Chapter 3 · Matrices

Exercise 3.3 · Q9: Find $\frac12(A+A')$ and $\frac12(A-A')$

Exercise 3.3Symmetric and skew symmetric matrices · Section 3.6

The question

Find $\frac{1}{2}(A + A')$ and $\frac{1}{2}(A - A')$ ⁠, when

A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}.

Exam answer

A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}

A + A' = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}

$\therefore\ \tfrac12(A + A') = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$

A - A' = \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}

$\therefore\ \tfrac12(A - A') = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$

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

The one idea this problem is built on

Look at $A$ before calculating: its diagonal is zero and each mirror pair is opposite ( $a$ and $-a$ ⁠, $b$ and $-b$ ⁠, $c$ and $-c$ ⁠). So $A$ is already skew symmetric: $A' = -A$ ⁠. Then $A + A' = A - A = O$ and $A - A' = A + A = 2A$ ⁠. Halving gives $O$ and $A$ ⁠. In the language of Theorem 2, the symmetric part of $A$ is zero and the skew symmetric part is all of $A$ ⁠.

Given, to find, and the plan

Given: A $3\times3$ matrix $A$ whose entries involve arbitrary numbers $a, b, c$ ⁠.
To find: $\tfrac12(A+A')$ and $\tfrac12(A-A')$ ⁠.
Plan: (1) Write $A'$ ⁠. (2) Compute $A + A'$ entry by entry and halve it. (3) Compute $A - A'$ entry by entry and halve it. (4) Interpret: $A$ is skew symmetric.

Basics you need

1The transpose

$(A')_{ij} = a_{ji}$ ⁠: row $i$ of $A$ becomes column $i$ of $A'$ ⁠. Basics: mirror patterns

2Sums, differences and scalar multiples

Same-order matrices add and subtract entry by entry; $\tfrac12 X$ halves every entry of $X$ ⁠. Basics: adding only like with like

3Skew symmetric and the split of Theorem 2

$A$ is skew symmetric if $A' = -A$ ⁠. Every square matrix is $\tfrac12(A+A')$ (symmetric) plus $\tfrac12(A-A')$ (skew symmetric). Lesson: symmetric and skew symmetric

Solution

Write $A'$

$A$ is $3\times3$ ⁠, so $A'$ is $3\times3$ and $A \pm A'$ are defined. The rows of $A$ are $0, a, b$ ⁠; $-a, 0, c$ ⁠; $-b, -c, 0$ ⁠. As columns:
$A' = \begin{bmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{bmatrix}.$
Comparing with $A$ ⁠: every entry of $A'$ is the negative of the entry of $A$ in the same place. So $A' = -A$ ⁠, and $A$ is skew symmetric. We still compute the sums in full, as the question asks.

Transpose, Definition 3 (Basic 1). Lesson: transpose
Compute $A + A'$ ⁠, then halve
$\begin{aligned} &A + A' \\ &= \begin{bmatrix} 0+0 & a+(-a) & b+(-b) \\ -a+a & 0+0 & c+(-c) \\ -b+b & -c+c & 0+0 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O. \end{aligned}$
Half of the zero matrix is still the zero matrix:
$\tfrac12(A + A') = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}.$
Each entry is a number plus its own negative, so it is 0 whatever $a, b, c$ are.

Sum entry by entry, Section 3.4.1; scalar multiple, Section 3.4.2 (Basic 2).
Compute $A - A'$ ⁠, then halve
$\begin{aligned} &A - A' \\ &= \begin{bmatrix} 0-0 & a-(-a) & b-(-b) \\ -a-a & 0-0 & c-(-c) \\ -b-b & -c-c & 0-0 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 0 & 2a & 2b \\ -2a & 0 & 2c \\ -2b & -2c & 0 \end{bmatrix}. \end{aligned}$
Halving every entry:
$\tfrac12(A - A') = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} = A.$
Subtracting a negative adds: $a - (-a) = 2a$ ⁠. So $A - A' = 2A$ ⁠, and half of it is $A$ itself.
Interpret the answer

By Theorem 2, $A = \tfrac12(A+A') + \tfrac12(A-A')$ ⁠, with the first part symmetric and the second skew symmetric. Here the symmetric part is $O$ and the skew symmetric part is $A$ ⁠: $A = O + A$ ⁠. That is exactly what we expect, because $A$ is skew symmetric to begin with (Step 1).

The zero matrix counts as symmetric ( $O' = O$ ⁠), so the split still has the form Theorem 2 promises. In general, if $A' = -A$ ⁠, then $A + A' = O$ and $A - A' = 2A$ ⁠, so the answer would be $O$ and $A$ for any skew symmetric matrix.

Theorem 2, Section 3.6 (Basic 3). Lesson: symmetric and skew symmetric

Result

\tfrac12(A + A') = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix},

\tfrac12(A - A') = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix} = A.

(Textbook answer: the same two matrices.)

Check it with numbers

Take $a = 2$ ⁠, $b = -3$ ⁠, $c = 5$ ⁠. Then $a_{12} = 2$ and $a_{21} = -2$ ⁠; in $A + A'$ the (1,2) entry is $2 + (-2) = 0$ ⁠, and in $A - A'$ it is $2 - (-2) = 4$ ⁠, whose half is $2 = a_{12}$ ⁠. ✓ The (3,2) entry of $A - A'$ is $-5 - 5 = -10$ ⁠, whose half is $-5 = a_{32}$ ⁠. ✓

Common mistakes

Sign slips. $a - (-a) = 2a$ ⁠, not 0; $-a - a = -2a$ ⁠.
Writing the number 0 instead of the zero matrix. $\tfrac12(A+A')$ is the $3\times3$ zero matrix $O$ ⁠.
Forgetting to halve. $A - A' = 2A$ ⁠; the question asks for half of it.

What to take away

If $A$ is skew symmetric, then $A + A' = O$ and $A - A' = 2A$ ⁠.
Recognising the structure first (zero diagonal, opposite mirror pairs) tells you the answer before computing.
In Theorem 2's split, a skew symmetric matrix has zero symmetric part; a symmetric matrix has zero skew part.

Every fact used, and where it comes from

Fact	Used in	Source
Transpose $(A')_{ij} = a_{ji}$	Step 1	Definition 3, Section 3.5 (Lesson: transpose)
Sum, difference, scalar multiple entry by entry	Steps 2, 3	Sections 3.4.1, 3.4.2
Skew symmetric: $A' = -A$	Steps 1, 4	Definition 5, Section 3.6 (Lesson)
$A = \tfrac12(A+A') + \tfrac12(A-A')$ ⁠, symmetric plus skew	Step 4	Theorem 2, Section 3.6

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Exercise 3.3 · Q9: Find 12(A+A′)\frac12(A+A')21​(A+A′) and 12(A−A′)\frac12(A-A')21​(A−A′)

Basics you need

Solution

Write A′A'A′

Compute A+A′A + A'A+A′⁠, then halve

Compute A−A′A - A'A−A′⁠, then halve

Interpret the answer

Every fact used, and where it comes from

Exercise 3.3 · Q9: Find $\frac12(A+A')$ and $\frac12(A-A')$

Write $A'$

Compute $A + A'$ ⁠, then halve

Compute $A - A'$ ⁠, then halve