## Linear Algebra and Its Applications, Review Exercise 1.12

Review exercise 1.12. State whether the following are true or false. If a statement is true explain why it is true. If a statement is false provide a counter-example.

(a) If $A$ is invertible and $B$ has the same rows as $A$ but in reverse order, then $B$ is invertible as well.

(b) If $A$ and $B$ are both symmetric matrices then their product $AB$ is also a symmetric matrix.

(c) If $A$ and $B$ are both invertible then their product $BA$ is also invertible.

(d) If $A$ is a nonsingular matrix then it can be factored into the product $A = LU$ of a lower triangular and upper triangular matrix.

Answer: (a) True. If $B$ has the same rows as $A$ but in reverse order then we have $B = PA$ where $P$ is the permutation matrix that reverses the order of rows. For example, for the 3 by 3 case we have

$P = \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix}$

If we apply $P$ twice then it restores the order of the rows back to the original order; in other words $P^2 = I$ so that $P^{-1} = P$.

If $A$ is invertible then $A^{-1}$ exists. Consider the product $A^{-1}P$. We have

$B(A^{-1}P) = (PA)(A^{-1}P) = P(A^{-1}A)P = PIP = P^2 = I$

so that $A^{-1}P$ is a right inverse for $B$. We also have

$(A^{-1}P)B = (A^{-1}P)(PA) = A^{-1}P^2A = A^{-1}IA = A^{-1}A = I$

so that $A^{-1}P$ is a left inverse for $B$ as well. Since $A^{-1}P$ is both a left and right inverse for $B$ we have $B^{-1} = A^{-1}P$ so that $B$ is invertible if $A$ is.

Incidentally, note that while multiplying by $P$ on the left reverses the order of the rows, multiplying by $P$ on the right reverse the order of the columns. For example, in the 3 by 3 case we have

$\begin{bmatrix} 1&2&3 \\ 4&5&6 \\ 7&8&9 \end{bmatrix} \begin{bmatrix} 0&0&1 \\ 0&1&0 \\ 1&0&0 \end{bmatrix} = \begin{bmatrix} 3&2&1 \\ 6&5&4 \\ 9&8&7 \end{bmatrix}$

Thus if $A^{-1}$ exists and $B = PA$ then $B^{-1} = A^{-1}P$ exists and consists of $A^{-1}$ with its columns reversed.

(b) False. The product of two symmetric matrices is not necessarily itself a symmetric matrix, as shown by the following counterexample:

$\begin{bmatrix} 2&3 \\ 3&1 \end{bmatrix} \begin{bmatrix} 3&5 \\ 5&1 \end{bmatrix} = \begin{bmatrix} 21&13 \\ 14&16 \end{bmatrix}$

(c) True. Suppose that both $A$ and $B$ are invertible; then both $A^{-1}$ and $B^{-1}$ exist. Consider the product matrices $BA$ and $A^{-1}B^{-1}$. We have

$(BA)(A^{-1}B{-1}) = B(AA^{-1})B{-1} = BIB^{-1} = BB^{-1} = I$

and also

$(A^{-1}B{-1})(BA) = A^{-1}(B{-1}B)A = A^{-1}IA = A^{-1}A = I$

So $A^{-1}B{-1}$ is both a left and right inverse for $BA$ and thus $(BA)^{-1} = A^{-1}B{-1}$. If both $A$ and $B$ are invertible then their product $BA$ is also.

(d) False. A matrix $A$ cannot necessarily be factored into the form $A = LU$ because you may need to do row exchanges in order for elimination to succeed. Consider the following counterexample:

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

This matrix requires exchanging the first and second rows before elimination can commence. We can do this by multiplying by an appropriate permutation matrix:

$PA = \begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} \begin{bmatrix} 0&1&2 \\ 1&1&1 \\ 1&2&1 \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 1&2&1 \end{bmatrix}$

We then multiply the (new) first row by 1 and subtract it from the third row (i.e., the multiplier $l_{31} = 1$):

$\begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 1&2&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&1&0 \end{bmatrix}$

and then multiply the second row by 1 and subtract it from the third ($l_{32} = 1$):

$\begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&1&0 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&0&-2 \end{bmatrix}$

We then have

$L = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&1&1 \end{bmatrix} \quad U = \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&0&-2 \end{bmatrix}$

and

$LU = \begin{bmatrix} 1&0&0 \\ 0&1&0 \\ 1&1&1 \end{bmatrix} \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 0&0&-2 \end{bmatrix} = \begin{bmatrix} 1&1&1 \\ 0&1&2 \\ 1&2&1 \end{bmatrix} = PA \ne A$

So a matrix $A$ cannot always be factored into the form $A = LU$.

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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