Linear Algebra · Solutions

Matrices

12 questions · about 107 minutes

Work this sheet on paper, showing every step — the method is what is being marked, and the solutions are written so you can check your route rather than just your answer. Check shapes before every product, say “on the left” or “on the right” every time you multiply by an inverse, and name every result you use (“by the reversal rule”, “by the Invertible Matrix Theorem, statement (d)”, “by the one-sided inverse corollary”). No calculator is needed anywhere: every answer is an integer or a small fraction. Full worked solutions are on a separate page; each question also has its own reveal, so you can check one answer without seeing the rest.

1.Computation
Let $A = (24 - 1 3)$ and $B = (10 5 - 2)$ . State the size of every intermediate matrix you form.
1. (a)Compute $3 A - 2 B^{T}$ .
2. (b)Write $A = S + K$ with $S$ symmetric and $K$ skew-symmetric, and check both types by inspecting entries.
3. (c)Compute $A B$ and $B A$ . Confirm that $tr (A B) = tr (B A)$ even though $A B \neq = B A$ , and check whether either equals $tr (A) tr (B)$ .
Solution
(a) Transpose, scale, then subtract. Transposing swaps the off-diagonal entries: $B^{T} = (15 0 - 2)$ , still $2 \times 2$ , so the combination is defined. Then
$3 A = (612 - 3 9), 2 B^{T} = (210 0 - 4), 3 A - 2 B^{T} = (42 - 3 13) .$
(b) Use the formulas $S = \frac{1}{2} (A + A^{T})$ and $K = \frac{1}{2} (A - A^{T})$ . With $A^{T} = (2 - 1 43)$ ,
$A + A^{T} = (4336), S = (2 \frac{3}{2} \frac{3}{2} 3), A - A^{T} = (05 - 5 0), K = (0 \frac{5}{2} - \frac{5}{2} 0) .$
$S$ is symmetric because $S_{12} = S_{21} = \frac{3}{2}$ . $K$ is skew-symmetric because $K_{12} = - \frac{5}{2} = - K_{21}$ and its diagonal is zero, as skew-symmetry forces. Sanity check: $S + K = (2 \frac{3}{2} + \frac{5}{2} \frac{3}{2} - \frac{5}{2} 3) = (24 - 1 3) = A$ .
(c) Row–column rule, four entries each. Both products are $2 \times 2$ :
$A B = (2 \cdot 1 + (- 1) \cdot 0 4 \cdot 1 + 3 \cdot 0 2 \cdot 5 + (- 1) (- 2) 4 \cdot 5 + 3 (- 2)) = (241214), B A = (1 \cdot 2 + 5 \cdot 4 0 \cdot 2 + (- 2) 4 1 (- 1) + 5 \cdot 3 0 (- 1) + (- 2) 3) = (22 - 8 14 - 6) .$
The two products differ in three of their four entries, so $A B \neq = B A$ . Their traces are $tr (A B) = 2 + 14 = 16$ and $tr (B A) = 22 - 6 = 16$ : equal, exactly as $tr (X Y) = tr (Y X)$ guarantees. By contrast $tr (A) tr (B) = 5 \cdot (- 1) = - 5$ , which matches neither — the trace is linear, not multiplicative.
2.Proof
Throughout, $A$ is an $n \times n$ matrix and $A = S + K$ is its symmetric–skew decomposition, $S = \frac{1}{2} (A + A^{T})$ and $K = \frac{1}{2} (A - A^{T})$ . Argue from the definitions and the transpose and trace laws; testing on examples earns nothing.
1. (a)Prove that a matrix which is both symmetric and skew-symmetric must be the zero matrix.
2. (b)Prove that $tr (K) = 0$ and hence $tr (S) = tr (A)$ . Give two different proofs that $tr (K) = 0$ : one entrywise, one using only the trace laws.
3. (c)Prove that $tr (S K) = 0$ . (Write $tr (S K)$ as a double sum using the row–column rule and swap the names of the two summation indices.) Deduce that $tr (A^{2}) = tr (S^{2}) + tr (K^{2})$ .
Solution
(a) Compare the two conditions entry by entry. Symmetric means $A_{ij} = A_{ji}$ for all $i, j$ ; skew-symmetric means $A_{ij} = - A_{ji}$ . If both hold then $A_{ji} = A_{ij} = - A_{ji}$ , so $2 A_{ji} = 0$ and $A_{ji} = 0$ . Every entry is zero, so $A = 0$ . $■$
(This is the reason the decomposition is unique: two decompositions $S + K = S^{'} + K^{'}$ give $S - S^{'} = K^{'} - K$ , a matrix that is symmetric by the left side and skew-symmetric by the right side, hence zero.)
(b) Entrywise proof. Setting $i = j$ in $K_{ij} = - K_{ji}$ gives $K_{ii} = - K_{ii}$ , so every diagonal entry of $K$ is $0$ and $tr (K) = \sum_{i} K_{ii} = 0$ .
Proof from the trace laws. The trace is linear and $tr (A^{T}) = tr (A)$ , so
$tr (K) = \frac{1}{2} (tr (A) - tr (A^{T})) = \frac{1}{2} (tr (A) - tr (A)) = 0.$
Either way, linearity then gives $tr (A) = tr (S + K) = tr (S) + tr (K) = tr (S)$ . $■$
(c) Write the trace as a double sum. By the row–column rule, $(S K)_{ii} = \sum_{j} S_{ij} K_{ji}$ , so
$tr (S K) = i = 1 \sum n j = 1 \sum n S_{ij} K_{ji} .$
The names of the summation indices are arbitrary, so rename $i \leftrightarrow j$ : the same number is $\sum_{j} \sum_{i} S_{ji} K_{ij}$ . Now use $S_{ji} = S_{ij}$ and $K_{ij} = - K_{ji}$ :
$tr (S K) = i, j \sum S_{ji} K_{ij} = i, j \sum S_{ij} (- K_{ji}) = - tr (S K) .$
Hence $2 tr (S K) = 0$ , so $tr (S K) = 0$ . $■$ (A second route: $tr (S K) = tr ((S K)^{T}) = tr (K^{T} S^{T}) = tr (- K S) = - tr (K S) = - tr (S K)$ , using $tr (M^{T}) = tr (M)$ , the reversal rule, and $tr (X Y) = tr (Y X)$ .)
Deduction. Expand without assuming anything commutes: $A^{2} = (S + K) (S + K) = S^{2} + S K + K S + K^{2}$ . Taking traces and using linearity, $tr (A^{2}) = tr (S^{2}) + tr (S K) + tr (K S) + tr (K^{2})$ . The middle terms are $tr (S K) = 0$ and $tr (K S) = tr (S K) = 0$ , leaving $tr (A^{2}) = tr (S^{2}) + tr (K^{2})$ . $■$
3.Computation
Let $A = (1 - 1 0321)$ and $B = 201 1 - 1 4$ .
1. (a)For each of $A B$ , $B A$ , $A^{2}$ , $A A^{T}$ , $A^{T} B^{T}$ and $B^{T} A^{T}$ , decide from the shapes alone whether it is defined, and give the size of each one that is.
2. (b)Compute $A B$ and $B A$ , and verify that their traces agree.
3. (c)Compute $B^{T} A^{T}$ directly and compare it with $(A B)^{T}$ . Then, without multiplying anything out, say which transpose $A^{T} B^{T}$ is equal to, and why.
Solution
(a) Match inner dimensions. $A$ is $2 \times 3$ and $B$ is $3 \times 2$ , so $A^{T}$ is $3 \times 2$ and $B^{T}$ is $2 \times 3$ .
$A B$ : $(2 \times \underline{3}) (\underline{3} \times 2)$ , defined, size $2 \times 2$ . $B A$ : $(3 \times \underline{2}) (\underline{2} \times 3)$ , defined, size $3 \times 3$ . $A^{2} = AA$ : $(2 \times 3) (2 \times 3)$ , inner dimensions $3 \neq = 2$ , undefined — only square matrices have powers. $A A^{T}$ : $(2 \times 3) (3 \times 2)$ , defined, $2 \times 2$ . $A^{T} B^{T}$ : $(3 \times 2) (2 \times 3)$ , defined, $3 \times 3$ . $B^{T} A^{T}$ : $(2 \times 3) (3 \times 2)$ , defined, $2 \times 2$ .
Note that $A^{T} B^{T}$ is defined here only because $A$ and $B$ have reversed shapes, and it cannot be $(A B)^{T}$ , which is $2 \times 2$ .
(b) Row–column rule. For $A B$ , row $1$ of $A$ is $(1, 0, 2)$ and row $2$ is $(- 1, 3, 1)$ :
$A B = (1 \cdot 2 + 0 \cdot 0 + 2 \cdot 1 - 1 \cdot 2 + 3 \cdot 0 + 1 \cdot 1 1 \cdot 1 + 0 (- 1) + 2 \cdot 4 - 1 \cdot 1 + 3 (- 1) + 1 \cdot 4) = (4 - 1 90) .$
For $B A$ use the row rule: row $i$ of $B A$ is row $i$ of $B$ times $A$ , a combination of the two rows of $A$ . Row $1$ of $B$ is $(2, 1)$ , giving $2 (1, 0, 2) + 1 (- 1, 3, 1) = (1, 3, 5)$ ; row $2$ is $(0, - 1)$ , giving $- (- 1, 3, 1) = (1, - 3, - 1)$ ; row $3$ is $(1, 4)$ , giving $(1, 0, 2) + 4 (- 1, 3, 1) = (- 3, 12, 6)$ . So
$B A = 11 - 3 3 - 3 12 5 - 1 6 .$
Traces: $tr (A B) = 4 + 0 = 4$ and $tr (B A) = 1 - 3 + 6 = 4$ . A $2 \times 2$ and a $3 \times 3$ matrix with the same trace, exactly as $tr (X Y) = tr (Y X)$ promises for any compatible pair.
(c) Compute, then compare. $B^{T} = (21 0 - 1 14)$ and $A^{T} = 102 - 1 31$ , so
$B^{T} A^{T} = (2 \cdot 1 + 0 \cdot 0 + 1 \cdot 2 1 \cdot 1 + (- 1) 0 + 4 \cdot 2 2 (- 1) + 0 \cdot 3 + 1 \cdot 1 1 (- 1) + (- 1) 3 + 4 \cdot 1) = (49 - 1 0) .$
Transposing the answer to (b) gives $(A B)^{T} = (49 - 1 0)$ — the same, confirming the reversal rule $(A B)^{T} = B^{T} A^{T}$ .
For $A^{T} B^{T}$ , apply the reversal rule to the other product: $(B A)^{T} = A^{T} B^{T}$ . So $A^{T} B^{T}$ is the transpose of $B A$ , namely the $3 \times 3$ matrix $135 1 - 3 - 1 - 3 126$ . Nothing new needs multiplying; the rule tells you which product to transpose.
4.Proof
Let $A = (3013)$ .
1. (a)Compute $A^{2}$ and $A^{3}$ .
2. (b)Conjecture a formula for $A^{n}$ valid for every integer $n \geq 1$ and prove it by induction on $n$ .
3. (c)Write $A = 3 I + N$ and compute $N^{2}$ . Explain exactly why the binomial theorem may be applied to $(3 I + N)^{n}$ even though matrix multiplication is not commutative, and use it to rederive your formula.
Solution
(a) Multiply directly.
$A^{2} = (3 \cdot 3 + 1 \cdot 0 0 3 \cdot 1 + 1 \cdot 3 3 \cdot 3) = (9069), A^{3} = A^{2} A = (270 9 \cdot 1 + 6 \cdot 3 27) = (2702727) .$
(b) Conjecture. The diagonal is $3^{n}$ ; the corner runs $1, 6, 27$ , which is $1 \cdot 3^{0}, 2 \cdot 3^{1}, 3 \cdot 3^{2}$ . So conjecture
$A^{n} = (3^{n} 0 n 3^{n - 1} 3^{n}) .$
Proof by induction. Base case $n = 1$ : the formula gives $(3013) = A$ . Inductive step: suppose the formula holds for some $n \geq 1$ . Since $A^{n + 1} = A^{n} A$ by the definition of powers (associativity makes the bracketing irrelevant),
$A^{n + 1} = (3^{n} 0 n 3^{n - 1} 3^{n}) (3013) = (3^{n + 1} 0 3^{n} + 3 n 3^{n - 1} 3^{n + 1}) = (3^{n + 1} 0 (n + 1) 3^{n} 3^{n + 1}),$
because $3^{n} + n 3^{n} = (n + 1) 3^{n}$ . This is the formula with $n + 1$ in place of $n$ , so it holds for every $n \geq 1$ . $■$
(c) Split off the identity. $N = A - 3 I = (0010)$ , and $N^{2} = (0 \cdot 0 + 1 \cdot 0 0 0 \cdot 1 + 1 \cdot 0 0) = 0$ . So $N$ is nilpotent and every power $N^{k}$ with $k \geq 2$ is zero.
The binomial expansion of $(X + Y)^{n}$ is obtained by multiplying out $n$ brackets and collecting terms, and collecting $X Y X$ together with $XX Y$ requires $X Y = Y X$ . So the theorem is valid for matrices exactly when the two summands commute. Here $X = 3 I$ and $Y = N$ , and $3 I$ commutes with everything: $(3 I) N = 3 N = N (3 I)$ by the product laws $r (A B) = (r A) B = A (r B)$ and $I N = N = N I$ . Hence
$(3 I + N)^{n} = k = 0 \sum n (k n) (3 I)^{n - k} N^{k} = 3^{n} I + n 3^{n - 1} N,$
since every term with $k \geq 2$ contains $N^{k} = 0$ . That is $(3^{n} 0 0 3^{n}) + (00 n 3^{n - 1} 0)$ , the formula from (b). Sanity check at $n = 3$ : $27 I + 3 \cdot 9 N = (2702727)$ , matching (a).
5.Computation
Let $A = (4172)$ and $B = (1201)$ . Every time you multiply by an inverse, say whether it is on the left or on the right.
1. (a)Show that $A$ is invertible, find $A^{- 1}$ , and verify $A^{- 1} A = I$ before using it.
2. (b)Solve $A X = B$ and $X A = B$ for the unknown $2 \times 2$ matrix $X$ . Check each answer by substitution, and explain why the two solutions differ.
3. (c)Now let $P$ and $Q$ be invertible $n \times n$ matrices. Solve $(PX Q^{- 1})^{T} = Q$ for $X$ , and verify your answer by substituting back.
Solution
(a) Determinant, then swap-and-negate. $a d - b c = 4 \cdot 2 - 7 \cdot 1 = 1 \neq = 0$ , so $A$ is invertible and
$A^{- 1} = \frac{1}{1} (2 - 1 - 7 4) = (2 - 1 - 7 4) .$
Verify: $A^{- 1} A = (2 \cdot 4 - 7 \cdot 1 - 1 \cdot 4 + 4 \cdot 1 2 \cdot 7 - 7 \cdot 2 - 1 \cdot 7 + 4 \cdot 2) = (1001)$ .
(b) Strip $A$ from the correct side. For $A X = B$ , multiply on the left by $A^{- 1}$ : $A^{- 1} A X = A^{- 1} B$ , so $X = A^{- 1} B$ :
$X_{1} = (2 - 1 - 7 4) (1201) = (2 - 14 - 1 + 8 - 7 4) = (- 12 7 - 7 4) .$
Check: $A X_{1} = (- 48 + 49 - 12 + 14 - 28 + 28 - 7 + 8) = (1201) = B$ .
For $X A = B$ , multiply on the right by $A^{- 1}$ : $X A A^{- 1} = B A^{- 1}$ , so $X = B A^{- 1}$ :
$X_{2} = (1201) (2 - 1 - 7 4) = (2 4 - 1 - 7 - 14 + 4) = (23 - 7 - 10) .$
Check: $X_{2} A = (8 - 7 12 - 10 14 - 14 21 - 20) = (1201) = B$ .
The solutions differ because $A^{- 1} B \neq = B A^{- 1}$ : multiplication is not commutative, so “divide by $A$ ” is meaningless until you say which side. Each solution is unique, by the same left- or right-multiplication that produced it.
(c) Undo the operations in reverse order. The outermost operation is the transpose, so remove it first: transposing both sides and using $(M^{T})^{T} = M$ gives $PX Q^{- 1} = Q^{T}$ . Next multiply on the left by $P^{- 1}$ : $X Q^{- 1} = P^{- 1} Q^{T}$ . Finally multiply on the right by $Q$ : $X = P^{- 1} Q^{T} Q$ .
Verify: $PX Q^{- 1} = P (P^{- 1} Q^{T} Q) Q^{- 1} = (P P^{- 1}) Q^{T} (Q Q^{- 1}) = Q^{T}$ , regrouping by associativity, and transposing gives $(Q^{T})^{T} = Q$ . The order $Q^{T} Q$ cannot be swapped to $Q Q^{T}$ : those are different matrices in general.
6.Computation
Let $A = 121011232$ .
1. (a)Compute $A^{- 1}$ by row-reducing $[A ∣ I]$ . Name every row operation, and say at which point you can already be certain that $A$ is invertible.
2. (b)Write each row operation you used as an elementary matrix $E_{1}, E_{2}, \dots$ in the order performed. State the identity relating $E_{p} \dots E_{1}$ , $A$ and $I$ , explain from it why $A^{- 1} = E_{p} \dots E_{1}$ , and hence express $A$ itself as a product of elementary matrices (you need not multiply that product out). Write $E_{1}^{- 1}$ explicitly.
3. (c)Use $A^{- 1}$ to solve $A x = (1, 0, 1)^{T}$ , and check the answer in the original system.
Solution
(a) Reduce, carrying $I$ along.
$121011232100010001 R_{2} - 2 R_{1}, R_{3} - R_{1} 100011 2 - 1 0 1 - 2 - 1 010001 R_{3} - R_{2} 100010 2 - 1 1 1 - 2 1 01 - 1 001 .$
The left block is now upper triangular with three nonzero pivots, so it has $3$ pivot positions and the reduction can never produce a zero row: $A$ is invertible (Invertible Matrix Theorem, statement (c)), before the upward sweep is even started. Now clear upward: $R_{2} + R_{3}$ and $R_{1} - 2 R_{3}$ give
$100010001 - 1 - 1 1 20 - 1 - 2 11, so A^{- 1} = - 1 - 1 1 20 - 1 - 2 11 .$
Sanity check two entries of $A A^{- 1}$ : row $2$ of $A$ times column $1$ of $A^{- 1}$ is $2 (- 1) + 1 (- 1) + 3 (1) = 0$ ; row $3$ times column $3$ is $1 (- 2) + 1 (1) + 2 (1) = 1$ .
(b) One elementary matrix per operation, first operation nearest $A$ . In the order performed,
$E_{1} = L_{21} (- 2) = 1 - 2 0 010001, E_{2} = L_{31} (- 1), E_{3} = L_{32} (- 1), E_{4} = L_{23} (1), E_{5} = L_{13} (- 2),$
where $L_{ij} (k)$ is $I$ with $k$ placed in position $(i, j)$ . Because a row operation is left multiplication by its elementary matrix, the reduction says
$E_{5} E_{4} E_{3} E_{2} E_{1} A = I .$
So $M = E_{5} E_{4} E_{3} E_{2} E_{1}$ satisfies $M A = I$ . Multiplying this on the right by $A^{- 1}$ , which exists by (a), gives $M = A^{- 1}$ ; alternatively, the one-sided inverse corollary says a left inverse of a square matrix is the inverse. That is exactly what the right block computed: it is $M I = M$ .
Inverting the product with the reversal rule and $L_{ij} (k)^{- 1} = L_{ij} (- k)$ ,
$A = E_{1}^{- 1} E_{2}^{- 1} E_{3}^{- 1} E_{4}^{- 1} E_{5}^{- 1} = L_{21} (2) L_{31} (1) L_{32} (1) L_{23} (- 1) L_{13} (2), E_{1}^{- 1} = 120010001 .$
(Multiplying the five factors out does return $A$ ; the first factor alone already restores the $2$ in position $(2, 1)$ .)
(c) Multiply on the left by $A^{- 1}$ .
$x = A^{- 1} 101 = - 1 + 0 - 2 - 1 + 0 + 1 1 + 0 + 1 = - 3 02 .$
Check: $A x = (- 3 + 0 + 4, - 6 + 0 + 6, - 3 + 0 + 4) = (1, 0, 1)$ , and this is the only solution because $A$ is invertible.
7.Critique
Three short arguments follow. Each contains an error. For each one, quote the exact sentence that is wrong, name the theorem or hypothesis being abused, and then give a correct treatment — which may or may not reach the same conclusion.
1. (a)“Let $A$ be a square matrix with $A^{2} = A$ . Then $AA = A I$ . Cancelling the factor $A$ on the left gives $A = I$ . So the identity is the only idempotent matrix.”
2. (b)“Let $A = (100111)$ and $B = 100010$ . A direct computation gives $A B = I_{2}$ . By the corollary that a one-sided inverse suffices, $B = A^{- 1}$ , and therefore $B A = I_{3}$ as well.”
3. (c)“Let $A$ be $3 \times 3$ and suppose that the system $A x = (1, 2, 3)^{T}$ has exactly one solution. Since $A x = b$ has a solution, statement (g) of the Invertible Matrix Theorem holds, so $A$ is invertible.”
Solution
(a) The wrong sentence is “Cancelling the factor $A$ on the left gives $A = I$ .” There is no cancellation law for matrices: from $AA = A I$ one may conclude $A = I$ only by multiplying on the left by $A^{- 1}$ , which requires $A$ to be invertible — and nothing in the hypothesis says it is. The argument silently assumes the very thing that decides the question.
Correct treatment: the deduction is valid if $A$ is invertible, so the true statement is “an invertible idempotent matrix is the identity”. A singular idempotent need not be: $A = (1000)$ satisfies $A^{2} = A$ and $A \neq = I$ . (It is singular, as it must be: $A x = 0$ has the nontrivial solution $(0, 1)^{T}$ , so statement (d) fails.) The conclusion of the argument is false.
(b) The wrong sentence is “By the corollary that a one-sided inverse suffices, $B = A^{- 1}$ .” That corollary is a statement about $n \times n$ matrices — it is a consequence of the Invertible Matrix Theorem, which is a square-matrix theorem. Here $A$ is $2 \times 3$ and $B$ is $3 \times 2$ ; neither is square, so neither can be invertible in the first place (the definition $A C = C A = I$ needs both products to have the same size), and the symbol $A^{- 1}$ has no meaning. The computation $A B = I_{2}$ is correct; the inference from it is not.
Correct treatment: compute $B A$ directly. Row $i$ of $B A$ is row $i$ of $B$ times $A$ , so rows $1$ and $2$ of $B A$ are rows $1$ and $2$ of $A$ and row $3$ is zero:
$B A = 100010110 \neq = I_{3} .$
Indeed $B A$ could never be $I_{3}$ : $A x = 0$ has the nonzero solution $x = (1, 1, - 1)^{T}$ (three unknowns, two equations), and then $B A x = B 0 = 0$ , whereas $I_{3} x = x \neq = 0$ .
(c) The wrong sentence is “Since $A x = b$ has a solution, statement (g) of the Invertible Matrix Theorem holds.” Statement (g) says $A x = b$ is solvable for every $b \in R^{3}$ ; solvability for one particular $b$ establishes nothing. (The matrix $A = diag (1, 1, 0)$ with $b = (1, 2, 0)^{T}$ is a solvable system on a singular matrix.) The argument quotes the wrong statement — but its conclusion happens to be true, and the unused word “exactly” is what saves it.
Correct treatment: use the uniqueness. Let $x_{0}$ be the unique solution and suppose $A v = 0$ . Then $A (x_{0} + v) = A x_{0} + A v = b + 0 = b$ , so $x_{0} + v$ is also a solution, and uniqueness forces $x_{0} + v = x_{0}$ , that is $v = 0$ . Hence $A x = 0$ has only the trivial solution, which is statement (d), and by the Invertible Matrix Theorem $A$ is invertible. In words: a unique solution to any one consistent square system already certifies invertibility; a merely existing one does not.
8.Computation
Let $M = 1000110021110323$ .
1. (a)Partition $M$ into four $2 \times 2$ blocks $(A 0 B D)$ , write down $A$ , $B$ and $D$ , and explain — without row-reducing $M$ — why $M$ is invertible.
2. (b)Compute $M^{- 1}$ using the block formula for a block upper triangular matrix. Keep the factors in the order the formula prescribes, and check one full row of $M M^{- 1}$ .
3. (c)Replace the zero block by $C = (0001)$ to obtain $N = (A C B D)$ . Compute the Schur complement $N / A = D - C A^{- 1} B$ and decide whether $N$ is invertible, stating the criterion you use.
Solution
(a) Read off the blocks.
$A = (1011), B = (2103), D = (1123),$
and the lower-left block is $0$ . A block upper triangular matrix with square diagonal blocks is invertible if and only if its diagonal blocks are. Here $det A = 1 \cdot 1 - 1 \cdot 0 = 1 \neq = 0$ and $det D = 1 \cdot 3 - 2 \cdot 1 = 1 \neq = 0$ , so both are invertible by the $2 \times 2$ criterion, and therefore $M$ is invertible.
(b) Assemble the formula $M^{- 1} = (A^{- 1} 0 - A^{- 1} B D^{- 1} D^{- 1})$ . Swap-and-negate (both determinants are $1$ ):
$A^{- 1} = (10 - 1 1), D^{- 1} = (3 - 1 - 2 1) .$
The corner block, left to right:
$A^{- 1} B = (10 - 1 1) (2103) = (11 - 3 3), A^{- 1} B D^{- 1} = (11 - 3 3) (3 - 1 - 2 1) = (60 - 5 1),$
so $- A^{- 1} B D^{- 1} = (- 6 0 5 - 1)$ and
$M^{- 1} = 1000 - 1 100 - 6 03 - 1 5 - 1 - 2 1 .$
Check row $1$ of $M M^{- 1}$ : row $1$ of $M$ is $(1, 1, 2, 0)$ , and against the four columns of $M^{- 1}$ it gives $1$ , $- 1 + 1 = 0$ , $- 6 + 0 + 6 = 0$ , $5 - 1 - 4 = 0$ . That is $(1, 0, 0, 0)$ , the first row of $I_{4}$ . (Row $2$ , $(0, 1, 1, 3)$ , gives $0$ , $1$ , $0 + 3 - 3 = 0$ , $- 1 - 2 + 3 = 0$ as well.)
(c) Block elimination. From (b), $A^{- 1} B = (11 - 3 3)$ , so
$C A^{- 1} B = (0001) (11 - 3 3) = (0103), N / A = D - C A^{- 1} B = (1123) - (0103) = (1020) .$
The criterion: since $A$ is invertible, $N = (I C A^{- 1} 0 I) (A 0 B N / A)$ . The first factor is block triangular with identity diagonal blocks, hence invertible, so $N$ is invertible exactly when the second factor is — that is, exactly when $A$ and $N / A$ are both invertible. Here $N / A$ has a zero row, so it is singular, and therefore $N$ is not invertible.
Sanity check on $N$ itself: the change put a single $1$ in position $(4, 2)$ , so row $4$ of $N$ is $(0, 1, 1, 3)$ , identical to row $2$ . Two equal rows means $R_{4} - R_{2}$ produces a zero row, so $N$ has fewer than $4$ pivots — singular by the Invertible Matrix Theorem, exactly as the Schur complement predicted.
9.Critique
Each argument below reaches a false conclusion. For each: identify the exact step that fails and name the law of numbers that is being applied to matrices without justification; produce an explicit counterexample to the conclusion, with the verifying computation; and state the correct theorem that survives, with proof.
1. (a)“Suppose $Q$ is both symmetric and orthogonal. Then $Q^{T} = Q$ and $Q^{T} = Q^{- 1}$ , so $Q = Q^{- 1}$ , hence $Q^{2} = I$ . Taking square roots, $Q = I$ . So the identity is the only symmetric orthogonal matrix.”
2. (b)“Let $N$ be nilpotent, say $N^{k} = 0$ . Taking $k$ -th roots of both sides gives $N = 0$ . So the only nilpotent matrix is the zero matrix.”
3. (c)“If $P$ and $Q$ are idempotent $n \times n$ matrices then so is $PQ$ , because $(PQ)^{2} = P^{2} Q^{2} = PQ$ .”
Solution
(a) The failing step is “Taking square roots, $Q = I$ .” Even for real numbers $x^{2} = 1$ has two solutions, and for matrices $X^{2} = I$ has infinitely many: there is no square-root law. Everything before that step is correct, so what has actually been proved is that a symmetric orthogonal matrix satisfies $Q^{2} = I$ .
Counterexample: $Q = (0110)$ . It is symmetric by inspection, and $Q^{T} Q = Q^{2} = (0 \cdot 0 + 1 \cdot 1 1 \cdot 0 + 0 \cdot 1 0 \cdot 1 + 1 \cdot 0 1 \cdot 1 + 0 \cdot 0) = I$ , so it is orthogonal — yet $Q \neq = I$ . (It is the reflection in the line $y = x$ .)
Correct theorem: a symmetric matrix is orthogonal if and only if $Q^{2} = I$ . Proof. If $Q^{T} = Q$ and $Q^{T} Q = I$ then $Q^{2} = Q^{T} Q = I$ . Conversely if $Q^{T} = Q$ and $Q^{2} = I$ then $Q^{T} Q = Q^{2} = I$ , which is the definition of orthogonal. $■$
(b) The failing step is “Taking $k$ -th roots of both sides gives $N = 0$ .” For numbers, $x^{k} = 0$ forces $x = 0$ because there are no zero divisors; matrices have zero divisors, so a power of a nonzero matrix can vanish and no root-extraction is available.
Counterexample: $N = (0010) \neq = 0$ , and $N^{2} = (0 \cdot 0 + 1 \cdot 0 0 0 \cdot 1 + 1 \cdot 0 0) = 0$ .
Correct theorem: nilpotent matrices are far from being zero in general, but they are always singular — a nilpotent matrix is never invertible. Proof. If $N^{k} = 0$ and $N$ were invertible, multiplying $N^{k} = 0$ on the left by $(N^{- 1})^{k}$ gives $I = 0$ , which is false for $n \geq 1$ . $■$
(c) The failing step is “ $(PQ)^{2} = P^{2} Q^{2}$ .” By definition $(PQ)^{2} = PQPQ$ , and rearranging the middle factors to $PPQQ$ uses commutativity, $QP = PQ$ , which is not given. This is the same error as writing $(A B)^{2} = A^{2} B^{2}$ .
Counterexample: $P = (1000)$ and $Q = \frac{1}{2} (1111)$ . Both are idempotent ( $P^{2} = P$ by inspection; $Q^{2} = \frac{1}{4} (2222) = Q$ ). Their product is $PQ = \frac{1}{2} (1010)$ , and
$(PQ)^{2} = \frac{1}{4} (1010) (1010) = \frac{1}{4} (1010) = \frac{1}{2} PQ \neq = PQ .$
So $PQ$ is not idempotent.
Correct theorem: if $P$ and $Q$ are idempotent and commute, then $PQ$ is idempotent. Proof. $(PQ)^{2} = PQPQ = P (QP) Q = P (PQ) Q = P^{2} Q^{2} = PQ$ , where the third equality is exactly the commuting hypothesis and the last uses idempotence. $■$
10.Proof
Let $A$ be an $m \times n$ matrix, so that $A^{T} A$ is square of size $n \times n$ . You may use that $v^{T} v = v_{1}^{2} + \dots + v_{m}^{2}$ for any $v \in R^{m}$ .
1. (a)Prove that the equations $A x = 0$ and $A^{T} A x = 0$ have exactly the same solutions $x \in R^{n}$ .
2. (b)Deduce that $A^{T} A$ is invertible if and only if the columns of $A$ are linearly independent. Say which statements of the Invertible Matrix Theorem you use and to which matrix you apply them.
3. (c)Let $A = 101210$ . Verify (b) in both directions for this $A$ . Then show that $A A^{T}$ is not invertible by exhibiting a nonzero $y$ with $A A^{T} y = 0$ , and explain why the same must happen for every $3 \times 2$ matrix.
Solution
(a) Two inclusions. If $A x = 0$ , then multiplying on the left by $A^{T}$ gives $A^{T} A x = A^{T} 0 = 0$ . So every solution of the first equation solves the second.
Conversely suppose $A^{T} A x = 0$ . Multiply on the left by the $1 \times n$ row $x^{T}$ : $x^{T} A^{T} A x = 0$ . By the reversal rule $x^{T} A^{T} = (A x)^{T}$ , so with $v = A x \in R^{m}$ this reads
$v^{T} v = v_{1}^{2} + v_{2}^{2} + \dots + v_{m}^{2} = 0.$
A sum of squares of real numbers is zero only when every term is zero, so $v = 0$ , that is $A x = 0$ . The two solution sets coincide. $■$
(b) Apply the theorem to the square matrix $A^{T} A$ . Statements (a) and (d) of the Invertible Matrix Theorem, applied to the $n \times n$ matrix $A^{T} A$ , say: $A^{T} A$ is invertible if and only if $A^{T} A x = 0$ has only the trivial solution. By part (a) that is the same as $A x = 0$ having only the trivial solution. Finally, $A x = x_{1} a_{1} + \dots + x_{n} a_{n}$ , so “ $A x = 0$ only for $x = 0$ ” is word for word the definition of the columns $a_{1}, \dots, a_{n}$ being linearly independent — the (d) $\Leftrightarrow$ (e) rewrite, which is valid for any shape of $A$ because it is a rewrite, not a theorem about square matrices. Chaining the three equivalences: $A^{T} A$ invertible $\Leftrightarrow$ columns of $A$ independent. $■$
Note that the theorem is applied only to $A^{T} A$ , which is square; $A$ itself may be any shape, and for $m \neq = n$ it is never invertible.
(c) Both directions, then the other product. The columns $(1, 0, 1)^{T}$ and $(2, 1, 0)^{T}$ are independent: neither is a multiple of the other (the second has a nonzero middle entry, the first does not). And
$A^{T} A = (120110) 101210 = (1 + 0 + 1 2 + 0 + 0 2 + 0 + 0 4 + 1 + 0) = (2225),$
with $det = 2 \cdot 5 - 2 \cdot 2 = 6 \neq = 0$ , so it is invertible — independent columns and invertible $A^{T} A$ , as (b) requires.
For $A A^{T}$ (size $3 \times 3$ ), first find $y \neq = 0$ with $A^{T} y = 0$ : the equations are $y_{1} + y_{3} = 0$ and $2 y_{1} + y_{2} = 0$ . Taking $y_{1} = 1$ gives $y = (1, - 2, - 1)^{T}$ . Then $A A^{T} y = A (A^{T} y) = A 0 = 0$ with $y \neq = 0$ , so statement (d) fails for $A A^{T}$ and it is singular. As a direct check,
$A A^{T} = 521210101, A A^{T} y = 5 - 4 - 1 2 - 2 + 0 1 + 0 - 1 = 0 .$
This is forced for every $3 \times 2$ matrix $A$ : $A^{T}$ is $2 \times 3$ , so $A^{T} y = 0$ is a homogeneous system of $2$ equations in $3$ unknowns. Its echelon form has at most $2$ pivots, hence at least one free variable, hence a nonzero solution $y$ ; and then $A A^{T} y = A (A^{T} y) = A 0 = 0$ by the same two-step argument. So $A A^{T}$ is always singular when $A$ has more rows than columns, while $A^{T} A$ is invertible exactly when the columns are independent — which is why least squares works with $A^{T} A$ and never with $A A^{T}$ .
11.Proof
Let $A$ be an invertible $n \times n$ matrix and $N$ a nilpotent $n \times n$ matrix with $N^{k} = 0$ , and suppose $A$ and $N$ commute: $A N = N A$ .
1. (a)Prove that $A^{- 1}$ also commutes with $N$ : $A^{- 1} N = N A^{- 1}$ .
2. (b)Let $M = A^{- 1} N$ . Prove by induction that $M^{j} = (A^{- 1})^{j} N^{j}$ for every $j \geq 1$ , and conclude that $M$ is nilpotent.
3. (c)Prove that $A - N$ is invertible, with $(A - N)^{- 1} = (I + M + M^{2} + \dots + M^{k - 1}) A^{- 1}$ .
4. (d)Apply this with $A = 2 I_{3}$ and $N = 000100010$ to compute $(2 I - N)^{- 1}$ exactly, and check the first row of the product $(2 I - N) (2 I - N)^{- 1}$ .
Solution
(a) Multiply the commuting relation by $A^{- 1}$ on both sides. Start from $A N = N A$ . Multiplying on the left by $A^{- 1}$ gives $N = A^{- 1} N A$ . Now multiply on the right by $A^{- 1}$ : $N A^{- 1} = A^{- 1} N (A A^{- 1}) = A^{- 1} N$ . $■$
(b) Induction on $j$ . The case $j = 1$ is the definition of $M$ . Suppose $M^{j} = (A^{- 1})^{j} N^{j}$ . Then
$M^{j + 1} = M^{j} M = (A^{- 1})^{j} N^{j} A^{- 1} N .$
By (a), $A^{- 1}$ commutes with $N$ , hence with $N^{j}$ (move $A^{- 1}$ past one factor of $N$ at a time, $j$ times): $N^{j} A^{- 1} = A^{- 1} N^{j}$ . So
$M^{j + 1} = (A^{- 1})^{j} A^{- 1} N^{j} N = (A^{- 1})^{j + 1} N^{j + 1},$
which completes the induction. In particular $M^{k} = (A^{- 1})^{k} N^{k} = (A^{- 1})^{k} \cdot 0 = 0$ , so $M$ is nilpotent. $■$ (Without the commuting hypothesis this fails: $M^{2} = A^{- 1} N A^{- 1} N$ cannot be regrouped.)
(c) Factor out $A$ and use the finite geometric series. Since $A$ is invertible, $A - N = A (I - A^{- 1} N) = A (I - M)$ . Let $S = I + M + M^{2} + \dots + M^{k - 1}$ . Multiplying out, the sum telescopes:
$(I - M) S = (I + M + \dots + M^{k - 1}) - (M + M^{2} + \dots + M^{k}) = I - M^{k} = I,$
and the same computation in the other order gives $S (I - M) = I$ (every factor here is a power of $M$ , and powers of one matrix commute with each other). So $I - M$ is invertible with inverse $S$ . Then $A - N = A (I - M)$ is a product of two invertible matrices, hence invertible, and by the reversal rule for inverses
$(A - N)^{- 1} = (I - M)^{- 1} A^{- 1} = (I + M + M^{2} + \dots + M^{k - 1}) A^{- 1} . ■$
(d) Specialise. $A = 2 I$ commutes with everything and $A^{- 1} = \frac{1}{2} I$ , so $M = \frac{1}{2} N$ . Powers of $N$ : $N^{2} = 000000100$ (the single $1$ moves one place further from the diagonal) and $N^{3} = 0$ , so $k = 3$ and $M^{2} = \frac{1}{4} N^{2}$ . Therefore
$(2 I - N)^{- 1} = (I + \frac{1}{2} N + \frac{1}{4} N^{2}) \frac{1}{2} I = \frac{1}{2} 100 \frac{1}{2} 10 \frac{1}{4} \frac{1}{2} 1 = \frac{1}{2} 00 \frac{1}{4} \frac{1}{2} 0 \frac{1}{8} \frac{1}{4} \frac{1}{2} .$
Check: $2 I - N = 200 - 1 20 0 - 1 2$ , and its first row $(2, - 1, 0)$ against the three columns of the candidate inverse gives $2 \cdot \frac{1}{2} = 1$ , $2 \cdot \frac{1}{4} - \frac{1}{2} = 0$ , $2 \cdot \frac{1}{8} - \frac{1}{4} = 0$ : the first row of $I$ , as required.
12.Sketch
Work in homogeneous coordinates, writing the point $(x, y)$ as $(x, y, 1)^{T}$ . Let $R$ be the $3 \times 3$ matrix of the counterclockwise rotation by $9 0^{\circ}$ about the origin, and $T$ the $3 \times 3$ matrix of the translation by $(2, 0)$ .
1. (a)Write down $R$ and $T$ , and compute both products $TR$ and $RT$ . State in words which motion each product performs, in which order.
2. (b)Let $U$ be the unit square with vertices $O = (0, 0)$ , $P = (1, 0)$ , $Q = (1, 1)$ , $S = (0, 1)$ . Compute the image of each vertex under $TR$ and under $RT$ , then draw one set of axes showing $U$ and both image squares, labelling each image vertex with the name of the vertex it came from.
3. (c)Each of $TR$ and $RT$ is a rotation by $9 0^{\circ}$ about some point. Find the fixed point of each by solving a linear system, and mark both centres on your sketch. Explain what the sketch shows about $TR$ versus $RT$ .
4. (d)Compute $(TR)^{- 1}$ by the reversal rule, without row reduction, and verify that it fixes the centre you found for $TR$ .
Solution
(a) The two matrices and their products. In homogeneous coordinates a linear map sits in the top-left $2 \times 2$ block and a translation in the last column:
$R = 010 - 1 00 001, T = 100010201 .$
Multiplying (the rightmost factor acts first):
$TR = 010 - 1 00 201, RT = 010 - 1 00 021 .$
For $TR$ the only new work is the last column: row $1$ of $T$ times column $3$ of $R$ is $1 \cdot 0 + 0 \cdot 0 + 2 \cdot 1 = 2$ . For $RT$ , row $2$ of $R$ times column $3$ of $T$ is $1 \cdot 2 + 0 + 0 = 2$ . So $TR$ is “rotate, then translate”: $(x, y) \mapsto (2 - y, x)$ . And $RT$ is “translate, then rotate”: $(x, y) \mapsto (- y, x + 2)$ . The products differ, so the two orders are different motions.
(b) Images of the vertices. Under $TR$ , $(x, y) \mapsto (2 - y, x)$ :
$O = (0, 0) \mapsto (2, 0), P = (1, 0) \mapsto (2, 1), Q = (1, 1) \mapsto (1, 1), S = (0, 1) \mapsto (1, 0) .$
Under $RT$ , $(x, y) \mapsto (- y, x + 2)$ :
$O \mapsto (0, 2), P \mapsto (0, 3), Q \mapsto (- 1, 3), S \mapsto (- 1, 2) .$
The sketch must show three unit squares on one grid. The original $U$ occupies $[0, 1] \times [0, 1]$ . The $TR$ -image occupies $[1, 2] \times [0, 1]$ , immediately to the right of $U$ and sharing the edge $x = 1$ ; its corner labels read $S$ at $(1, 0)$ , $O$ at $(2, 0)$ , $P$ at $(2, 1)$ , $Q$ at $(1, 1)$ — the labels go round in the same counterclockwise sense as on $U$ , because a rotation preserves orientation, but they have been turned a quarter turn. The $RT$ -image occupies $[- 1, 0] \times [2, 3]$ , up and to the left, with $O$ at $(0, 2)$ , $P$ at $(0, 3)$ , $Q$ at $(- 1, 3)$ , $S$ at $(- 1, 2)$ . The two image squares do not overlap; the gap between them is the whole content of $TR \neq = RT$ .
(c) Fixed points. A fixed point of $TR$ satisfies $(2 - y, x) = (x, y)$ , i.e. $x = 2 - y$ and $y = x$ , so $x = y = 1$ : the centre is $(1, 1)$ , which is the vertex $Q$ — consistent with (b), where $Q$ stayed put. So $TR$ is the rotation by $9 0^{\circ}$ about $(1, 1)$ ; the image square is $U$ spun a quarter turn about its top-right corner.
A fixed point of $RT$ satisfies $(- y, x + 2) = (x, y)$ , i.e. $x = - y$ and $y = x + 2$ , so $y = - y + 2$ , $y = 1$ , $x = - 1$ : the centre is $(- 1, 1)$ . Check: $RT$ sends $(- 1, 1)$ to $(- 1, - 1 + 2) = (- 1, 1)$ . So $RT$ is the rotation by $9 0^{\circ}$ about $(- 1, 1)$ , a point outside $U$ , which is why the image landed far away.
Mark both centres. The sketch shows that the two composites are the same kind of motion — a quarter turn — about different centres, two units apart. That is the geometry behind non-commutativity here: which operation runs first changes where the rotation centre ends up, not the amount of rotation.
(d) Reverse the order and invert each factor. By the reversal rule $(TR)^{- 1} = R^{- 1} T^{- 1}$ . Undoing a translation by $(2, 0)$ is translating by $(- 2, 0)$ , and undoing a $9 0^{\circ}$ rotation is rotating by $- 9 0^{\circ}$ :
$T^{- 1} = 100010 - 2 01, R^{- 1} = R^{T} = 0 - 1 0 100001$
(the columns of $R$ are orthonormal, so $R$ is orthogonal and its inverse is its transpose), and
$(TR)^{- 1} = R^{- 1} T^{- 1} = 0 - 1 0 100021,$
the $(2, 3)$ entry being $- 1 \cdot (- 2) + 0 + 0 = 2$ . Check that it is the inverse: $(TR) (R^{- 1} T^{- 1})$ has first row $(0, - 1, 2)$ times the columns $(0, - 1, 0)^{T}$ , $(1, 0, 0)^{T}$ , $(0, 2, 1)^{T}$ , giving $(1, 0, - 2 + 2) = (1, 0, 0)$ ; the other rows work the same way, so the product is $I_{3}$ . As a map, $(TR)^{- 1}$ sends $(x, y) \mapsto (y, 2 - x)$ , and at the centre $(1, 1)$ this gives $(1, 2 - 1) = (1, 1)$ : the inverse fixes the same point, as it must — undoing a rotation about $(1, 1)$ is rotating the other way about $(1, 1)$ .

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