Chapter 3 · Matrices

Miscellaneous · Q2: $B'AB$ symmetric or skew symmetric

Miscellaneous ExerciseSymmetric and skew symmetric matrices · Section 3.6

The question

Show that the matrix B′AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Exam answer

Let $A$ be of order $n$ and $B$ of order $n\times m$ ⁠. Then $B'AB$ is a square matrix of order $m$ ⁠.

Using $(XYZ)' = Z'Y'X'$ and $(B')' = B$ ⁠,

\begin{aligned} (B'AB)' &= B'A'(B')' \\ &= B'A'B \quad\ldots(\text{i}) \end{aligned}

Case 1. $A$ is symmetric, so $A' = A$ ⁠. By (i),

(B'AB)' = B'AB

So $B'AB$ is symmetric.

Case 2. $A$ is skew symmetric, so $A' = -A$ ⁠. By (i),

\begin{aligned} (B'AB)' &= B'(-A)B \\ &= -(B'AB) \end{aligned}

So $B'AB$ is skew symmetric.

Hence $B'AB$ is symmetric or skew symmetric according as $A$ is symmetric or skew symmetric.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Transpose the whole product $B'AB$ ⁠. The reversal rule turns it into $B'A'(B')'$ ⁠, and $(B')' = B$ ⁠, so the outer factors $B'$ and $B$ come back exactly where they were. Only the middle factor changes, from A to $A'$ ⁠. So whatever A does under transposing (stays the same, or changes sign), $B'AB$ does too.

Given, to show, and the plan

Given: A is a square matrix, say $n\times n$ (symmetric and skew symmetric matrices are square). B is any matrix for which $B'AB$ is defined.
To show: (a) If $A' = A$ ⁠, then $(B'AB)' = B'AB$ ⁠. (b) If $A' = -A$ ⁠, then $(B'AB)' = -B'AB$ ⁠.
Plan: (1) Work out the order of B and of $B'AB$ ⁠. (2) Extend the rule $(XY)' = Y'X'$ to three factors. (3) Transpose $B'AB$ once, for any A. (4) Put in $A' = A$ ⁠, then $A' = -A$ ⁠. (5) Check with numbers.

Basics you need

1Transpose and its rules

$(A')' = A$ ⁠, $(kA)' = kA'$ ⁠, and $(AB)' = B'A'$ ⁠: the transpose of a product is the product of the transposes in reverse order. Lesson: transpose

2Symmetric and skew symmetric

Square matrix with $A' = A$ (symmetric) or $A' = -A$ (skew symmetric). Lesson: symmetric and skew symmetric matrices

3Associativity

$(XY)Z = X(YZ)$ whenever both sides are defined, so a product of three matrices can be written $XYZ$ without brackets. Lesson: properties of the operations

4Orders in a product

$(m\times n)(n\times p)$ is defined and gives $m\times p$ ⁠. Basics: pair up, multiply, add

Solution

Orders: what B can be, and what $B'AB$ is

Let A be $n\times n$ and B be $n\times m$ for some $m$ ⁠. Then $B'$ is $m\times n$ ⁠, and:
- $B'A$ ⁠: $(m\times n)(n\times n)$ ⁠, defined, order $m\times n$ ⁠;
- $(B'A)B$ ⁠: $(m\times n)(n\times m)$ ⁠, defined, order $m\times m$ ⁠.
So $B'AB$ is a square matrix of order $m$ ⁠, and B itself need not be square. (If B had a number of rows other than $n$ ⁠, the product $AB$ would not exist.)

Symmetric and skew symmetric are only defined for square matrices, so before we can ask whether $B'AB$ is one of them, we must know it is square. The order count shows it always is: the $m$ rows come from $B'$ and the $m$ columns from B.

Order of a product: rows of the first factor by columns of the last. Basics: pair up, multiply, add
The reversal rule for three factors

For matrices X, Y, Z whose product is defined, group the product as $(XY)Z$ and apply $(PQ)' = Q'P'$ twice:
$\begin{aligned} (XYZ)' &= \big((XY)Z\big)' \\ &= Z'(XY)' \\ &= Z'Y'X'. \end{aligned}$
The book states the rule only for two factors. The first use treats $XY$ as a single matrix P and Z as Q; the second applies the rule to $XY$ itself. Associativity lets us drop the brackets at the end. The pattern is the same as taking off socks and shoes: the last thing put on is the first taken off.

Property (iv) of Section 3.5.1, and the associative law of Section 3.4.6. Lesson: properties of the operations
Transpose $B'AB$ once, for any square A

Use Step 2 with $X = B'$ ⁠, $Y = A$ ⁠, $Z = B$ ⁠:
$\begin{aligned} (B'AB)' &= B'A'(B')' \\ &= B'A'B. \end{aligned}$
In the first line, $Z' = B'$ goes in front, $Y' = A'$ stays in the middle, and $X' = (B')'$ goes at the back. The second line uses $(B')' = B$ ⁠: transposing twice turns rows into columns and back again. This one identity covers both halves of the question, because so far we have used nothing about A.

Property (i) of Section 3.5.1: $(A')' = A$ ⁠. Lesson: transpose
Case 1: A is symmetric

If $A' = A$ ⁠, substitute into Step 3:
$(B'AB)' = B'A'B = B'AB.$
$B'AB$ is square (Step 1) and equals its own transpose, so $B'AB$ is symmetric.

Only the middle letter changed under transposing, and for a symmetric A that change does nothing.

Definition 4 (Section 3.6). Lesson: symmetric and skew symmetric matrices
Case 2: A is skew symmetric

If $A' = -A$ ⁠, substitute into Step 3:
$\begin{aligned} (B'AB)' &= B'(-A)B \\ &= -(B'AB). \end{aligned}$
$B'AB$ is square and its transpose is its negative, so $B'AB$ is skew symmetric.

A scalar factor can be moved to the front of a product: $B'\big((-1)A\big)B = (-1)\,B'AB$ ⁠, because each entry of the product is a sum of terms, each containing exactly one entry of the middle factor, so the factor $-1$ comes out of every term.

Definition 5 (Section 3.6); $k(XY) = (kX)Y = X(kY)$ for a scalar $k$ ⁠. Lesson: symmetric and skew symmetric matrices

Result

For any B with $B'AB$ defined, $(B'AB)' = B'A'B$ ⁠. Hence $B'AB$ is symmetric when A is symmetric, and skew symmetric when A is skew symmetric. (Proofs have no entry in the answer key.)

Check it with numbers

Take

B = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}, \quad B' = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}

Symmetric A. With the symmetric matrix below, the entries of $B'A$ are

A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}

\begin{aligned} (1,1) &: 1\cdot1 + 0\cdot2 = 1 \\ (1,2) &: 1\cdot2 + 0\cdot3 = 2 \\ (2,1) &: 2\cdot1 + 1\cdot2 = 4 \\ (2,2) &: 2\cdot2 + 1\cdot3 = 7 \end{aligned}

so $B'A$ has rows $(1, 2)$ and $(4, 7)$ ⁠. Then the entries of $(B'A)B$ are

\begin{aligned} (1,1) &: 1\cdot1 + 2\cdot0 = 1 \\ (1,2) &: 1\cdot2 + 2\cdot1 = 4 \\ (2,1) &: 4\cdot1 + 7\cdot0 = 4 \\ (2,2) &: 4\cdot2 + 7\cdot1 = 15 \end{aligned}

B'AB = \begin{bmatrix} 1 & 4 \\ 4 & 15 \end{bmatrix},

which is symmetric (4 opposite 4) ✓.

Skew symmetric A. With the skew symmetric matrix below, the entries of $B'K$ are

K = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}

\begin{aligned} (1,1) &: 1\cdot0 + 0\cdot(-1) = 0 \\ (1,2) &: 1\cdot1 + 0\cdot0 = 1 \\ (2,1) &: 2\cdot0 + 1\cdot(-1) = -1 \\ (2,2) &: 2\cdot1 + 1\cdot0 = 2 \end{aligned}

so $B'K$ has rows $(0, 1)$ and $(-1, 2)$ ⁠. Then the entries of $(B'K)B$ are

\begin{aligned} (1,1) &: 0\cdot1 + 1\cdot0 = 0 \\ (1,2) &: 0\cdot2 + 1\cdot1 = 1 \\ (2,1) &: (-1)\cdot1 + 2\cdot0 = -1 \\ (2,2) &: (-1)\cdot2 + 2\cdot1 = 0 \end{aligned}

B'KB = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},

which is skew symmetric (zero diagonal, 1 opposite $-1$ ⁠) ✓. The verification script repeats this for 300 random cases, including non-square B (for example a $2\times3$ B, which gives $3\times3$ results).

Common mistakes

Transposing without reversing, writing $(B'AB)' = (B')'A'B' = BA'B'$ ⁠. That is a different matrix (often of a different order) and the argument collapses.
Forgetting $(B')' = B$ , and leaving $(B')'$ in the answer, so the result does not visibly match $B'AB$ ⁠.
Assuming B is square or symmetric. Nothing about B is needed beyond the product being defined.
Proving only one case. The words "according as" ask for both: symmetric gives symmetric, skew gives skew.

What to take away

$(XYZ)' = Z'Y'X'$ ⁠: the reversal rule extends to any number of factors.
Sandwiching A between $B'$ and B preserves its type. In particular $B'B = B'IB$ is always symmetric, since I is symmetric.
Do the transpose once for a general A, then substitute each case. That is shorter and clearer than two separate proofs.

Every fact used, and where it comes from

Fact	Used in	Source
Order of a product	Step 1	Section 3.4.5 (Lesson: matrix multiplication)
$(A')' = A$ ⁠, $(AB)' = B'A'$	Steps 2, 3	Section 3.5.1, properties (i) and (iv) (Lesson: transpose)
Associative law $(AB)C = A(BC)$	Step 2	Section 3.4.6 (Lesson: properties of the operations)
A scalar factor moves through a product	Step 5	Follows from the entry formula of Section 3.4.5
Definitions of symmetric and skew symmetric	Steps 4, 5	Definitions 4 and 5, Section 3.6 (Lesson: symmetric and skew symmetric)
All matrices in the check; random tests	Check	Recomputed with sympy in the site's verification script

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Miscellaneous · Q2: B′ABB'ABB′AB symmetric or skew symmetric

Basics you need

Solution

Orders: what B can be, and what B′ABB'ABB′AB is

The reversal rule for three factors

Transpose B′ABB'ABB′AB once, for any square A

Case 1: A is symmetric

Case 2: A is skew symmetric

Every fact used, and where it comes from

Miscellaneous · Q2: $B'AB$ symmetric or skew symmetric

Orders: what B can be, and what $B'AB$ is

Transpose $B'AB$ once, for any square A