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Transpose and symmetric matrices

The transpose of a matrix is obtained by interchanging its rows and columns. Although the operation is simple, it interacts beautifully with addition and multiplication, and it singles out two structurally important families: symmetric matrices, where AT=AA^T = A, and skew-symmetric matrices, where AT=AA^T = -A. Every square matrix can be uniquely decomposed into a symmetric and a skew-symmetric part , a fact you will use throughout the chapter.

Definition

If A=(aij)m×nA = (a_{ij})_{m \times n}, its transpose ATA^T (also written AA') is the n×mn \times m matrix

(AT)ij=aji.(A^T)_{ij} = a_{ji}.

So row ii of AA becomes column ii of ATA^T.

Example

A=(123456)AT=(142536)A = \begin{pmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{pmatrix} \Rightarrow A^T = \begin{pmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{pmatrix}.

Algebraic properties of transpose

  1. (AT)T=A(A^T)^T = A.
  2. (A+B)T=AT+BT(A + B)^T = A^T + B^T.
  3. (λA)T=λAT(\lambda A)^T = \lambda A^T.
  4. (AB)T=BTAT(AB)^T = B^T A^T.

The reversal in (4) is the most-tested fact. It mirrors the inverse formula (AB)1=B1A1(AB)^{-1} = B^{-1} A^{-1}.

Proof of (4). Both sides are n×mn \times m matrices (if AA is m×pm \times p and BB is p×np \times n). The (i,j)(i, j) entry of (AB)T(AB)^T is the (j,i)(j, i) entry of ABAB, which is kajkbki\sum_k a_{jk} b_{ki}. The (i,j)(i, j) entry of BTATB^T A^T is k(BT)ik(AT)kj=kbkiajk\sum_k (B^T)_{ik}(A^T)_{kj} = \sum_k b_{ki} a_{jk}. Same expression. \blacksquare

Symmetric matrices

A square matrix AA is symmetric if AT=AA^T = A, i.e., aij=ajia_{ij} = a_{ji} for all i,ji, j. Symmetric matrices are precisely those that look the same when reflected across the main diagonal.

Examples: any diagonal matrix, (1223)\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}, the identity.

The set of n×nn \times n symmetric matrices is closed under addition and scalar multiplication, but not under multiplication: A,BA, B symmetric does not imply ABAB symmetric.

Skew-symmetric matrices

A square matrix AA is skew-symmetric if AT=AA^T = -A, i.e., aij=ajia_{ij} = -a_{ji}. Setting i=ji = j shows aii=0a_{ii} = 0: all diagonal entries are zero.

Examples: (0220)\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix}, (0aba0cbc0)\begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}.

The decomposition theorem

Theorem. Every square matrix AA can be written uniquely as A=P+QA = P + Q, where PP is symmetric and QQ is skew-symmetric.

Proof. Set P=12(A+AT)P = \tfrac{1}{2}(A + A^T) and Q=12(AAT)Q = \tfrac{1}{2}(A - A^T). Then A=P+QA = P + Q. Check PT=12(AT+A)=PP^T = \tfrac{1}{2}(A^T + A) = P, so PP symmetric. Check QT=12(ATA)=QQ^T = \tfrac{1}{2}(A^T - A) = -Q, so QQ skew-symmetric.

Uniqueness: suppose A=P1+Q1=P2+Q2A = P_1 + Q_1 = P_2 + Q_2. Then P1P2=Q2Q1P_1 - P_2 = Q_2 - Q_1. The left side is symmetric, the right side skew-symmetric. The only matrix that is both is the zero matrix. So P1=P2,Q1=Q2P_1 = P_2, Q_1 = Q_2. \blacksquare

Worked decomposition

Let A=(1234)A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}. Then AT=(1324)A^T = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}.

P=12(A+AT)=12(2558)=(15/25/24)P = \tfrac{1}{2}(A + A^T) = \tfrac{1}{2}\begin{pmatrix} 2 & 5 \\ 5 & 8 \end{pmatrix} = \begin{pmatrix} 1 & 5/2 \\ 5/2 & 4 \end{pmatrix}. Symmetric.

Q=12(AAT)=12(0110)=(01/21/20)Q = \tfrac{1}{2}(A - A^T) = \tfrac{1}{2}\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & -1/2 \\ 1/2 & 0 \end{pmatrix}. Skew-symmetric.

Verify P+Q=AP + Q = A. \checkmark

Worked examples

Example 1. Find ATA^T for A=(213402)A = \begin{pmatrix} 2 & -1 & 3 \\ 4 & 0 & -2 \end{pmatrix}.

AT=(241032)A^T = \begin{pmatrix} 2 & 4 \\ -1 & 0 \\ 3 & -2 \end{pmatrix}.

Example 2. Verify (AB)T=BTAT(AB)^T = B^T A^T for A=(12),B=(34)A = \begin{pmatrix} 1 & 2 \end{pmatrix}, B = \begin{pmatrix} 3 \\ 4 \end{pmatrix}.

AB=(3+8)=(11)AB = (3 + 8) = (11), so (AB)T=(11)(AB)^T = (11). BT=(3,4)B^T = (3, 4), AT=(12)A^T = \begin{pmatrix} 1 \\ 2 \end{pmatrix}, so BTAT=3+8=(11)B^T A^T = 3 + 8 = (11). \checkmark

Example 3. Show that A=(123254347)A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 4 \\ 3 & 4 & 7 \end{pmatrix} is symmetric.

Compute ATA^T: (123254347)\begin{pmatrix} 1 & 2 & 3 \\ 2 & 5 & 4 \\ 3 & 4 & 7 \end{pmatrix}, same as AA.

Example 4. Decompose (2310)\begin{pmatrix} 2 & 3 \\ 1 & 0 \end{pmatrix} into symmetric and skew-symmetric parts.

AT=(2130)A^T = \begin{pmatrix} 2 & 1 \\ 3 & 0 \end{pmatrix}. P=12(4440)=(2220)P = \tfrac{1}{2}\begin{pmatrix} 4 & 4 \\ 4 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 2 \\ 2 & 0 \end{pmatrix}. Q=12(0220)=(0110)Q = \tfrac{1}{2}\begin{pmatrix} 0 & 2 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}. Verify P+Q=AP + Q = A.

Example 5. Show: if AA is square then A+ATA + A^T is symmetric and AATA - A^T is skew-symmetric.

(A+AT)T=AT+A=A+AT(A + A^T)^T = A^T + A = A + A^T. (AAT)T=ATA=(AAT)(A - A^T)^T = A^T - A = -(A - A^T). \checkmark

Example 6. Show that the product of two symmetric matrices is symmetric iff they commute.

(AB)T=BTAT=BA(AB)^T = B^T A^T = BA. So ABAB is symmetric iff AB=BAAB = BA.

Try it yourself

  1. Write the transpose of (101234)\begin{pmatrix} 1 & 0 & -1 \\ 2 & 3 & 4 \end{pmatrix}.
  2. Determine whether (012103230)\begin{pmatrix} 0 & 1 & 2 \\ -1 & 0 & 3 \\ -2 & -3 & 0 \end{pmatrix} is symmetric or skew-symmetric.
  3. Show: AA is symmetric iff A+AT=2AA + A^T = 2A.
  4. Show: AA is skew-symmetric iff A+AT=OA + A^T = O.
  5. Decompose (1512)\begin{pmatrix} 1 & 5 \\ -1 & 2 \end{pmatrix} into symmetric and skew-symmetric parts.
  6. Verify (AB)T=BTAT(AB)^T = B^T A^T for A=(1001)A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} and B=(2345)B = \begin{pmatrix} 2 & 3 \\ 4 & 5 \end{pmatrix}.
  7. Show: if AA is symmetric and invertible, then A1A^{-1} is symmetric.
  8. Find aa if (0a20)\begin{pmatrix} 0 & a \\ -2 & 0 \end{pmatrix} is to be skew-symmetric.
  9. Show: every diagonal entry of a skew-symmetric matrix is zero.
  10. If AA is skew-symmetric of order nn and nn is odd, what is detA\det A? (Preview from Chapter 4.)
  11. Find ATAA^T A for A=(100111)A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{pmatrix}.
  12. Show that ATAA^T A is always symmetric, for any matrix AA.
  13. Decompose (324325112)\begin{pmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{pmatrix} into symmetric and skew-symmetric parts.
  14. Prove: if AA and BB are both skew-symmetric, then so is αA+βB\alpha A + \beta B for any scalars.

Pitfalls / Tricks

  • (AB)T=BTAT(AB)^T = B^T A^T , the order reverses. Forgetting this is a classical error.
  • A skew-symmetric matrix has zero diagonal; a symmetric matrix can have anything on the diagonal.
  • The decomposition A=P+QA = P + Q is unique. The 1/21/2 factor is essential.
  • Product of two symmetric matrices is symmetric only if they commute.
  • ATAA^T A and AATAA^T are always symmetric, but they are different matrices in general (and need not be equal even in order).

Next, elementary row operations and inverses.

Practice quiz

Quick check on this topic.

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Quick check : Transpose and symmetric
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