Corrected exercises: matrix algebra, transpose and the matrix product (MATH 133)

The test is always the same: write the two sizes side by side and look at the two numbers in the MIDDLE. $AB$ is $(2 \times 3)(3 \times 2)$, inner $3 = 3$, so it exists and is $2 \times 2$. $BA$ is $(3 \times 2)(2 \times 3)$, inner $2 = 2$, so it exists and is $3 \times 3$. $AC$ is $(2 \times 3)(3 \times 3)$, which is $2 \times 3$. $CA$ is $(3 \times 3)(2 \times 3)$, inner $3 \neq 2$: it does not exist, even though $AC$ does. $BC$ is $(3 \times 2)(3 \times 3)$, inner $2 \neq 3$: undefined. $A + B^T$ is a sum, and a sum needs IDENTICAL sizes, not matching inner ones: $B^T$ is $2 \times 3$ like $A$, so the sum exists and is $2 \times 3$, while $A + B$ would not. $u^T C$ is $(1 \times 3)(3 \times 3) = 1 \times 3$, and $uC$ is $(3 \times 1)(3 \times 3)$, inner $1 \neq 3$, undefined. Writing out the two sizes costs three seconds and it is the only line that catches a product that does not exist before you have spent a page computing it.

Row $i$ of $A$ against column $j$ of $B$. Row 1 of $A$ is $(1, 2, 0)$: against column 1 of $B$, $(2, 0, 1)$, it gives $2 + 0 + 0 = 2$; against column 2, $(1, -1, 4)$, it gives $1 - 2 + 0 = -1$. Row 2, $(-1, 3, 1)$, gives $-2 + 0 + 1 = -1$ and $-1 - 3 + 4 = 0$. So $AB = \begin{pmatrix} 2 & -1 \\ -1 & 0 \end{pmatrix}$. For $BA$, row 1 of $B$ is $(2, 1)$ and the columns of $A$ are $(1, -1)$, $(2, 3)$, $(0, 1)$, giving $1$, $7$, $1$; row 2, $(0, -1)$, gives $1$, $-3$, $-1$; row 3, $(1, 4)$, gives $-3$, $14$, $4$. So $BA = \begin{pmatrix} 1 & 7 & 1 \\ 1 & -3 & -1 \\ -3 & 14 & 4 \end{pmatrix}$. The first thing to notice needs no entry at all: $AB$ is $2 \times 2$ and $BA$ is $3 \times 3$. Non-commutativity starts here, in its crudest form. For non-square factors the two orders are not even comparable.

$u^T u$ is $(1 \times 3)(3 \times 1)$, a $1 \times 1$ matrix: $1 + 4 + 1 = 6$. $u u^T$ is $(3 \times 1)(1 \times 3)$, a $3 \times 3$ matrix whose $(i,j)$ entry is $u_i u_j$, so $u u^T = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ -1 & -2 & 1 \end{pmatrix}$. The same two factors in the two orders give a number in one case and nine numbers in the other. A student who writes $u u^T = 6$ has computed the wrong product, and nothing in the arithmetic warns them: only the sizes do. Notice also that $u u^T$ is symmetric and that its diagonal adds up to $6$, a first sighting of the trace identity of exercise 7.

If $A$ is $m \times n$, then $AX$ needs $X$ to have $n$ rows and $XA$ needs $X$ to have $m$ columns, so $X$ is $n \times m$. Then $AX$ is $m \times m$ and $XA$ is $n \times n$. The equation $AX = XA$ compares two matrices, so it only makes sense when they have the same size, that is $m = n$: commuting is a question about SQUARE matrices, and exercise 6 only asks it there.

$BAB$ is $(3 \times 2)(2 \times 3)(3 \times 2)$, a $3 \times 2$ matrix, and associativity lets us bracket it either way. $(BA)B$ first builds the $3 \times 3$ matrix $BA$, nine entries; $B(AB)$ first builds the $2 \times 2$ matrix $AB$, four entries, already found in b). Take the cheaper one. The $(3,1)$ entry of $B(AB)$ is row 3 of $B$ times column 1 of $AB$: $(1, 4) \cdot (2, -1) = 2 - 4 = -2$. The check through the other bracketing, row 3 of $BA$ times column 1 of $B$, gives $(-3)(2) + 14(0) + 4(1) = -2$ as well. What you may NOT do is move a factor: $B(AB) = (BA)B$ is associativity, allowed, while $BAB = ABB$ would be commutativity, forbidden.

Row 2 of $A$ is $(-1, 3, 1)$ and column 3 of $B$ is $(3, 1, 0)$, so $(AB)_{23} = (-1)(3) + 3(1) + 1(0) = -3 + 3 + 0 = 0$. Three multiplications and two additions, not twenty-seven multiplications. The classic slip is to take column 2 of $A$ with row 3 of $B$, the indices read backwards: that gives $0(1) + 3(2) + 1(0) = 6$, a wrong answer that looks perfectly reasonable. Say it aloud once: the FIRST index picks a row of the LEFT factor, the SECOND index picks a column of the RIGHT factor.

Column 2 of $B$ is $(1, -1, 2)^T$. Then $A(1, -1, 2)^T = (1 - 0 + 4,\; -1 - 3 + 2,\; 2 - 1 + 0)^T = (5, -2, 1)^T$. Read the same computation column by column: $A$ times a vector is the combination of the COLUMNS of $A$ with the entries of that vector as weights, $1 \cdot (1, -1, 2)^T - 1 \cdot (0, 3, 1)^T + 2 \cdot (2, 1, 0)^T = (5, -2, 1)^T$. This reading is the bridge to the previous chapter: every column of $AB$ is a linear combination of the columns of $A$, so every column of $AB$ lives wherever the columns of $A$ live.

Row 3 of $A$ is $(2, 1, 0)$, and a row times $B$ is the combination of the ROWS of $B$ with those weights: $2(2, 1, 3) + 1(0, -1, 1) + 0(1, 2, 0) = (4, 2, 6) + (0, -1, 1) = (4, 1, 7)$. Its first entry, $4$, is the entry the figure computed, which is the free check. The two readings are mirror images: columns of $AB$ are built from the columns of the LEFT factor, rows of $AB$ from the rows of the RIGHT factor. Mixing them up, combining rows of $A$ for instance, is the error that turns a correct method into a wrong matrix.

Twice column 1 of $B$ minus column 2 is $2(2, 0, 1)^T - (1, -1, 2)^T = (3, 1, 0)^T$, which is column 3: checked. Column 1 of $AB$ is $A(2, 0, 1)^T = (2 + 0 + 2,\; -2 + 0 + 1,\; 4 + 0 + 0)^T = (4, -1, 4)^T$. Now $A$ times a combination is the same combination of the products, because multiplying by $A$ is linear: $A(2b_1 - b_2) = 2Ab_1 - Ab_2$. So column 3 of $AB$ is $2(4, -1, 4)^T - (5, -2, 1)^T = (3, 0, 7)^T$, with no new row-times-column product. It agrees with a), whose entry $0$ sits in position 2, and with c), whose row ends in $7$. A relation among the columns of $B$ is inherited by the columns of $AB$: that is the column reading doing real work, not just giving a second route.

By the row reading, row 2 of $AB$ is row 2 of $A$ times $B$. If that row of $A$ is zero, row 2 of $AB$ is zero, and the other two rows, which never used row 2 of $A$, do not move: $AB$ becomes $\begin{pmatrix} 4 & 5 & 3 \\ 0 & 0 & 0 \\ 4 & 1 & 7 \end{pmatrix}$. By the column reading, column 1 of $AB$ is $A$ times column 1 of $B$; a zero column there gives a zero column in $AB$, and nothing else changes. So a product can acquire a zero row or a zero column without either factor being zero, and pushing the idea one step further, the whole product can vanish while both factors are non-zero. Exercise 3 builds exactly such a pair.

Row 1 of $A$, $(1, 2)$, against the columns $(2, -1)$ and $(-4, 2)$ of $B$ gives $2 - 2 = 0$ and $-4 + 4 = 0$; row 2, $(2, 4)$, gives $4 - 4 = 0$ and $-8 + 8 = 0$. So $AB = 0$. In the other order, row 1 of $B$, $(2, -4)$, against the columns $(1, 2)$ and $(2, 4)$ of $A$ gives $2 - 8 = -6$ and $4 - 16 = -12$; row 2, $(-1, 2)$, gives $-1 + 4 = 3$ and $-2 + 8 = 6$. So $BA = \begin{pmatrix} -6 & -12 \\ 3 & 6 \end{pmatrix}$, which is not zero. Neither factor is zero, and the two orders do not even agree on whether the product is zero. Any argument that starts with 'since $AB = 0$, also $BA = 0$' is already lost.

Column $j$ of $AB$ is $A$ times column $j$ of $B$. Both columns of $AB$ are zero, so both columns of $B$ are solutions of the homogeneous system $Ax = 0$. That system is the single equation $x_1 + 2x_2 = 0$, since the second row of $A$ is twice the first, and its solutions are the multiples of $(2, -1)^T$. Indeed column 1 of $B$ is $(2, -1)^T$ and column 2 is $(-4, 2)^T = -2(2, -1)^T$. The product vanishes because $B$ was built entirely out of directions that $A$ kills. Nothing of the kind holds for $BA$: its columns are $B$ times the columns of $A$, and $(1, 2)^T$ is not killed by $B$.

By associativity, $(BA)^2 = (BA)(BA) = B(AB)A = B \cdot 0 \cdot A = 0$. The brackets move, the factors do not: the four letters stay in the order $B, A, B, A$, and the pair $AB$ that sits in the middle is zero. Check: $(BA)^2 = \begin{pmatrix} -6 & -12 \\ 3 & 6 \end{pmatrix}\begin{pmatrix} -6 & -12 \\ 3 & 6 \end{pmatrix}$, whose $(1,1)$ entry is $36 - 36 = 0$, $(1,2)$ entry $72 - 72 = 0$, $(2,1)$ entry $-18 + 18 = 0$ and $(2,2)$ entry $-36 + 36 = 0$. So $BA$ is a non-zero matrix whose square is zero, a nilpotent matrix, obtained without a single guess.

By b), $AX = 0$ says exactly that each column of $X$ solves $x_1 + 2x_2 = 0$, so each column is a multiple of $(2, -1)^T$, with its own multiple. The answer is $X = \begin{pmatrix} 2s & 2t \\ -s & -t \end{pmatrix}$ for all real $s$, $t$: a two-parameter family, and $B$ is the member $s = 1$, $t = -2$. The trap is to give one example and stop, or to force both columns to be the same multiple. The question says ALL, and the column reading turns a matrix equation into two independent vector equations.

$A + B = \begin{pmatrix} 3 & -2 \\ 1 & 6 \end{pmatrix}$ and $A - B = \begin{pmatrix} -1 & 6 \\ 3 & 2 \end{pmatrix}$, so $(A + B)(A - B) = \begin{pmatrix} -9 & 14 \\ 17 & 18 \end{pmatrix}$. Also $A^2 = \begin{pmatrix} 5 & 10 \\ 10 & 20 \end{pmatrix}$ and $B^2 = \begin{pmatrix} 8 & -16 \\ -4 & 8 \end{pmatrix}$, so $A^2 - B^2 = \begin{pmatrix} -3 & 26 \\ 14 & 12 \end{pmatrix}$. Expanding while keeping every factor in its place, $(A + B)(A - B) = A^2 - AB + BA - B^2$, so the difference with $A^2 - B^2$ is $BA - AB$. Here $AB = 0$, so the difference should be $BA = \begin{pmatrix} -6 & -12 \\ 3 & 6 \end{pmatrix}$, and indeed $\begin{pmatrix} -9 & 14 \\ 17 & 18 \end{pmatrix} - \begin{pmatrix} -3 & 26 \\ 14 & 12 \end{pmatrix} = \begin{pmatrix} -6 & -12 \\ 3 & 6 \end{pmatrix}$. The difference of squares survives for matrices exactly when $AB = BA$, never by default.

$AB$ is $(2 \times 3)(3 \times 1) = 2 \times 1$: row 1 of $A$ gives $2 - 2 + 0 = 0$ and row 2 gives $6 + 1 - 2 = 5$, so $AB = (0, 5)^T$ and $(AB)^T = (0 \;\; 5)$, a $1 \times 2$ row. On the other side $B^T = (2 \;\; 1 \;\; -1)$ is $1 \times 3$ and $A^T$ is $3 \times 2$, so $B^T A^T$ is $1 \times 2$: against the columns of $A^T$, which are the rows of $A$, it gives $2 - 2 + 0 = 0$ and $6 + 1 - 2 = 5$. The two agree. Now try the order a student writes by reflex: $A^T B^T$ is $(3 \times 2)(1 \times 3)$, inner $2 \neq 1$, so it does not exist at all. The size test settles the question before any entry is computed, and it is the fastest way to remember that the transpose REVERSES the order.