Chapter 3 · Matrices

Exercise 3.3 · Q10: Symmetric plus skew symmetric, four matrices

Exercise 3.3Symmetric and skew symmetric matrices · Section 3.6

The question

Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

(i)

\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}

(ii)

\begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix}

(iii)

\begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix}

(iv)

\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}

Exam answer

For a square matrix $A$ ⁠, $A = P + Q$ ⁠, where $P = \tfrac12(A + A')$ is symmetric and $Q = \tfrac12(A - A')$ is skew symmetric.

(i)

A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix},\ A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}

\begin{aligned} P &= \tfrac12(A + A') \\ &= \tfrac12\begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} \end{aligned}

$P' = P$ ⁠, so $P$ is symmetric.

\begin{aligned} Q &= \tfrac12(A - A') \\ &= \tfrac12\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \end{aligned}

$Q' = -Q$ ⁠, so $Q$ is skew symmetric.

$\therefore\ A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$

(ii)

A' = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} = A

So $P = \tfrac12(A + A') = A$ ⁠, which is symmetric, and $Q = \tfrac12(A - A') = O$ ⁠, which is skew symmetric as $O' = -O$ ⁠.

$\begin{aligned} \therefore\ A &= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} \\ &\quad + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{aligned}$

(iii)

A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}

A + A' = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}

\begin{aligned} P &= \tfrac12(A + A') \\ &= \begin{bmatrix} 3 & \tfrac12 & -\tfrac52 \\ \tfrac12 & -2 & -2 \\ -\tfrac52 & -2 & 2 \end{bmatrix} \end{aligned}

$P' = P$ ⁠, so $P$ is symmetric.

A - A' = \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix}

\begin{aligned} Q &= \tfrac12(A - A') \\ &= \begin{bmatrix} 0 & \tfrac52 & \tfrac32 \\ -\tfrac52 & 0 & 3 \\ -\tfrac32 & -3 & 0 \end{bmatrix} \end{aligned}

$Q' = -Q$ ⁠, so $Q$ is skew symmetric.

$\begin{aligned} \therefore\ A &= \begin{bmatrix} 3 & \tfrac12 & -\tfrac52 \\ \tfrac12 & -2 & -2 \\ -\tfrac52 & -2 & 2 \end{bmatrix} \\ &\quad + \begin{bmatrix} 0 & \tfrac52 & \tfrac32 \\ -\tfrac52 & 0 & 3 \\ -\tfrac32 & -3 & 0 \end{bmatrix} \end{aligned}$

(iv)

A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix},\ A' = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}

\begin{aligned} P &= \tfrac12(A + A') \\ &= \tfrac12\begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} \end{aligned}

$P' = P$ ⁠, so $P$ is symmetric.

\begin{aligned} Q &= \tfrac12(A - A') \\ &= \tfrac12\begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix} \end{aligned}

$Q' = -Q$ ⁠, so $Q$ is skew symmetric.

$\therefore\ A = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}$

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

The one idea this problem is built on

Theorem 2 gives a recipe that works for every square matrix:

A = \underbrace{\tfrac12(A + A')}_{P\ \text{symmetric}} + \underbrace{\tfrac12(A - A')}_{Q\ \text{skew symmetric}}.

In words: the symmetric part $P$ replaces each mirror pair by its average, $p_{ij} = \tfrac12(a_{ij} + a_{ji})$ ⁠, and the skew part $Q$ holds half the difference, $q_{ij} = \tfrac12(a_{ij} - a_{ji})$ ⁠. Adding them gives back $a_{ij}$ ⁠. This split is also the only one possible (see the last step).

Given, to find, and the plan

Given: Four square matrices: (i), (iv) of order $2\times2$ ⁠; (ii), (iii) of order $3\times3$ ⁠.
To find: For each, a symmetric $P$ and a skew symmetric $Q$ with $A = P + Q$ ⁠.
Plan: For each part: (1) write $A'$ ⁠; (2) compute $A + A'$ and $P = \tfrac12(A+A')$ ⁠, check $P' = P$ ⁠; (3) compute $A - A'$ and $Q = \tfrac12(A-A')$ ⁠, check $Q' = -Q$ ⁠; (4) check $P + Q = A$ ⁠.

Basics you need

1The transpose

$(A')_{ij} = a_{ji}$ ⁠: rows become columns. Basics: mirror patterns

2Sums, differences and halves

Same-order matrices combine entry by entry; $\tfrac12X$ halves every entry. Basics: adding only like with like

3Symmetric and skew symmetric

$P' = P$ (mirror pairs equal); $Q' = -Q$ (mirror pairs opposite, diagonal zero). Lesson: symmetric and skew symmetric

4Theorems 1 and 2

$A + A'$ is symmetric and $A - A'$ is skew symmetric, so $P$ and $Q$ above always have the right type. Example 22 of the textbook does this for one $3\times3$ matrix. Example 22

Part (i)

(i) Transpose, sum and difference
$A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}, \qquad A' = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}.$ $\begin{aligned} A + A' &= \begin{bmatrix} 3+3 & 5+1 \\ 1+5 & -1+(-1) \end{bmatrix} \\[1mm] &= \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}, \\[2mm] A - A' &= \begin{bmatrix} 3-3 & 5-1 \\ 1-5 & -1-(-1) \end{bmatrix} \\[1mm] &= \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}. \end{aligned}$
All four matrices are $2\times2$ ⁠, so every sum and difference is defined.

Transpose (Basic 1); entry-by-entry sums (Basic 2).
(i) Halve to get $P$ and $Q$ ⁠, and check their types
$\begin{aligned} P &= \tfrac12(A + A') = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}, \\[1mm] Q &= \tfrac12(A - A') = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}. \end{aligned}$
$P$ ⁠: its mirror pair is $3, 3$ ⁠, so $P' = P$ ⁠, symmetric. $Q$ ⁠: diagonal 0, mirror pair $2, -2$ ⁠, so $Q' = -Q$ ⁠, skew symmetric.
(i) Check $P + Q = A$
$\begin{aligned} P + Q &= \begin{bmatrix} 3+0 & 3+2 \\ 3+(-2) & -1+0 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = A. \end{aligned}$
The sum is the original matrix. ✓

Part (ii)

(ii) Transpose: $A$ is already symmetric

The rows of $A$ are $6, -2, 2$ ⁠; $-2, 3, -1$ ⁠; $2, -1, 3$ ⁠. Written as columns they give
$A' = \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} = A,$
because each mirror pair is equal: $a_{12} = a_{21} = -2$ ⁠, $a_{13} = a_{31} = 2$ ⁠, $a_{23} = a_{32} = -1$ ⁠.

Noticing this saves work: if $A' = A$ ⁠, then $A + A' = 2A$ and $A - A' = O$ ⁠.
(ii) $P$ and $Q$
$\begin{aligned} P &= \tfrac12(A + A') = \tfrac12(2A) = A, \\ Q &= \tfrac12(A - A') = \tfrac12\,O = O. \end{aligned}$
So
$\begin{aligned} A &= \begin{bmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{bmatrix} \\[1mm] &\quad + \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \end{aligned}$
$P = A$ is symmetric (Part (ii), Step 1), and the zero matrix is skew symmetric because $O' = O = -O$ ⁠.

The zero matrix is both symmetric and skew symmetric. That is why a symmetric matrix can still be written as "symmetric + skew symmetric".

Part (iii)

(iii) Transpose

Rows $3, 3, -1$ ⁠; $-2, -2, 1$ ⁠; $-4, -5, 2$ become columns:
$A' = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix}.$
(iii) $A + A'$ and $P$

The $(i,j)$ entry of $A + A'$ is $a_{ij} + a_{ji}$ ⁠. Both entries of a mirror pair are the same sum, so it is enough to compute the diagonal and one entry of each pair:
$\begin{aligned} (1,1)&:\ 3 + 3 = 6, \\ (2,2)&:\ -2 + (-2) = -4, \\ (3,3)&:\ 2 + 2 = 4, \\ (1,2), (2,1)&:\ 3 + (-2) = 1, \\ (1,3), (3,1)&:\ -1 + (-4) = -5, \\ (2,3), (3,2)&:\ 1 + (-5) = -4. \end{aligned}$ $A + A' = \begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix}.$ $P = \tfrac12(A + A') = \begin{bmatrix} 3 & \tfrac12 & -\tfrac52 \\ \tfrac12 & -2 & -2 \\ -\tfrac52 & -2 & 2 \end{bmatrix}.$
Mirror pairs of $P$ ⁠: $\tfrac12, \tfrac12$ ⁠; $-\tfrac52, -\tfrac52$ ⁠; $-2, -2$ ⁠. All equal, so $P' = P$ ⁠: symmetric.

Each off-diagonal entry of $P$ is the average of a mirror pair of $A$ ⁠; for instance $p_{12} = \tfrac12(a_{12} + a_{21})$ $= \tfrac12(3 + (-2)) = \tfrac12$ ⁠. Averages of the same two numbers are the same, so $P$ is symmetric.
(iii) $A - A'$ and $Q$

The $(i,j)$ entry of $A - A'$ is $a_{ij} - a_{ji}$ ⁠. On the diagonal this is $a_{ii} - a_{ii} = 0$ ⁠. Off the diagonal, swapping $i$ and $j$ reverses the subtraction, so the two entries of a mirror pair are opposite:
$\begin{aligned} (1,2)&:\ 3 - (-2) = 5, \\ (2,1)&:\ -2 - 3 = -5, \\ (1,3)&:\ -1 - (-4) = 3, \\ (3,1)&:\ -4 - (-1) = -3, \\ (2,3)&:\ 1 - (-5) = 6, \\ (3,2)&:\ -5 - 1 = -6. \end{aligned}$ $A - A' = \begin{bmatrix} 0 & 5 & 3 \\ -5 & 0 & 6 \\ -3 & -6 & 0 \end{bmatrix}.$ $Q = \tfrac12(A - A') = \begin{bmatrix} 0 & \tfrac52 & \tfrac32 \\ -\tfrac52 & 0 & 3 \\ -\tfrac32 & -3 & 0 \end{bmatrix}.$
Diagonal of $Q$ is 0; mirror pairs $\tfrac52, -\tfrac52$ ⁠; $\tfrac32, -\tfrac32$ ⁠; $3, -3$ are opposite. So $Q' = -Q$ ⁠: skew symmetric.
(iii) Check $P + Q = A$

Row by row:
- row 1: $3 + 0 = 3$ ⁠, $\tfrac12 + \tfrac52 = 3$ ⁠, $-\tfrac52 + \tfrac32 = -1$ ⁠;
- row 2: $\tfrac12 - \tfrac52 = -2$ ⁠, $-2 + 0 = -2$ ⁠, $-2 + 3 = 1$ ⁠;
- row 3: $-\tfrac52 - \tfrac32 = -4$ ⁠, $-2 - 3 = -5$ ⁠, $2 + 0 = 2$ ⁠.
These are the rows of $A$ ⁠. ✓

Part (iv)

(iv) Transpose, sum and difference
$A = \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}, \qquad A' = \begin{bmatrix} 1 & -1 \\ 5 & 2 \end{bmatrix}.$ $\begin{aligned} A + A' &= \begin{bmatrix} 1+1 & 5+(-1) \\ -1+5 & 2+2 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 2 & 4 \\ 4 & 4 \end{bmatrix}, \\[2mm] A - A' &= \begin{bmatrix} 1-1 & 5-(-1) \\ -1-5 & 2-2 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 0 & 6 \\ -6 & 0 \end{bmatrix}. \end{aligned}$
(iv) $P$ ⁠, $Q$ and the check
$P = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix}, \qquad Q = \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}.$
$P$ has equal mirror pair $2, 2$ ⁠, so it is symmetric; $Q$ has zero diagonal and opposite pair $3, -3$ ⁠, so it is skew symmetric. Check:
$\begin{aligned} P + Q &= \begin{bmatrix} 1+0 & 2+3 \\ 2+(-3) & 2+0 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} = A. \end{aligned}$
The sum is the original matrix. ✓
Why this is the only possible split

Suppose $A = P + Q$ with $P' = P$ and $Q' = -Q$ ⁠. Transposing both sides gives $A' = P' + Q' = P - Q$ ⁠. Adding the two equations, $A + A' = 2P$ ⁠; subtracting, $A - A' = 2Q$ ⁠. So necessarily $P = \tfrac12(A + A')$ and $Q = \tfrac12(A - A')$ ⁠.

This shows our answers are not just one choice among many: any symmetric-plus-skew split of a square matrix must be this one. So the split found in each of the four parts is the only one possible.

$(P+Q)' = P' + Q'$ ⁠, Section 3.5.1 (iii); Theorems 1 and 2, Section 3.6. Lesson: symmetric and skew symmetric

Result

(i)

\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}

(ii) The matrix is symmetric, so it equals itself plus the $3\times3$ zero matrix.

(iii)

$A = P + Q$ with the symmetric part

P = \begin{bmatrix} 3 & \tfrac12 & -\tfrac52 \\ \tfrac12 & -2 & -2 \\ -\tfrac52 & -2 & 2 \end{bmatrix}

and the skew symmetric part

Q = \begin{bmatrix} 0 & \tfrac52 & \tfrac32 \\ -\tfrac52 & 0 & 3 \\ -\tfrac32 & -3 & 0 \end{bmatrix}.

(iv)

\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & 3 \\ -3 & 0 \end{bmatrix}

(Textbook answer: the same in all four parts.)

Check it with numbers

A quick test for any split: the diagonal of $Q$ must be 0, so the diagonal of $P$ must equal the diagonal of $A$ ⁠. In (iii), $P$ has diagonal $3, -2, 2$ and $A$ has diagonal $3, -2, 2$ ⁠. ✓ In (i), $3, -1$ and $3, -1$ ⁠. ✓ In (iv), $1, 2$ and $1, 2$ ⁠. ✓

Common mistakes

Forgetting the $\tfrac12$ ⁠. $A + A'$ and $A - A'$ add up to $2A$ ⁠, not $A$ ⁠.
Sign slips in $A - A'$ ⁠. In (iii), $-1-(-4) = 3$ and $-4-(-1) = -3$ ⁠.
Swapping the roles. The symmetric part comes from the sum, the skew part from the difference.
Leaving out the check. $P + Q = A$ ⁠, $P' = P$ and $Q' = -Q$ take a few seconds and catch most slips.

What to take away

$P = \tfrac12(A + A')$ averages mirror pairs; $Q = \tfrac12(A - A')$ takes half their difference.
The split exists for every square matrix and is unique.
A symmetric matrix has $Q = O$ ⁠; a skew symmetric matrix has $P = O$ (as in Question 9).

Every fact used, and where it comes from

Fact	Used in	Source
Transpose $(A')_{ij} = a_{ji}$	First step of each part	Definition 3, Section 3.5 (Lesson: transpose)
Sums, differences, scalar multiples entry by entry	Every part	Sections 3.4.1, 3.4.2
Symmetric $P' = P$ ⁠, skew symmetric $Q' = -Q$	Type checks in every part	Definitions 4 and 5, Section 3.6 (Lesson)
$\tfrac12(A+A')$ symmetric, $\tfrac12(A-A')$ skew, sum $A$	Every part	Theorems 1 and 2, Section 3.6; Example 22
Uniqueness of the split	Part (iv), Step 3	Proved on this page from Section 3.5.1 (iii)

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Exercise 3.3 · Q10: Symmetric plus skew symmetric, four matrices

Basics you need

Part (i)

(i) Transpose, sum and difference

(i) Halve to get PPP and QQQ⁠, and check their types

(i) Check P+Q=AP + Q = AP+Q=A

Part (ii)

(ii) Transpose: AAA is already symmetric

(ii) PPP and QQQ

Part (iii)

(iii) Transpose

(iii) A+A′A + A'A+A′ and PPP

(iii) A−A′A - A'A−A′ and QQQ

(iii) Check P+Q=AP + Q = AP+Q=A

Part (iv)

(iv) Transpose, sum and difference

(iv) PPP⁠, QQQ and the check

Why this is the only possible split

Every fact used, and where it comes from

(i) Halve to get $P$ and $Q$ ⁠, and check their types

(i) Check $P + Q = A$

(ii) Transpose: $A$ is already symmetric

(ii) $P$ and $Q$

(iii) $A + A'$ and $P$

(iii) $A - A'$ and $Q$

(iii) Check $P + Q = A$

(iv) $P$ ⁠, $Q$ and the check