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Chapter 3 · Matrices

Exercise 3.3 · Q8: A+A′A+A' is symmetric and A−A′A-A' is skew symmetric

Exercise 3.3Symmetric and skew symmetric matrices · Section 3.6
The question

For the matrix

A=[1567],A = \begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix},

verify that

(i) (A+A′)(A + A') is a symmetric matrix
(ii) (A−A′)(A - A') is a skew symmetric matrix

Exam answer A′=[1657]A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}

(i)

A+A′=[2111114]A + A' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} (A+A′)′=[2111114]=A+A′(A + A')' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} = A + A'

So A+A′A + A' is a symmetric matrix.

(ii)

A−A′=[0−110]A - A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} (A−A′)′=[01−10]=−(A−A′)\begin{aligned} (A - A')' &= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \\ &= -(A - A') \end{aligned}

So A−A′A - A' is a skew symmetric matrix.

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

The one idea this problem is built on

AA itself is neither symmetric nor skew symmetric: its off-diagonal entries are 5 and 6. Adding A′A' puts 5+65 + 6 in both off-diagonal places, so the mirror pairs become equal. Subtracting A′A' gives 5−65 - 6 in one place and 6−56 - 5 in the other, so the mirror pairs become opposite, and the diagonal entries cancel to 0. This is Theorem 1 of the chapter, seen in one example.

Given, to find, and the plan
Given
A 2×22\times2 matrix AA⁠.
To show
(i) (A+A′)′=A+A′(A+A')' = A + A'⁠; (ii) (A−A′)′=−(A−A′)(A-A')' = -(A-A')⁠.
Plan
(1) Write A′A'⁠. (2) Compute A+A′A+A'⁠, transpose it, compare. (3) Compute A−A′A-A'⁠, transpose it, compare with its negative. (4) Explain why this always happens.

Basics you need

1The transpose

(A′)ij=aji(A')_{ij} = a_{ji}⁠. For a 2×22\times2 matrix the off-diagonal entries swap. Basics: mirror patterns

2Symmetric and skew symmetric

Square BB is symmetric if B′=BB' = B⁠, and skew symmetric if B′=−BB' = -B⁠. Lesson: symmetric and skew symmetric

3Adding and subtracting

Matrices of the same order are added or subtracted entry by entry. Basics: adding only like with like

Solution

  1. Write A′A'

    Rows 1,51, 5 and 6,76, 7 become columns:

    A′=[1657].A' = \begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}.

    AA and A′A' are both 2×22\times2⁠, so A+A′A + A' and A−A′A - A' are defined and are 2×22\times2 (square, so symmetry can be tested).

    Transpose, Definition 3 (Basic 1).

  2. (i) Compute A+A′A + A'

    A+A′=[1+15+66+57+7]=[2111114].\begin{aligned} A + A' &= \begin{bmatrix} 1+1 & 5+6 \\ 6+5 & 7+7 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}. \end{aligned}
    The two off-diagonal entries are 5+65+6 and 6+56+5⁠. Addition of numbers is commutative, so they are the same number, 11. That is the whole reason A+A′A+A' comes out symmetric.
    Sum entry by entry, Section 3.4.1 (Basic 3).

  3. (i) Transpose and compare

    Let P=A+A′P = A + A'⁠. Swapping its off-diagonal entries (both 11) changes nothing:

    P′=[2111114]=P.P' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix} = P.

    So P′=PP' = P and A+A′A + A' is symmetric.

    Definition 4: symmetric means P′=PP' = P⁠. Lesson: symmetric matrices

  4. (ii) Compute A−A′A - A'

    A−A′=[1−15−66−57−7]=[0−110].\begin{aligned} A - A' &= \begin{bmatrix} 1-1 & 5-6 \\ 6-5 & 7-7 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}. \end{aligned}
    Each diagonal entry is a number minus itself, so it is 0. The off-diagonal entries are 5−6=−15-6 = -1 and 6−5=16-5 = 1⁠: the same subtraction done in opposite orders, which gives opposite numbers.

  5. (ii) Transpose and compare with the negative

    Let Q=A−A′Q = A - A'⁠. Then

    Q′=[01−10],−Q=[01−10].Q' = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \qquad -Q = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}.

    So Q′=−QQ' = -Q and A−A′A - A' is skew symmetric.

    Definition 5: skew symmetric means Q′=−QQ' = -Q⁠. Lesson: skew symmetric matrices

  6. Why it works for every square matrix

    For any square matrix AA⁠, using the rules of Section 3.5.1:

    (A+A′)′=A′+(A′)′=A′+A=A+A′,(A−A′)′=A′−(A′)′=A′−A=−(A−A′).\begin{aligned} (A + A')' &= A' + (A')' \\ &= A' + A = A + A', \\[1mm] (A - A')' &= A' - (A')' \\ &= A' - A = -(A - A'). \end{aligned}

    The first line says A+A′A + A' is always symmetric, the second that A−A′A - A' is always skew symmetric.

    The computation in Steps 2 to 5 checks one matrix; these two lines prove the statement for all square matrices. It is Theorem 1 of Section 3.6. The facts used are (X+Y)′=X′+Y′(X+Y)' = X'+Y'⁠, (A′)′=A(A')' = A⁠, and that matrix addition is commutative.
    Theorem 1, Section 3.6; properties (i), (iii) of Section 3.5.1. Lesson: symmetric and skew symmetric
Result A+A′=[2111114],A + A' = \begin{bmatrix} 2 & 11 \\ 11 & 14 \end{bmatrix}, A−A′=[0−110].A - A' = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.

The first equals its transpose, so it is symmetric; the transpose of the second is its negative, so it is skew symmetric. Both are verified.

Check it with numbers

Half the sum plus half the difference must give back AA⁠: 12(2)+12(0)=1\tfrac12(2) + \tfrac12(0) = 1⁠, 12(11)+12(−1)=5\tfrac12(11) + \tfrac12(-1) = 5⁠, 12(11)+12(1)=6\tfrac12(11) + \tfrac12(1) = 6⁠, 12(14)+12(0)=7\tfrac12(14) + \tfrac12(0) = 7⁠. These are the entries of AA⁠. ✓ (This split is exactly what Question 10 asks for.)

Common mistakes
  1. Transposing AA wrongly. Only the off-diagonal entries 5 and 6 swap; the diagonal 1, 7 stays.
  2. Comparing Q′Q' with QQ instead of −Q-Q⁠. Skew symmetric means Q′=−QQ' = -Q⁠.
  3. Stopping at the sums. "Verify symmetric" needs the transpose to be written and compared.
What to take away
  1. For every square AA⁠: A+A′A + A' is symmetric and A−A′A - A' is skew symmetric (Theorem 1).
  2. Off-diagonal: aij+ajia_{ij} + a_{ji} is the same in both mirror places; aij−ajia_{ij} - a_{ji} and aji−aija_{ji} - a_{ij} are opposite.
  3. 12(A+A′)+12(A−A′)=A\tfrac12(A + A') + \tfrac12(A - A') = A⁠, the starting point of Theorem 2.

Every fact used, and where it comes from

FactUsed inSource
Transpose (A′)ij=aji(A')_{ij} = a_{ji}Steps 1, 3, 5Definition 3, Section 3.5 (Lesson: transpose)
Sum and difference entry by entrySteps 2, 4Sections 3.4.1, 3.4.2
Symmetric P′=PP' = P⁠; skew symmetric Q′=−QQ' = -QSteps 3, 5Definitions 4 and 5, Section 3.6 (Lesson)
(A′)′=A(A')' = A⁠, (A+B)′=A′+B′(A+B)' = A'+B'Step 6Section 3.5.1, properties (i), (iii)
A+A′A + A' symmetric, A−A′A - A' skew symmetricStep 6Theorem 1, Section 3.6
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