Transpose; symmetric & skew-symmetric matrices

MATTranspose, Symmetric & Skew-Symmetric

The transpose $A'$ (also written $A^{T}$ ) is obtained by interchanging rows and columns. A square matrix is symmetric if $A'=A$ and skew-symmetric if $A'=-A$ , and a key board result is that every square matrix splits uniquely into a symmetric plus a skew-symmetric part.

These ideas test the transpose laws and the standard decomposition, both frequent 3-4 mark questions.

Transpose definition

(A')_{ij} = a_{ji}

A'

(transpose of

A

) swaps rows and columns; if

A

m\times n

then

A'

n\times m

Transpose laws

(A')' = A, \quad (kA)' = kA', \quad (A+B)' = A'+B', \quad (AB)' = B'A'

k

a scalar,

A,B

conformable; note the reversal of order in the product law.

Symmetric and skew-symmetric conditions

A' = A \ (\text{symmetric}), \qquad A' = -A \ (\text{skew-symmetric})

Both require

A

square. Skew-symmetric forces every diagonal entry

a_{ii}=0

since

a_{ii}=-a_{ii}

Unique decomposition of a square matrix

A = \underbrace{\tfrac{1}{2}(A + A')}_{\text{symmetric}} + \underbrace{\tfrac{1}{2}(A - A')}_{\text{skew-symmetric}}

\frac{1}{2}(A+A')

is symmetric,

\frac{1}{2}(A-A')

is skew-symmetric; valid for any square matrix

A

$A'$ exists for any matrix; symmetric/skew-symmetric properties are defined only for SQUARE matrices.
$A+A'$ is always symmetric and $A-A'$ is always skew-symmetric — proven via $(A+A')'=A'+A=A+A'$ .
Every diagonal element of a skew-symmetric matrix is 0; its off-diagonal entries satisfy $a_{ij}=-a_{ji}$ .
The decomposition $A=\frac{1}{2}(A+A')+\frac{1}{2}(A-A')$ is UNIQUE; don't forget the factor $\frac{1}{2}$ .
For the product law $(AB)'=B'A'$ , the order reverses; this generalises to $(ABC)'=C'B'A'$ .
To check your symmetric part $P$ , verify $P'=P$ ; for the skew part $Q$ , verify $Q'=-Q$ and $P+Q=A$ .
A symmetric matrix needs $a_{ij}=a_{ji}$ ; if it holds for all off-diagonal pairs the matrix is symmetric.

Where the marks go

Dropping the $\frac{1}{2}$ factor in the decomposition, so $P+Q$ no longer equals $A$ .
Writing $(AB)'=A'B'$ instead of the correct reversed order $B'A'$ .
Calling a non-square matrix symmetric or skew-symmetric.
Forgetting that skew-symmetric diagonal entries must be 0, leaving nonzero diagonals in $Q$ .

How the board asks it

Conversionunique split into symmetric and skew-symmetric parts
Express the matrix $A=\begin{bmatrix} 3 & -2 & 4 \\ 5 & 1 & 0 \\ -1 & 6 & 2 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.
Derive / provetranspose laws and the definitions $A'=A$ , $A'=-A$
If $A$ is any square matrix, prove that $A+A'$ is symmetric and $A-A'$ is skew-symmetric.
Numericalskew-symmetric condition $a_{ij}=-a_{ji}$
For what values of $x$ and $y$ is the matrix $\begin{bmatrix} 0 & 2x-3 & 7 \\ -1 & 0 & y \\ -7 & -4 & 0 \end{bmatrix}$ skew-symmetric?
Multiple choicediagonal entries of a skew-symmetric matrix are zero
If $A=[a_{ij}]$ is a skew-symmetric matrix of order $3$ , then the value of $a_{11}+a_{22}+a_{33}$ is $(a)$ $1$ $(b)$ $0$ $(c)$ $-1$ $(d)$ cannot be determined.
Give reasonsproduct law $(AB)'=B'A'$
If $A$ and $B$ are symmetric matrices of the same order, show that $AB$ is symmetric if and only if $AB=BA$ .

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.