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Chapter 3 · Matrices

Exercise 3.3 · Q11 (MCQ): AB−BAAB-BA for symmetric A,BA, B

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

If AA⁠, BB are symmetric matrices of same order, then AB−BAAB - BA is a

(A) Skew symmetric matrix
(B) Symmetric matrix
(C) Zero matrix
(D) Identity matrix

Exam answer

Here A′=AA' = A⁠, B′=BB' = B⁠. Using (XY)′=Y′X′(XY)' = Y'X'⁠,

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

So AB−BAAB - BA is skew symmetric.

Answer: (A) Skew symmetric matrix

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

The one idea this problem is built on

Transpose AB−BAAB - BA using the reversal rule (XY)′=Y′X′(XY)' = Y'X'⁠. Since A′=AA' = A and B′=BB' = B⁠, the transpose of ABAB is BABA⁠, and the transpose of BABA is ABAB⁠. So transposing AB−BAAB - BA swaps the two terms and gives BA−ABBA - AB⁠, which is the negative of what we started with. A matrix whose transpose is its negative is skew symmetric.

Given, to find, and the plan
Given
Symmetric matrices AA⁠, BB of the same order: A′=AA' = A⁠, B′=BB' = B⁠.
To find
Which of the four descriptions always fits AB−BAAB - BA⁠.
Plan
(1) Check that AB−BAAB - BA is defined. (2) Compute its transpose in general. (3) Conclude. (4) Test a specific pair to rule out the other options.

Basics you need

1Transpose rules

(X−Y)′=X′−Y′(X - Y)' = X' - Y' and (XY)′=Y′X′(XY)' = Y'X' (note the reversed order). Basics: reading a table sideways

2Symmetric and skew symmetric

Symmetric: X′=XX' = X⁠. Skew symmetric: X′=−XX' = -X⁠, which forces zeros on the diagonal. Lesson: symmetric and skew symmetric

3Multiplying matrices

Entry (i,k)(i,k) of a product is row ii times column kk⁠. In general AB≠BAAB \ne BA⁠. Basics: pair up, multiply, add

Solution

  1. The expression is defined

    A symmetric matrix is square (it must have the same order as its transpose). So AA and BB are both n×nn\times n for the same nn⁠. Then ABAB and BABA are both defined and both n×nn\times n⁠, so AB−BAAB - BA is defined and is n×nn\times n⁠, a square matrix.

    Symmetric and skew symmetric only make sense for square matrices, so we first confirm that AB−BAAB - BA is square.
    Product of n×nn\times n by n×nn\times n is n×nn\times n⁠, Section 3.4.5. Basics: pair up, multiply, add

  2. Transpose AB−BAAB - BA

    Let C=AB−BAC = AB - BA⁠. Using the rules of Basic 1 and then A′=AA' = A⁠, B′=BB' = B⁠:

    C′=(AB)′−(BA)′=B′A′−A′B′=BA−AB=−(AB−BA)=−C.\begin{aligned} C' &= (AB)' - (BA)' \\ &= B'A' - A'B' \\ &= BA - AB \\ &= -(AB - BA) = -C. \end{aligned}
    Line 1 uses (X−Y)′=X′−Y′(X - Y)' = X' - Y'⁠. Line 2 uses the reversal rule (XY)′=Y′X′(XY)' = Y'X' on each product. Line 3 replaces A′A' by AA and B′B' by BB⁠, because both are symmetric. Line 4 takes out a minus sign. Every step holds for all symmetric AA⁠, BB of the same order, so this is a general proof.
    Section 3.5.1, properties (ii), (iii), (iv). Lesson: transpose

  3. Conclude

    CC is square and C′=−CC' = -C⁠, so AB−BAAB - BA is skew symmetric, option (A).

    Definition 5, Section 3.6 (Basic 2).

  4. A concrete pair, to test the other options

    Take the symmetric 2×22\times2 matrices

    A=[1000],B=[0110].A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}.

    Entries of ABAB (row of AA times column of BB⁠):

    (AB)11=1⋅0+0⋅1=0,(AB)12=1⋅1+0⋅0=1,(AB)21=0⋅0+0⋅1=0,(AB)22=0⋅1+0⋅0=0.\begin{aligned} (AB)_{11} &= 1\cdot0 + 0\cdot1 = 0, \\ (AB)_{12} &= 1\cdot1 + 0\cdot0 = 1, \\ (AB)_{21} &= 0\cdot0 + 0\cdot1 = 0, \\ (AB)_{22} &= 0\cdot1 + 0\cdot0 = 0. \end{aligned}

    Entries of BABA⁠:

    (BA)11=0⋅1+1⋅0=0,(BA)12=0⋅0+1⋅0=0,(BA)21=1⋅1+0⋅0=1,(BA)22=1⋅0+0⋅0=0.\begin{aligned} (BA)_{11} &= 0\cdot1 + 1\cdot0 = 0, \\ (BA)_{12} &= 0\cdot0 + 1\cdot0 = 0, \\ (BA)_{21} &= 1\cdot1 + 0\cdot0 = 1, \\ (BA)_{22} &= 1\cdot0 + 0\cdot0 = 0. \end{aligned}

    So

    AB−BA=[0100]−[0010]=[01−10].\begin{aligned} AB - BA &= \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \\[1mm] &= \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}. \end{aligned}

    This is skew symmetric (zero diagonal, mirror pair 1,−11, -1⁠), as Step 3 says it must be.

    A general proof shows (A) is always true; a single example is enough to show that a description is not always true. This one pair will rule out (B), (C) and (D) at once.

  5. Examine every option

    • (A) Skew symmetric: correct. Proved for all symmetric AA⁠, BB of the same order in Step 2.
    • (B) Symmetric: wrong. In Step 4, the (1,2) entry is 1 and the (2,1) entry is −1-1⁠, so AB−BAAB - BA is not equal to its transpose. (A matrix that is both symmetric and skew symmetric satisfies C=−CC = -C⁠, so it is OO⁠; symmetric would only happen when AB−BAAB - BA is the zero matrix.)
    • (C) Zero matrix: wrong. It happens for some pairs, for instance A=BA = B (then AB−BA=A2−A2=OAB - BA = A^2 - A^2 = O⁠), but Step 4 gives a nonzero result, so it is not always true.
    • (D) Identity matrix: wrong. A skew symmetric matrix has every diagonal entry 0, while II has diagonal entries 1. So AB−BAAB - BA can never be II⁠.
    In a multiple-choice question about a property that holds "always", the correct option must hold for every allowed pair, and each wrong option needs only one pair where it fails, or a reason it can never hold.
Result

(A): AB−BAAB - BA is a skew symmetric matrix, because (AB−BA)′=BA−AB=−(AB−BA)(AB - BA)' = BA - AB = -(AB - BA)⁠. (Textbook answer: A.)

Check it with numbers

Another pair, both symmetric:

A=[1223],B=[0114].A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 1 \\ 1 & 4 \end{bmatrix}.

Then (AB)12=1⋅1+2⋅4=9(AB)_{12} = 1\cdot1 + 2\cdot4 = 9 and (AB)21=2⋅0+3⋅1=3(AB)_{21} = 2\cdot0 + 3\cdot1 = 3⁠; (BA)12=0⋅2+1⋅3=3(BA)_{12} = 0\cdot2 + 1\cdot3 = 3 and (BA)21=1⋅1+4⋅2=9(BA)_{21} = 1\cdot1 + 4\cdot2 = 9⁠. So AB−BAAB - BA has (1,2) entry 9−3=69 - 3 = 6 and (2,1) entry 3−9=−63 - 9 = -6⁠: opposite, as skew symmetry requires. ✓

Common mistakes
  1. Writing (AB)′=A′B′(AB)' = A'B'⁠. The order reverses: (AB)′=B′A′(AB)' = B'A'⁠. Without the reversal you would wrongly get C′=CC' = C⁠.
  2. Assuming AB=BAAB = BA⁠. Then you would pick (C). Matrix products usually do not commute.
  3. Treating one example as a proof. The example rules options out; the general argument of Step 2 is what proves (A).
What to take away
  1. If AA⁠, BB are symmetric, then (AB)′=BA(AB)' = BA⁠.
  2. So AB−BAAB - BA is always skew symmetric (and, by the same steps, AB+BAAB + BA is always symmetric).
  3. A skew symmetric matrix has zero diagonal, so it can never be the identity matrix.

Every fact used, and where it comes from

FactUsed inSource
Symmetric matrices are square; products of n×nn\times n matrices are n×nn\times nStep 1Definition 4, Section 3.6; Section 3.4.5
(X−Y)′=X′−Y′(X - Y)' = X' - Y'⁠, (XY)′=Y′X′(XY)' = Y'X'Step 2Section 3.5.1, properties (ii)–(iv) (Lesson: transpose)
Skew symmetric: C′=−CC' = -C⁠, zero diagonalSteps 3, 5Definition 5, Section 3.6 (Lesson)
Product entries, row times columnStep 4Section 3.4.5 (Lesson: multiplication)
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