_k.$ Then, we have \begin{align*} (A_1 \cdots A_k)^{-1}A_1\cdots A_k &= I \end{align*} But, by the associative law of matrix multiplication, we have \begin{align*} [(A_1 \cdots A_k)^{-1}A_1\cdots A_{k-1}]A_k &= I \end{align*} which means there exists a matrix $B$ such that $BA_k = I.$ Thus, $A_k$ is nonsingular. But since $A_k$ is nonsingular, its left inverse is also a right inverse, giving us \begin{align*} A_k[(A_1 \cdots A_k)^{-1}A_1\cdots A_{k-1}] &= I \end{align*} Applying the associative law once more, we find that $A_{k-1}$ is nonsingular \begin{align*} [A_k(A_1 \cdots A_k)^{-1}A_1\cdots A_{k-2}]A_{k-1} &= I. \end{align*} Continuing this process for the remaining $k - 2$ matrices in the product, we find that if the matrix product $A_1 \cdots A_k$ is nonsingular, then each matrix $A_i$ is nonsingular. $\quad \blacksquare$