MATH 133 Linear Algebra and Geometry • McGill University, Montreal

Revision sheet: matrix algebra, transpose and the matrix product (MATH 133)

This sheet is not a summary of sections 2.1 and 2.3 of Nicholson: you already have the course notes. It answers one question only, what makes students lose marks on matrix algebra in MATH 133 at McGill University, and which precise gesture avoids each loss.

Almost every trap on this chapter has the same root. Matrices are added like numbers but multiplied like compositions, so the ORDER of the factors carries information, and the rules of high-school algebra that quietly swap two factors stop being true. The sheet sorts them: the ones that survive, the ones that die, and the counterexample with small integers that kills each one in ten seconds.

The thread of the chapter

A matrix product remembers the order of its factors: a row of the LEFT factor meets a column of the RIGHT one, so check the sizes before any entry, and never use a rule of ordinary algebra that swaps two factors, $ab = ba$ , $(ab)^T = a^T b^T$ , $ab = 0$ so $a = 0$ or $b = 0$ , until it has been proved for matrices.

This chapter is part of MATH 133, Linear Algebra and Geometry (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Matrix addition
Scalar multiplication
Matrix product
Compatible matrix sizes
Transpose of a matrix
Symmetric matrix
Skew-symmetric matrix
Powers of a matrix
Nilpotent matrix
Idempotent matrix
Commuting matrices
Trace of a matrix

The essentials

The product: sizes first, then a row meets a column

• $A$ is $m \times n$ and $B$ is $n \times p$ : $AB$ exists because the INNER sizes agree, and it is $m \times p$ , the OUTER sizes. A sum $A + B$ needs identical sizes.
• $(AB)_{ij}$ is row $i$ of $A$ against column $j$ of $B$ : $\sum_{k} a_{ik} b_{kj}$ . The first index picks a row of the LEFT factor, the second a column of the RIGHT factor.
• Column $j$ of $AB$ is $A$ times column $j$ of $B$ , a combination of the columns of $A$ . Row $i$ of $AB$ is row $i$ of $A$ times $B$ , a combination of the rows of $B$ .
• Always true: $(AB)C = A(BC)$ , $A(B + C) = AB + AC$ , $(B + C)A = BA + CA$ , $k(AB) = (kA)B = A(kB)$ , $AI = IA = A$ .
• Not true in general: $AB = BA$ , and ' $AB = 0$ so $A = 0$ or $B = 0$ '. Both fail already for $2 \times 2$ matrices.

The same two matrices in the two orders:

AB

2 \times 2

and

BA

3 \times 3

. For non-square factors,

AB = BA

cannot even be asked.

Write the two sizes side by side before the first entry. It takes three seconds and catches both a product that does not exist and a result of the wrong shape, before a page of arithmetic is spent on it.

Transpose, symmetric parts, powers and trace

• $(A^T)_{ij} = a_{ji}$ . Rules: $(A^T)^T = A$ , $(A + B)^T = A^T + B^T$ , $(kA)^T = kA^T$ , and $(AB)^T = B^T A^T$ , with the order REVERSED.
• Symmetric: $S^T = S$ , mirror pairs equal. Skew-symmetric: $K^T = -K$ , mirror pairs opposite and diagonal all zero. Every square $A$ is $\frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$ , in exactly one way.
• Powers exist for SQUARE matrices only. Nilpotent: $A^k = 0$ for some $k$ . Idempotent: $A^2 = A$ , hence $A^n = A$ for every $n \ge 1$ .
• $\operatorname{tr} A$ is the sum of the diagonal entries. $\operatorname{tr}(A + B) = \operatorname{tr} A + \operatorname{tr} B$ and $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ , but $\operatorname{tr}(AB) \neq \operatorname{tr} A \cdot \operatorname{tr} B$ in general.

Same mirror, two behaviours: on the left each coloured pair is equal, on the right each pair is opposite and the diagonal is forced to

0

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

Rules of ordinary algebra: which ones survive for matrices

Each line is an identity you use without thinking for numbers. The last column says whether it holds for square matrices of the same size. Red means false in general, and the example proves it with small integers: that example is the counterexample to write on a true or false question.

Numbers	Matrices	Verdict
$a(bc) = (ab)c$	$A(BC) = (AB)C$	holds
Example: Brackets may move: with $A = B = C = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ , both bracketings give $\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}$ .
$a(b + c) = ab + ac$	$A(B + C) = AB + AC$	holds
Example: $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 1 \end{pmatrix}\right) = \begin{pmatrix} 1 \\ 3 \end{pmatrix} + \begin{pmatrix} 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$ .
$ab = ba$	$AB = BA$	fails false for matrices
Example: $\begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ , but the other order gives $\begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}$ . What to do: Keep every factor in its place. Swap two factors only after checking $AB = BA$ , or when one of them is $cI$ .
$(a + b)(a - b)$	$(A + B)(A - B)$	$A^2 - B^2 + BA - AB$ two extra terms
Example: With $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ , the extra part $BA - AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ is not zero. What to do: Expand term by term, keeping the order: $A^2 - AB + BA - B^2$ . The middle cancels only if $AB = BA$ .
$ab = 0$	$AB = 0$	one factor is $0$ false for matrices
Example: $\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} = 0$ with both factors non-zero. What to do: Read $AB = 0$ column by column: each column of $B$ solves $Ax = 0$ . Nothing more.
$(ab)^2 = a^2 b^2$	$(AB)^2 = A^2 B^2$	fails false for matrices
Example: With $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ : $(AB)^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ but $A^2 B^2 = 0$ . What to do: $(AB)^2 = ABAB$ : the $BA$ in the middle cannot be turned into $AB$ .

The two lines that survive are exactly the ones that never swap two factors. That is the whole table in one sentence.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Multiplying two matrices entry by entry

the whole question

What not to write

“ $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 3 & 0 \end{pmatrix}$ ”

What to write

“Row against column: $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}$ , the two columns swapped.”

Why: The entry-by-entry product agrees with the real one only for diagonal matrices, which is why it survives so long. Here the right factor swaps the columns of the left one, and the column reading says so before any arithmetic: column 1 of the product is $A(0, 1)^T$ , the second column of $A$ .

2. Squaring a matrix that is not square

1 to 2 marks, and every part built on it

What not to write

“ $A$ is $2 \times 3$ , so $A^2$ is $2 \times 3$ as well.”

What to write

“ $A^2 = AA$ is $(2 \times 3)(2 \times 3)$ , inner sizes $3 \neq 2$ : it does not exist. The products that do exist are $AA^T$ , $2 \times 2$ , and $A^T A$ , $3 \times 3$ .”

Why: Powers, commuting and the trace are questions about SQUARE matrices only. A size written next to each factor would have stopped the line before it was written.

3. Reading the indices of an entry backwards

the entry asked for, and it looks perfectly plausible

What not to write

“ $(AB)_{23}$ is column 2 of $A$ against row 3 of $B$ .”

What to write

“ $(AB)_{23}$ is row 2 of $A$ against column 3 of $B$ : the first index is a row of the LEFT factor, the second a column of the RIGHT factor.”

Why: A question asking for one entry is testing exactly this, and the wrong reading still produces a sensible number, so nothing flags it. With $A = \begin{pmatrix} 1 & 0 & 2 \\ -1 & 3 & 1 \\ 2 & 1 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 1 & 3 \\ 0 & -1 & 1 \\ 1 & 2 & 0 \end{pmatrix}$ , the right reading gives $0$ and the backward one gives $6$ .

4. Transposing a product without reversing it

2 marks, and every symmetry proof that uses it

What not to write

“ $(AB)^T = A^T B^T$ ”

What to write

“ $(AB)^T = B^T A^T$ . With $A$ of size $2 \times 3$ and $B$ of size $3 \times 1$ , $A^T B^T$ is $(3 \times 2)(1 \times 3)$ and does not even exist.”

Why: Entry $(i,j)$ of $(AB)^T$ is $(AB)_{ji} = \sum_k a_{jk} b_{ki}$ , which is row $i$ of $B^T$ against column $j$ of $A^T$ . Undoing a product means undoing the last factor first, so the order reverses; for square factors the wrong version exists and is simply false, the more dangerous case.

5. Using the difference of squares, or the binomial formula, as for numbers

2 marks

What not to write

“ $(A - B)(A + B) = A^2 - B^2$ ”

What to write

“ $(A - B)(A + B) = A^2 + AB - BA - B^2$ , which reduces to $A^2 - B^2$ only if $AB = BA$ .”

Why: With $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ , the leftover $AB - BA$ is $\begin{pmatrix} 0 & -1 \\ 0 & 0 \end{pmatrix}$ . The binomial formula is legitimate in one important case, a sum like $cI + N$ , because $cI$ commutes with everything.

6. Concluding that a matrix is zero because a power of it is

the whole question

What not to write

“ $N^3 = 0$ , so $N = 0$ .”

What to write

“ $N = \begin{pmatrix} 0 & 1 & 4 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$ is not zero and $N^3 = 0$ : $N$ is nilpotent. From $AB = 0$ , conclude only that each column of $B$ solves $Ax = 0$ .”

Each power pushes the non-zero entries one diagonal further from the main diagonal; at

N^3

there is no diagonal left and the matrix is zero.

Why: The rule “a product of non-zero numbers is non-zero” is exactly what fails for matrices. A strictly upper triangular $n \times n$ matrix always has $N^n = 0$ , and the figure shows why.

7. Raising a matrix to a power by raising its entries

the whole question, and the induction that follows

What not to write

“ $\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 9 \\ 0 & 1 \end{pmatrix}$ ”

What to write

“ $\begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}^2 = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix}$ : the corner is $1 \cdot 3 + 3 \cdot 1 = 6$ .”

Why: $A^2 = AA$ is a matrix product like any other. Entry-by-entry powers are correct only for DIAGONAL matrices, where no row ever meets a non-zero off-diagonal entry.

8. Assuming that a product of symmetric matrices is symmetric

2 marks on a true or false question

What not to write

“ $A$ and $B$ are symmetric, so $AB$ is symmetric.”

What to write

“ $(AB)^T = B^T A^T = BA$ , so $AB$ is symmetric exactly when $AB = BA$ . For instance $\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 2 & 0 \end{pmatrix}$ is not.”

Why: The reversal rule is the whole argument, one line long. Sums of symmetric matrices, $A^2$ and $AA^T$ are symmetric; a product of two different symmetric matrices is not, unless they commute.

9. Believing that diagonal matrices commute with everything

the whole commuting question

What not to write

“ $D$ is diagonal, so $DX = XD$ for every $X$ .”

What to write

“Only the scalar matrices $cI$ commute with everything: $\begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ but in the other order $\begin{pmatrix} 0 & 3 \\ 0 & 0 \end{pmatrix}$ .”

Why: A diagonal matrix on the LEFT scales the rows, on the RIGHT it scales the columns, and the two agree only when the diagonal entries are equal. Diagonal matrices do commute with EACH OTHER, which is the source of the confusion.

10. Treating the trace as multiplicative

2 marks

What not to write

“ $\operatorname{tr}(AB) = \operatorname{tr} A \cdot \operatorname{tr} B$ ”

What to write

“The trace is additive, and $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ ; it is not multiplicative: $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ , $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ give $\operatorname{tr}(AB) = 0$ but $\operatorname{tr} A \cdot \operatorname{tr} B = 1$ .”

Why: The one product rule the trace has is the swap rule, and it only allows cyclic shifts: $\operatorname{tr}(ABC) = \operatorname{tr}(CAB)$ , but not $\operatorname{tr}(ACB)$ in general. It is also what proves that $XY - YX = I$ is impossible.

Which method to choose

Which reading of the product the question is asking for

Read what the question wants FROM the product before multiplying anything

What each reading takes from each factor: a row and a column for one entry, all of

A

and one column for a column, one row and all of

B

for a row.

If one entry $(i,j)$ → row $i$ of the left factor against column $j$ of the right one, and nothing else

Example: for two $3 \times 3$ factors: 3 multiplications instead of 27
If one column of $AB$ → $A$ times that column of $B$ , a combination of the columns of $A$

Example: $A(1, -1, 2)^T = a_1 - a_2 + 2a_3$
If one row of $AB$ → that row of $A$ times $B$ , a combination of the rows of $B$

Example: $(2, 1, 0)\,B = 2r_1 + r_2$
If a trace → only the diagonal entries of the product, or $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ on the smaller product

Example: $A$ is $2 \times 3$ , $B$ is $3 \times 2$ : two entries of $AB$ , not three of $BA$
If a product of three factors → bracket so that the intermediate product is the smallest

Example: $B(AB)$ with $AB$ of size $2 \times 2$ , rather than $(BA)B$ with $BA$ of size $3 \times 3$
If a relation between the columns of $B$ → the same relation holds between the columns of $AB$

Example: $b_3 = 2b_1 - b_2$ gives $(AB)_3 = 2(AB)_1 - (AB)_2$

Computing the whole product to read one entry is not wrong, only slow, and on a midterm the time goes to the last question. The one reading that is WRONG is combining the rows of the left factor or the columns of the right one.

Before raising a matrix to a power

Look at the SHAPE of the matrix first, not at $n$

If the matrix is diagonal → raise each diagonal entry to the power $n$

Example: $\operatorname{diag}(2, -1)^5 = \operatorname{diag}(32, -1)$
If it is strictly upper triangular, $n \times n$ → it is nilpotent: $A^n = 0$

Example: $N = \begin{pmatrix} 0 & 1 & 4 \\ 0 & 0 & 2 \\ 0 & 0 & 0 \end{pmatrix}$ gives $N^3 = 0$
If $A^2 = A$ → idempotent: $A^n = A$ for every $n \ge 1$

Example: $\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}^{10} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$
If $A = cI + N$ with $N$ nilpotent → binomial theorem, legal because $cI$ commutes with $N$

Example: $\begin{pmatrix} 3 & 1 \\ 0 & 3 \end{pmatrix}^n = \begin{pmatrix} 3^{n} & n\,3^{n-1} \\ 0 & 3^{n} \end{pmatrix}$
If none of the above → compute $A^2$ and $A^3$ , conjecture, then prove by induction

Example: $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}$ , proved with $A^{n+1} = A^n A$

A pattern read off three powers is a conjecture, not an answer: without the induction step the question is worth half its marks.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Finding all matrices that commute with a given matrix

When to use it: “Find all $2 \times 2$ matrices $X$ such that $AX = XA$ .”

1 Write the unknown in full: $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ .
2 Compute $AX$ . Then compute $XA$ as a SEPARATE product, row of $X$ against column of $A$ ; it is not the same computation.
3 Equate the four entries and write the four equations, even the ones that repeat.
4 Solve the homogeneous system and name the free parameters.
5 Write the general $X$ and check one numerical member in both products.

Concluding sentence

“The matrices that commute with $A$ are exactly the matrices $X = \begin{pmatrix} 3b + d & b \\ 0 & d \end{pmatrix}$ with $b$ and $d$ real, that is $X = (b + d)I + bA$ .”

The trap: Stopping at one example that commutes. The question says ALL, so the answer is a family with its parameters, and the check with one member proves nothing about the others.

Marking: Typically 2 marks for the two products written separately, 1 for the equations, 1 for the solution, 1 for the final sentence.

Proving an identity with the transpose

When to use it: “Show that $AA^T$ is symmetric”, “show that $A + A^T$ is symmetric”, “prove $(A^n)^T = (A^T)^n$ ”

1 Start from ONE side, usually the transpose of the expression, never from the conclusion.
2 Transform it with named rules only: the reversal rule $(XY)^T = Y^T X^T$ , $(X^T)^T = X$ , $(X + Y)^T = X^T + Y^T$ .
3 Arrive at the expression itself, or at its opposite for a skew-symmetric claim.
4 For a statement about every $n$ , use induction: $(A^{n+1})^T = (A^n A)^T = A^T (A^n)^T$ .

Concluding sentence

“ $(AA^T)^T = (A^T)^T A^T = AA^T$ by the reversal rule and $(A^T)^T = A$ , so $AA^T$ is symmetric.”

The trap: Checking the identity on one numerical matrix. It illustrates the claim, it does not prove it, and it is worth no more than the illustration mark.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The size test, three seconds

Write the size under each factor. The inner sizes must agree and the outer sizes give the answer's shape; a sum needs identical sizes.

$(2 \times 3)(3 \times 2)$ gives $2 \times 2$ . A $3 \times 3$ answer for $AB$ means you computed $BA$ .
A matrix that must be symmetric, is

$AA^T$ , $A^T A$ , $A + A^T$ and the part $S$ of a decomposition must come out symmetric, and $K$ must have a zero diagonal. Any asymmetry is an arithmetic slip, found before the marker finds it.

$A = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ gives $AA^T = \begin{pmatrix} 5 & 2 \\ 2 & 1 \end{pmatrix}$ ; a $2$ and a $3$ off the diagonal would betray an error.
The trace test

When both $AB$ and $BA$ have been computed, their traces must agree, even if their sizes differ.

$\operatorname{tr}(AB) = 6$ and $\operatorname{tr}(BA) = 5$ means one of the two products is wrong.
One column recomputed another way

Pick one column of your product and recompute it as $A$ times the matching column of $B$ , as a combination of the columns of $A$ . It checks a third of a $3 \times 3$ product in one line.

$A(1, -1, 2)^T = a_1 - a_2 + 2a_3$ must equal the second column you wrote down.
A found family, checked on one member

After solving $AX = XA$ or $AX = 0$ , pick simple values of the parameters and compute both sides. A family that fails the check was solved with a wrong product.

$b = 1$ , $d = 0$ gives $X = \begin{pmatrix} 3 & 1 \\ 0 & 0 \end{pmatrix}$ and $AX = XA = \begin{pmatrix} 6 & 2 \\ 0 & 0 \end{pmatrix}$ .

The typical problem, taken apart

Every matrix that commutes with a given one, solved step by step

Find all $2 \times 2$ matrices $X$ that commute with $A = \begin{pmatrix} 2 & 1 \\ 0 & -1 \end{pmatrix}$ , and show that each of them is a combination of $I$ and $A$ .

Step 1

Write $X = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and compute $AX = \begin{pmatrix} 2a + c & 2b + d \\ -c & -d \end{pmatrix}$ , row of $A$ against column of $X$ .

Why

The unknown is written in full so that every entry has a name: the question becomes a system in four unknowns, which is a chapter you already master.

Step 2

Compute $XA$ separately: $XA = \begin{pmatrix} 2a & a - b \\ 2c & c - d \end{pmatrix}$ , row of $X$ against column of $A$ .

Why

This is the step that is skipped. $XA$ is a different product, with the roles of rows and columns exchanged, and its entries are not a rearrangement of those of $AX$ .

Step 3

Equate entry by entry: $2a + c = 2a$ , $2b + d = a - b$ , $-c = 2c$ and $-d = c - d$ . Three of them give $c = 0$ , the fourth gives $a = 3b + d$ .

Why

All four equations are written, even the repetitive ones: that is where the marker looks for the method, and a missing equation is how a free parameter gets lost.

Step 4

So $X = \begin{pmatrix} 3b + d & b \\ 0 & d \end{pmatrix}$ with $b$ , $d$ real. Since $bA + (b + d)I = \begin{pmatrix} 2b + b + d & b \\ 0 & -b + b + d \end{pmatrix} = X$ , every such $X$ is $(b + d)I + bA$ .

Why

Two free parameters, and the family is exactly the combinations of $I$ and $A$ . Matrices of the form $\alpha I + \beta A$ obviously commute with $A$ , so this is also the sanity check on the size of the answer.

Step 5

Check with $b = 1$ , $d = 0$ : $X = \begin{pmatrix} 3 & 1 \\ 0 & 0 \end{pmatrix}$ , and $AX = \begin{pmatrix} 6 & 2 \\ 0 & 0 \end{pmatrix} = XA$ .

Why

One member of the family, computed in both orders. It catches a sign error in the system, which is the usual slip of the third step.

The conclusion, written out

“The matrices that commute with $A$ are exactly $X = \begin{pmatrix} 3b + d & b \\ 0 & d \end{pmatrix}$ , $b, d \in \mathbb{R}$ , that is $X = (b + d)I + bA$ .”

The classic mistake on this problem: Writing $XA$ as the same matrix as $AX$ turns all four equations into $0 = 0$ and concludes that every matrix commutes with $A$ . One test refutes it: $X = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ gives $AX = \begin{pmatrix} 1 & 0 \\ -1 & 0 \end{pmatrix}$ but $XA = \begin{pmatrix} 0 & 0 \\ 2 & 1 \end{pmatrix}$ .

Learn by heart

• $(m \times n)(n \times p) = m \times p$ : the inner sizes agree, the outer sizes give the shape.
• $(AB)_{ij}$ = row $i$ of $A$ against column $j$ of $B$ . Column $j$ of $AB$ = $A b_j$ ; row $i$ of $AB$ = $r_i B$ .
• $AB \neq BA$ in general; $AB = 0$ , $A^2 = 0$ and $A^2 = I$ with $A \neq \pm I$ all happen with non-zero $2 \times 2$ matrices.
• $(AB)^T = B^T A^T$ , order REVERSED; $AA^T$ and $A^T A$ are always symmetric.
• $A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)$ : symmetric plus skew-symmetric, in exactly one way.
• Binomial theorem for $(A + B)^n$ ONLY when $AB = BA$ ; $cI$ commutes with every matrix, and only $cI$ does.
• Powers entry by entry: diagonal matrices ONLY.
• $\operatorname{tr}(AB) = \operatorname{tr}(BA)$ , cyclic shifts only; hence $XY - YX = I$ is impossible.

Frequently asked questions

Why is matrix multiplication not commutative?

Because a product combines rows of the left factor with columns of the right factor, and swapping the factors swaps which matrix gives the rows. For non-square matrices the two products do not even have the same size. For square ones they usually differ too: even two simple two by two matrices, one diagonal and one with a single one in the corner, give different products in the two orders.

How do I know if two matrices can be multiplied?

Write the size of each matrix, rows by columns, side by side. The product exists when the number of columns of the left matrix equals the number of rows of the right one, the two inner numbers. The result has the rows of the left matrix and the columns of the right one, the two outer numbers. A two by three times a three by four gives a two by four.

What is the transpose of a product of two matrices?

The transpose of AB is B transpose times A transpose, in the reverse order. The order must reverse because each entry of the transposed product is a row of B transpose against a column of A transpose. Writing A transpose times B transpose is a classic loss of marks, and for non-square matrices that wrong product often does not even exist.

How do I find all matrices that commute with a given matrix?

Write the unknown matrix with four letters, compute the product in both orders as two separate computations, and set the four entries equal. You get a homogeneous linear system in the four letters. Solve it, keep the free parameters, and write the general solution. Finish by checking one numerical member of the family in both products.

Can the product of two non-zero matrices be the zero matrix?

Yes, and it happens with two by two matrices. If every column of the right factor is a solution of the homogeneous system of the left factor, the product is zero although neither factor is. So from a zero product you may not conclude that one factor is zero, and a matrix can even have a zero square without being zero itself. Such a matrix is called nilpotent.

Practise it

Corrected exercises: Matrix algebra and the matrix product, MATH 133 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Homogeneous systems and the rank theorem Next sheet Matrix inverses

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Revision sheet: matrix algebra, transpose and the matrix product (MATH 133)

The essentials

The product: sizes first, then a row meets a column

Transpose, symmetric parts, powers and trace

The rules in table form

Rules of ordinary algebra: which ones survive for matrices

The mistakes that cost marks

1. Multiplying two matrices entry by entry

2. Squaring a matrix that is not square

3. Reading the indices of an entry backwards

4. Transposing a product without reversing it

5. Using the difference of squares, or the binomial formula, as for numbers

6. Concluding that a matrix is zero because a power of it is

7. Raising a matrix to a power by raising its entries

8. Assuming that a product of symmetric matrices is symmetric

9. Believing that diagonal matrices commute with everything

10. Treating the trace as multiplicative

Which method to choose

Which reading of the product the question is asking for

Before raising a matrix to a power

How the answer is expected to be written

Finding all matrices that commute with a given matrix

Proving an identity with the transpose

Check before you hand in

The typical problem, taken apart

Every matrix that commutes with a given one, solved step by step

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Frequently asked questions

Why is matrix multiplication not commutative?

How do I know if two matrices can be multiplied?

What is the transpose of a product of two matrices?

How do I find all matrices that commute with a given matrix?

Can the product of two non-zero matrices be the zero matrix?

Corrected exercises: Matrix algebra and the matrix product, MATH 133 at McGill

See also

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