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

a) 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.

b) 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.

c) $u^T u$ 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.

d) 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.

e) $BAB$ 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.

a) 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.

b) 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.

c) 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.

d) 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.

e) 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.

a) 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.

b) 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 $B$.

c) 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.

d) 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.

e) $A + B = \begin{pmatrix} 3 & -2 \\ 1 & 6 \end{pmatrix}$ 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.

a) $AB$ 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.

b) Take $1 \le i \le p$ and $1 \le j \le m$. On the left, $((AB)^T)_{ij} = (AB)_{ji} = \sum_{k=1}^{n} a_{jk} b_{ki}$. On the right, $(B^T A^T)_{ij} = \sum_{k=1}^{n} (B^T)_{ik} (A^T)_{kj} = \sum_{k=1}^{n} b_{ki} a_{jk}$. Each term is a product of two NUMBERS, $b_{ki} a_{jk} = a_{jk} b_{ki}$, so the two sums are equal term by term. Note where commutativity was used: between numbers, inside one term, never between the matrices. That is exactly why the matrices come out in the reverse order. A proof on one $2 \times 2$ example proves nothing; the marker wants the general $(i,j)$ entry.

c) By the rule of b) and $(A^T)^T = A$: $(AA^T)^T = (A^T)^T A^T = AA^T$, and $(A^T A)^T = A^T (A^T)^T = A^T A$. Both are symmetric, for every $A$, square or not. For our $2 \times 3$ matrix $A$, $AA^T$ is $2 \times 2$ and $A^T A$ is $3 \times 3$. Entry $(i,j)$ of $AA^T$ is row $i$ of $A$ against row $j$ of $A$: $1 + 4 + 0 = 5$, $3 - 2 + 0 = 1$, and $9 + 1 + 4 = 14$, so $AA^T = \begin{pmatrix} 5 & 1 \\ 1 & 14 \end{pmatrix}$. The two off-diagonal entries come out equal, which is the built-in check: an $AA^T$ that is not symmetric contains an arithmetic error, found before the marker finds it.

d) Put $S = \frac{1}{2}(M + M^T)$ and $K = \frac{1}{2}(M - M^T)$. Reading $M^T$ off the figure, each coloured pair swapped: $M^T = \begin{pmatrix} 1 & 0 & 7 \\ 4 & 2 & 1 \\ -3 & 5 & 3 \end{pmatrix}$. Then $M + M^T = \begin{pmatrix} 2 & 4 & 4 \\ 4 & 4 & 6 \\ 4 & 6 & 6 \end{pmatrix}$ and $M - M^T = \begin{pmatrix} 0 & 4 & -10 \\ -4 & 0 & 4 \\ 10 & -4 & 0 \end{pmatrix}$, which halve to $S = \begin{pmatrix} 1 & 2 & 2 \\ 2 & 2 & 3 \\ 2 & 3 & 3 \end{pmatrix}$ and $K = \begin{pmatrix} 0 & 2 & -5 \\ -2 & 0 & 2 \\ 5 & -2 & 0 \end{pmatrix}$. Checks in five seconds: $S$ is symmetric, $K$ has a zero diagonal and opposite mirror pairs, and $S + K$ gives back $M$, for instance $2 + 2 = 4$ and $2 - 2 = 0$ in the first coloured pair. $S$ averages each pair, $K$ takes half their difference.

e) If $K^T = -K$, the diagonal entries satisfy $k_{ii} = (K^T)_{ii} = -k_{ii}$, so $2k_{ii} = 0$ and $k_{ii} = 0$: the diagonal is the mirror, and a number equal to its own opposite is zero. Uniqueness: suppose $M = S + K$ with $S$ symmetric and $K$ skew-symmetric. Transposing, $M^T = S^T + K^T = S - K$. Adding and subtracting the two equations gives $M + M^T = 2S$ and $M - M^T = 2K$, so $S$ and $K$ have no choice: they are the matrices of d). The proof does not guess a formula, it DERIVES it from the two conditions, which is what makes it a proof of uniqueness rather than a verification of one example.