Matrices · Lesson 6 of 8
Transpose, Symmetry, and Decomposition
“Exchange rows and columns, use transpose laws, and split square matrices into symmetry parts.”
• Form a transpose and track the exchange of entry indices and dimensions. • Apply transpose laws, including reversal of the order in a product. • Identify symmetric and skew-symmetric matrices from entry relationships. • Prove and use the symmetric/skew-symmetric decomposition of a square matrix. • Relate symmetry of products to commutativity and justify related transpose results.
Turning rows into columns
Earlier we recorded the same data with people labelling either rows or columns. Transpose gives this systematic rearrangement a precise name. It preserves every entry’s value while exchanging its row and column addresses.
The transpose of A, denoted Aᵀ or A′, is obtained by interchanging rows and columns. If A has order m × n, Aᵀ has order n × m, with (Aᵀ)ⱼᵢ = aᵢⱼ.
A row matrix becomes a column matrix and a column matrix becomes a row matrix. A square matrix stays square. Its diagonal entries stay at their original positions because exchanging (i,i) leaves the same address. The off-diagonal entries exchange places across the main diagonal; this will give us a way to describe symmetry.
Transpose laws and reversed product order
Two transposes undo the exchange of addresses, and entrywise operations interact naturally with that exchange. A product requires more care: transposing it reverses the factor order. Checking dimensions first helps explain why the reversed order is necessary.
For double transpose, an entry at (i,j) first moves to (j,i) and then back to (i,j). Scaling before or after moving it gives the same number kaᵢⱼ. For a sum, moving aᵢⱼ+bᵢⱼ to its new address gives the same value as moving both entries first and adding. These are entrywise explanations of the first three laws.
Now let A be m × n and B be n × p. The transpose of AB has order p × m. BᵀAᵀ has order (p × n)(n × m), exactly p × m. By contrast, AᵀBᵀ may not even be defined. For the entries, (BᵀAᵀ)ₖᵢ sums bⱼₖaᵢⱼ over j, which equals the sum of aᵢⱼbⱼₖ defining (AB)ᵢₖ. The factors inside each scalar product commute, but the matrix factors have changed order as the law requires.
Problem
Let A = [3,√3,2;4,2,0] and B = [2,−1,2;1,2,4]. Verify double transpose, the sum law, and the scalar law for B.
- 1.Aᵀ = [3,4;√3,2;2,0]. Transposing this again gives [3,√3,2;4,2,0] = A.
- 2.Add first: A+B = [5,√3−1,4;5,4,4]. Its transpose is [5,5;√3−1,4;4,4].
- 3.Separately Bᵀ = [2,1;−1,2;2,4]. Adding this to Aᵀ gives [5,5;√3−1,4;4,4], the same array.
- 4.For any scalar k, kB = [2k,−k,2k;k,2k,4k]. Its transpose is [2k,k;−k,2k;2k,4k].
- 5.Multiplying Bᵀ by k gives exactly the same result, verifying (kB)ᵀ = kBᵀ. The positions change under transpose, but the factor k does not.
Problem
For A = [−2;4;5] and B = [1,3,−6], verify (AB)ᵀ = BᵀAᵀ.
- 1.The orders 3 × 1 and 1 × 3 make AB a 3 × 3 matrix. Multiplication gives AB = [−2,−6,12;4,12,−24;5,15,−30].
- 2.Transpose this product: (AB)ᵀ = [−2,4,5;−6,12,15;12,−24,−30].
- 3.Now Aᵀ = [−2,4,5] and Bᵀ = [1;3;−6]. Their reversed product is BᵀAᵀ = [−2,4,5;−6,12,15;12,−24,−30].
- 4.The results agree. Multiplying AᵀBᵀ instead would give a 1 × 1 matrix, so it could not equal the transpose of the 3 × 3 product.
Symmetric and skew-symmetric patterns
For square matrices we can compare a matrix directly with its transpose. Some matrices remain unchanged; others have every entry negated. These two patterns are described by matching pairs reflected across the main diagonal.
A square matrix A is symmetric when Aᵀ = A. Equivalently, aⱼᵢ = aᵢⱼ at every pair of reflected positions.
A square matrix A is skew-symmetric when Aᵀ = −A. Equivalently, aⱼᵢ = −aᵢⱼ for every pair of indices.
The symmetric matrix [2,3,−1;3,4,5;−1,5,6] has matching values across the diagonal. In the skew-symmetric matrix [0,3,−1;−3,0,5;1,−5,0], the corresponding values have opposite signs. For a skew-symmetric diagonal entry, the rule says aᵢᵢ = −aᵢᵢ. Thus 2aᵢᵢ = 0 and aᵢᵢ = 0. Every diagonal entry must vanish.
The matrix [0,1;2,0] has a zero diagonal but is not skew-symmetric because 2 is not −1. Check each reflected pair as well as the diagonal. In a symmetric matrix the diagonal values may be any real numbers.
Constructing both kinds from any square matrix
A matrix need not already have either symmetry pattern. Combining it with its transpose produces the patterns automatically. We will prove this before using the result to split a general square matrix into two simpler parts.
Let P = A+Aᵀ. Transpose and use the sum law: Pᵀ = Aᵀ+(Aᵀ)ᵀ = Aᵀ+A = A+Aᵀ = P. Hence A+Aᵀ is symmetric. Now let Q = A−Aᵀ. Then Qᵀ = Aᵀ−A = −(A−Aᵀ) = −Q, so A−Aᵀ is skew-symmetric. These calculations require A to be square so that the original sums are defined.
Adding those two expressions gives (A+Aᵀ)+(A−Aᵀ) = 2A. Taking half of each therefore recovers A, not 2A. A scalar multiple of a symmetric matrix is symmetric, and a scalar multiple of a skew-symmetric matrix is skew-symmetric, because transpose commutes with scalar multiplication.
Problem
Express B = [2,−2,−4;−1,3,4;1,−2,−3] as a symmetric matrix plus a skew-symmetric matrix.
- 1.First transpose: Bᵀ = [2,−1,1;−2,3,−2;−4,4,−3]. Keep reflected addresses paired carefully.
- 2.The sum is B+Bᵀ = [4,−3,−3;−3,6,2;−3,2,−6]. Divide by 2 to get S = [2,−3/2,−3/2;−3/2,3,1;−3/2,1,−3].
- 3.The difference is B−Bᵀ = [0,−1,−5;1,0,6;5,−6,0]. Divide by 2 to get K = [0,−1/2,−5/2;1/2,0,3;5/2,−3,0].
- 4.Check Sᵀ = S by its reflected equal entries, and Kᵀ = −K by its opposite reflected entries and zero diagonal.
- 5.Finally S+K = [2,−2,−4;−1,3,4;1,−2,−3] = B. For example, the top-right entry is −3/2−5/2 = −4, while its reflected entry is −3/2+5/2 = 1.
The decomposition gives the original matrix back even when the two reflected entries are different. If A is already symmetric, its skew-symmetric part is O. If A is already skew-symmetric, its symmetric part is O. A matrix that is both kinds must satisfy A = −A, so 2A = O and A = O. The square zero matrix is indeed both symmetric and skew-symmetric.
Symmetry of products needs a condition
Two symmetric factors do not automatically give a symmetric product. The reversed order in the transpose law tells us precisely what extra condition is required. This links symmetry to the distinction between commuting and noncommuting matrix products.
Problem
Let A and B be symmetric matrices of the same order. Prove that AB is symmetric if and only if AB = BA.
- 1.Since Aᵀ = A and Bᵀ = B, the product law gives (AB)ᵀ = BᵀAᵀ = BA.
- 2.First assume AB is symmetric. Its definition gives (AB)ᵀ = AB. Combining this with (AB)ᵀ = BA gives BA = AB.
- 3.Conversely, assume AB = BA. Then (AB)ᵀ = BA = AB, so AB is symmetric.
- 4.Both directions are needed for “if and only if”. A pair such as A = [1,0;0,2], B = [0,1;1,0] shows why the condition matters: AB = [0,1;2,0] is not symmetric, and BA = [0,2;1,0] is different.
A related calculation gives (AB−BA)ᵀ = BA−AB = −(AB−BA) when A and B are symmetric, so AB−BA is skew-symmetric. Another useful expression is BᵀAB: its transpose is BᵀAᵀB. Thus it is symmetric if Aᵀ = A and skew-symmetric if Aᵀ = −A, provided the dimensions allow the product. These conclusions come from transpose laws, not from swapping noncommuting factors.
Quiz
If A is 2 × 5, what is the order of Aᵀ?
For compatible A and B, which expression equals (AB)ᵀ?
Which matrix is skew-symmetric?
What is the symmetric part of A = [1,4;2,3]?
If a real square matrix is both symmetric and skew-symmetric, what is it?
For symmetric A and B of the same order, when is AB symmetric?
For a square A, which expression is always skew-symmetric?
Practice Problems
- Find the transpose of [5;1/2;−1], [1,−1;2,3], and [−1,5,6;√3,5,6;2,3,−1]. State each order before and after transposing.
- For A = [−1,2,3;5,7,9;−2,1,1] and B = [−4,1,−5;1,2,0;1,3,1], verify both (A+B)ᵀ = Aᵀ+Bᵀ and (A−B)ᵀ = Aᵀ−Bᵀ.
- Given Aᵀ = [3,4;−1,2;0,1] and B = [−1,2,1;1,2,3], first recover A and then verify the sum and difference transpose laws.
- Given Aᵀ = [−2,3;1,2] and B = [−1,0;1,2], find (A+2B)ᵀ using the laws and check by direct calculation.
- Verify (AB)ᵀ = BᵀAᵀ for (a) A = [1;−4;3], B = [−1,2,1] and (b) A = [0;1;2], B = [1,5,7].
- Show that [1,−1,5;−1,2,1;5,1,3] is symmetric and [0,1,−1;−1,0,1;1,−1,0] is skew-symmetric by comparing reflected entries.
- For A = [1,5;6,7], calculate A+Aᵀ and A−Aᵀ and identify their symmetry types. Then calculate the half-sum and half-difference and check that they reconstruct A.
- For A = [0,a,b;−a,0,c;−b,−c,0], calculate ½(A+Aᵀ) and ½(A−Aᵀ). Explain the result without assuming a, b, c are nonzero.
- Decompose each into a symmetric and a skew-symmetric part: [3,5;1,−1]; [6,−2,2;−2,3,−1;2,−1,3]; [3,3,−1;−2,−2,1;−4,−5,2]; [1,5;−1,2].
- For symmetric A and B of the same order, prove AB−BA is skew-symmetric. Explain why the expression need not be O.
- If A = [cos α,−sin α;sin α,cos α], find which of α = π/6, π/3, π, 3π/2 satisfies A+Aᵀ = I. Start by simplifying A+Aᵀ.
- Prove that BᵀAB is symmetric or skew-symmetric according as A is symmetric or skew-symmetric. State compatible dimensions and use the transpose of a three-factor product.
- Prove that the zero matrix is the only real square matrix that is both symmetric and skew-symmetric. Do not conclude that every diagonal matrix has both properties.
Key Takeaways
• Transpose exchanges rows and columns and changes order m × n to n × m. • Double transpose restores A; transpose respects scaling, sums, and differences. • The transpose of a product reverses factor order: (AB)ᵀ = BᵀAᵀ. • Symmetric entries match across the diagonal; skew-symmetric entries are negatives and their diagonal is zero. • Every real square matrix is ½(A+Aᵀ)+½(A−Aᵀ), a symmetric part plus a skew-symmetric part. • For symmetric factors, AB is symmetric exactly when AB = BA; the only matrix of both symmetry types is O.