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Chapter 3 · Matrices

Example 22: Symmetric plus skew symmetric

ExamplesSymmetric and skew symmetric matrices · Section 3.6
The question

Express the matrix

B=[2−2−4−1341−2−3]B = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}

as the sum of a symmetric and a skew symmetric matrix.

Exam answer B′=[2−11−23−2−44−3]B' = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix} P=12(B+B′)=12[4−3−3−362−32−6]=[2−32−32−3231−321−3]\begin{aligned} P &= \tfrac12(B + B') \\ &= \tfrac12\begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix} \\ &= \begin{bmatrix} 2 & -\frac32 & -\frac32 \\[2pt] -\frac32 & 3 & 1 \\[2pt] -\frac32 & 1 & -3 \end{bmatrix} \end{aligned} P′=[2−32−32−3231−321−3]=PP' = \begin{bmatrix} 2 & -\frac32 & -\frac32 \\[2pt] -\frac32 & 3 & 1 \\[2pt] -\frac32 & 1 & -3 \end{bmatrix} = P

So PP is a symmetric matrix.

Q=12(B−B′)=12[0−1−51065−60]=[0−12−52120352−30]\begin{aligned} Q &= \tfrac12(B - B') \\ &= \tfrac12\begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix} \\ &= \begin{bmatrix} 0 & -\frac12 & -\frac52 \\[2pt] \frac12 & 0 & 3 \\[2pt] \frac52 & -3 & 0 \end{bmatrix} \end{aligned} Q′=[01252−120−3−5230]=−QQ' = \begin{bmatrix} 0 & \frac12 & \frac52 \\[2pt] -\frac12 & 0 & -3 \\[2pt] -\frac52 & 3 & 0 \end{bmatrix} = -Q

So QQ is a skew symmetric matrix.

P+Q=[2−2−4−1341−2−3]=BP + Q = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = B

Hence B=P+QB = P + Q⁠, where PP is symmetric and QQ is skew symmetric.

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

The one idea this problem is built on

Every square matrix splits as

B=12(B+B′)+12(B−B′).B = \tfrac12(B + B') + \tfrac12(B - B').

The first piece equals its own transpose (symmetric) and the second piece equals minus its own transpose (skew symmetric). This is Theorem 2 of Section 3.6. The work is just computing B′B'⁠, the sum and difference, halving, and then checking the two properties and the sum.

Given, to find, and the plan
Given
The 3×33\times3 matrix BB⁠.
To find
A symmetric matrix PP and a skew symmetric matrix QQ⁠, both 3×33\times3⁠, with P+Q=BP + Q = B⁠.
Plan
(1) Find B′B'⁠. (2) Form P=12(B+B′)P = \frac12(B + B') and check P′=PP' = P⁠. (3) Form Q=12(B−B′)Q = \frac12(B - B') and check Q′=−QQ' = -Q⁠. (4) Check P+Q=BP + Q = B⁠.

Basics you need

1Transpose

B′B' has the rows of BB as its columns: its (i,j)(i,j) entry is bjib_{ji}⁠. Basics: reading a table sideways; mirror patterns

2Symmetric and skew symmetric

A square matrix PP is symmetric if P′=PP' = P⁠, that is pji=pijp_{ji} = p_{ij} for all i,ji, j⁠: entries mirror each other across the main diagonal. QQ is skew symmetric if Q′=−QQ' = -Q⁠, that is qji=−qijq_{ji} = -q_{ij}⁠; putting i=ji = j gives qii=−qiiq_{ii} = -q_{ii}⁠, so every diagonal entry is 0.

3Adding, subtracting and halving matrices

Matrices of the same order are added or subtracted entry by entry; 12X\frac12 X halves every entry. Basics: adding only like with like

4Transpose rules

(X+Y)′=X′+Y′(X+Y)' = X'+Y'⁠, (kX)′=kX′(kX)' = kX' and (X′)′=X(X')' = X (Section 3.5.1). These are what make the general split work.

Solution

  1. Find B′B'

    BB is 3×33\times3 (square), so B′B' is also 3×33\times3⁠. Row 1 of BB (2,−2,−42, -2, -4⁠) becomes column 1, row 2 (−1,3,4-1, 3, 4⁠) becomes column 2, and row 3 (1,−2,−31, -2, -3⁠) becomes column 3:

    B′=[2−11−23−2−44−3].B' = \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}.

    The diagonal entries 2,3,−32, 3, -3 stay where they are; each off-diagonal entry swaps with its mirror image, for example b12=−2b_{12} = -2 and b21=−1b_{21} = -1 trade places.

    Symmetry and skew symmetry are both defined by comparing a matrix with its transpose, so B′B' is the first thing we need. The split only makes sense for a square matrix, since only then do BB and B′B' have the same order and can be added.
    Definition 3, Section 3.5. Lesson: transpose

  2. Compute B+B′B + B' and halve it to get PP

    Add entry by entry:

    row 1: 2+2=4,  −2+(−1)=−3,−4+1=−3;row 2: −1+(−2)=−3,  3+3=6,4+(−2)=2;row 3: 1+(−4)=−3,  −2+4=2,−3+(−3)=−6.\begin{aligned} \text{row 1: }& 2+2 = 4,\ \ -2+(-1) = -3,\\ & -4+1 = -3;\\ \text{row 2: }& -1+(-2) = -3,\ \ 3+3 = 6,\\ & 4+(-2) = 2;\\ \text{row 3: }& 1+(-4) = -3,\ \ -2+4 = 2,\\ & -3+(-3) = -6. \end{aligned} B+B′=[4−3−3−362−32−6].B + B' = \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}.

    Now halve every entry:

    P=12(B+B′)=[2−32−32−3231−321−3].P = \frac12(B + B') = \begin{bmatrix} 2 & -\frac32 & -\frac32 \\[3pt] -\frac32 & 3 & 1 \\[3pt] -\frac32 & 1 & -3 \end{bmatrix}.
    We halve because B+B′B + B' counts BB "twice": adding 12(B+B′)\frac12(B+B') and 12(B−B′)\frac12(B-B') gives exactly 12B+12B=B\frac12 B + \frac12 B = B⁠. Without the halves, the two pieces would add up to 2B2B⁠.

  3. Check that PP is symmetric

    Compare each pair of mirror entries of PP⁠:

    p12=−32=p21,p13=−32=p31,p23=1=p32.\begin{aligned} &p_{12} = -\tfrac32 = p_{21},\\ &p_{13} = -\tfrac32 = p_{31},\\ &p_{23} = 1 = p_{32}. \end{aligned}

    Every entry equals its mirror image, so transposing PP changes nothing: P′=PP' = P⁠, and PP is symmetric.

    For a 3×33\times3 matrix there are only three mirror pairs above and below the diagonal; the diagonal entries are their own mirror images. Checking the three pairs is therefore a complete check of P′=PP' = P⁠. In general P′=12(B′+B′′)=12(B′+B)=PP' = \frac12(B' + B'') = \frac12(B' + B) = P⁠, so this always works.
    Definition 4, Section 3.6: PP is symmetric if P′=PP' = P⁠. Lesson: symmetric and skew symmetric matrices

  4. Compute B−B′B - B' and halve it to get QQ

    Subtract entry by entry (entry of BB minus the same entry of B′B'⁠):

    row 1: 2−2=0,  −2−(−1)=−1,−4−1=−5;row 2: −1−(−2)=1,  3−3=0,4−(−2)=6;row 3: 1−(−4)=5,  −2−4=−6,−3−(−3)=0.\begin{aligned} \text{row 1: }& 2-2 = 0,\ \ -2-(-1) = -1,\\ & -4-1 = -5;\\ \text{row 2: }& -1-(-2) = 1,\ \ 3-3 = 0,\\ & 4-(-2) = 6;\\ \text{row 3: }& 1-(-4) = 5,\ \ -2-4 = -6,\\ & -3-(-3) = 0. \end{aligned} B−B′=[0−1−51065−60].B - B' = \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}.

    Halve every entry:

    Q=12(B−B′)=[0−12−52120352−30].Q = \frac12(B - B') = \begin{bmatrix} 0 & -\frac12 & -\frac52 \\[3pt] \frac12 & 0 & 3 \\[3pt] \frac52 & -3 & 0 \end{bmatrix}.
    Subtracting a negative number adds it: −2−(−1)=−1-2-(-1) = -1 and 4−(−2)=64-(-2) = 6⁠. Most errors in this example happen here, so each subtraction is written out. The diagonal is automatically 0, because a diagonal entry of B′B' equals the same diagonal entry of BB⁠.

  5. Check that QQ is skew symmetric

    Each mirror pair must consist of opposite numbers, and the diagonal must be 0:

    q12=−12,  q21=12=−q12,q13=−52,  q31=52=−q13,q23=3,  q32=−3=−q23,q11=q22=q33=0.\begin{aligned} &q_{12} = -\tfrac12,\ \ q_{21} = \tfrac12 = -q_{12},\\ &q_{13} = -\tfrac52,\ \ q_{31} = \tfrac52 = -q_{13},\\ &q_{23} = 3,\ \ q_{32} = -3 = -q_{23},\\ &q_{11} = q_{22} = q_{33} = 0. \end{aligned}

    So

    Q′=[01252−120−3−5230]=−Q,Q' = \begin{bmatrix} 0 & \frac12 & \frac52 \\[3pt] -\frac12 & 0 & -3 \\[3pt] -\frac52 & 3 & 0 \end{bmatrix} = -Q,

    and QQ is skew symmetric.

    Q′=−QQ' = -Q means qji=−qijq_{ji} = -q_{ij} for every i,ji, j⁠; the list above checks all nine positions. In general Q′=12(B′−B′′)=12(B′−B)=−QQ' = \frac12(B' - B'') = \frac12(B' - B) = -Q⁠.
    Definition 5, Section 3.6. A note on the printed book: in its display of Q′Q'⁠, the (1,3)(1,3) entry is printed as 53\frac53⁠; the correct value is 52\frac52⁠, since it is q31=52q_{31} = \frac52⁠. The book's QQ itself and its conclusion are correct.

  6. Check that P+Q=BP + Q = B

    Add entry by entry:

    row 1: 2+0=2,  −32−12=−2,−32−52=−4;row 2: −32+12=−1,  3+0=3,1+3=4;row 3: −32+52=1,  1−3=−2,−3+0=−3.\begin{aligned} \text{row 1: }& 2+0 = 2,\ \ -\tfrac32-\tfrac12 = -2,\\ & -\tfrac32-\tfrac52 = -4;\\ \text{row 2: }& -\tfrac32+\tfrac12 = -1,\ \ 3+0 = 3,\\ & 1+3 = 4;\\ \text{row 3: }& -\tfrac32+\tfrac52 = 1,\ \ 1-3 = -2,\\ & -3+0 = -3. \end{aligned} P+Q=[2−2−4−1341−2−3]=B.P + Q = \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix} = B.
    P (symmetric) 2 −3/2 −3/2 −3/2 3 1 −3/2 1 −3 mirror entries equal Q (skew symmetric) 0 −1/2 −5/2 1/2 0 3 5/2 −3 0 mirror entries opposite dashed: the main diagonal, the mirror line
    The mirror patterns. In PP⁠, each coloured entry equals its partner across the dashed diagonal. In QQ⁠, each coloured entry is the negative of its partner, and the diagonal is all zeros.
    This final check confirms both the arithmetic and the method: the two pieces really do rebuild BB⁠. The fractions combine to whole numbers because, for every position (i,j)(i,j)⁠, pij+qij=12(bij+bji)+12(bij−bji)=bij.\begin{aligned} p_{ij} + q_{ij} &= \tfrac12(b_{ij} + b_{ji}) + \tfrac12(b_{ij} - b_{ji})\\ &= b_{ij}. \end{aligned}
    Theorem 2, Section 3.6: any square matrix is the sum of a symmetric and a skew symmetric matrix.

  7. This is the only possible split

    Suppose B=S+KB = S + K with SS symmetric and KK skew symmetric. Transposing both sides and using (S+K)′=S′+K′(S+K)' = S' + K'⁠:

    B′=S′+K′=S−K.B' = S' + K' = S - K.

    Adding B=S+KB = S + K and B′=S−KB' = S - K gives B+B′=2SB + B' = 2S⁠; subtracting gives B−B′=2KB - B' = 2K⁠. So S=12(B+B′)=PS = \frac12(B+B') = P and K=12(B−B′)=QK = \frac12(B-B') = Q⁠.

    The question asks for "the sum of a symmetric and a skew symmetric matrix", which could suggest many answers. This step shows there is exactly one, so any correct method must arrive at the same PP and QQ⁠.
    Uses (X+Y)′=X′+Y′(X+Y)' = X'+Y' (Section 3.5.1) and the two definitions S′=SS' = S⁠, K′=−KK' = -K⁠.
Result B=P+Q,B = P + Q, P=[2−32−32−3231−321−3] (symmetric),P = \begin{bmatrix} 2 & -\frac32 & -\frac32 \\[3pt] -\frac32 & 3 & 1 \\[3pt] -\frac32 & 1 & -3 \end{bmatrix}\ \text{(symmetric)}, Q=[0−12−52120352−30] (skew symmetric).Q = \begin{bmatrix} 0 & -\frac12 & -\frac52 \\[3pt] \frac12 & 0 & 3 \\[3pt] \frac52 & -3 & 0 \end{bmatrix}\ \text{(skew symmetric)}.
Check it with numbers

Two quick tests. The diagonal of PP must equal the diagonal of BB (since QQ has zeros there): 2,3,−32, 3, -3 in both. And for any mirror pair, pijp_{ij} is the average and qijq_{ij} is half the difference of bijb_{ij} and bjib_{ji}⁠: for (i,j)=(1,3)(i,j) = (1,3)⁠, b13=−4b_{13} = -4 and b31=1b_{31} = 1⁠, so p13=−4+12=−1.5p_{13} = \frac{-4+1}{2} = -1.5 and q13=−4−12=−2.5q_{13} = \frac{-4-1}{2} = -2.5⁠, matching −32-\frac32 and −52-\frac52⁠.

Common mistakes
  1. Forgetting the 12\frac12⁠. (B+B′)+(B−B′)=2B(B+B') + (B-B') = 2B⁠, not BB⁠.
  2. Sign errors in B−B′B - B'⁠. Subtract the whole entry of B′B'⁠: 4−(−2)=64 - (-2) = 6⁠, not 2.
  3. Expecting a skew symmetric matrix to have a non-zero diagonal. Its diagonal is always 0, which is a useful built-in check on QQ⁠.
  4. Stopping before the checks. The question is answered only once P′=PP' = P⁠, Q′=−QQ' = -Q and P+Q=BP + Q = B have all been shown.
What to take away
  1. For any square BB⁠: B=12(B+B′)+12(B−B′)B = \frac12(B+B') + \frac12(B-B')⁠, symmetric part plus skew symmetric part.
  2. Entry by entry, the symmetric part averages each mirror pair and the skew part takes half their difference.
  3. The split is unique, and the diagonal of BB goes entirely into the symmetric part.

Every fact used, and where it comes from

FactUsed inSource
Transpose B′=[bji]B' = [b_{ji}]Step 1Definition 3, Section 3.5 (Lesson: transpose)
Sum, difference, scalar multiple, entry by entrySteps 2, 4, 6Sections 3.4.1, 3.4.2 (Lesson: addition)
Symmetric: P′=PP' = P⁠; skew symmetric: Q′=−QQ' = -Q⁠, zero diagonalSteps 3, 5, 7Definitions 4 and 5, Section 3.6 (Lesson: symmetric and skew symmetric)
(X′)′=X(X')' = X⁠, (kX)′=kX′(kX)' = kX'⁠, (X+Y)′=X′+Y′(X+Y)' = X'+Y'Steps 3, 5, 7Section 3.5.1
B=12(B+B′)+12(B−B′)B = \frac12(B+B') + \frac12(B-B')Steps 2 to 6Theorems 1 and 2, Section 3.6
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