Why Every Square Matrix Splits Into a Symmetric and a Skew-Symmetric Part

A matrix is a rectangular array of numbers, and it can store a mark sheet, the coefficients of a system of equations, or the rotation of an image on a screen. Learning the types of matrices and how to combine them is the first step towards solving equations and transforming shapes with them.

This part covers types and order, symmetric and skew-symmetric matrices, addition and scalar multiplication, and the transpose.

**A matrix with m rows and n columns has order $m\times n$, and square matrices are classified as diagonal, scalar, identity, upper or lower triangular, symmetric or skew-symmetric according to the pattern of their entries.

Types of matrices:

- Row and column matrices** — orders $1\times n$ and $m\times 1$
- Diagonal — square, with every entry off the main diagonal equal to zero
- Scalar — diagonal, with all diagonal entries equal
- Identity — scalar, with every diagonal entry 1
- Upper or lower triangular — zeros below or above the main diagonal
- Symmetric — $A^T=A$, so $a_{ij}=a_{ji}$
- Skew-symmetric — $A^T=-A$, so $a_{ij}=-a_{ji}$ and every diagonal entry is 0

Worked example. Construct the $2\times 3$ matrix with $a_{ij}=2i-j$.

$$A=\left[\begin{array}{ccc}1& 0& -1\\ 3& 2& 1\end{array}\right]$$

For instance, $a_{21}=2(2)-1=3$.

An everyday example. A school timetable with 6 days as rows and 8 periods as columns is a $6\times 8$ matrix.

The substance. Every identity matrix is scalar and every scalar matrix is diagonal, but not the reverse — $\left[\begin{array}{cc}2& 0\\ 0& 3\end{array}\right]$ is diagonal but not scalar.

**For any square matrix A, $A+A^T$ is symmetric and $A-A^T$ is skew-symmetric, so every square matrix can be written uniquely as $A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)$, the sum of a symmetric and a skew-symmetric matrix.

Proofs:**

- $(A+A^T{)}^T=A^T+(A^T{)}^T=A^T+A$, so $A+A^T$ is symmetric
- $(A-A^T{)}^T=A^T-A=-(A-A^T)$, so $A-A^T$ is skew-symmetric
- In a skew-symmetric matrix $a_{ii}=-a_{ii}$, which forces every diagonal entry to be 0

Worked example. Express $A=\left[\begin{array}{cc}3& 5\\ 1& -1\end{array}\right]$ as such a sum.

$$P=\frac{1}{2}(A+A^T)=\left[\begin{array}{cc}3& 3\\ 3& -1\end{array}\right],Q=\frac{1}{2}(A-A^T)=\left[\begin{array}{cc}0& 2\\ -2& 0\end{array}\right]$$

Check: $P+Q=A$, P is symmetric and Q is skew-symmetric.

An everyday example. A table of road distances between cities is symmetric, since the distance from Pune to Nashik equals the distance from Nashik to Pune.

The substance. The only matrix that is both symmetric and skew-symmetric is the zero matrix — $A=A^T=-A$ forces every entry to be 0.

Matrices of the same order are added or subtracted by combining corresponding entries, and multiplying a matrix by a scalar k multiplies every entry by k; matrices of different orders cannot be added.

Properties:

- $A+B=B+A$ and $(A+B)+C=A+(B+C)$
- $A+O=A$ and $A+(-A)=O$
- $k(A+B)=kA+kB$ and $(k+l)A=kA+lA$

Worked example. Let $A=\left[\begin{array}{cc}2& -1\\ 4& 3\end{array}\right]$ and $B=\left[\begin{array}{cc}1& 5\\ -2& 0\end{array}\right]$.

$$2A-3B=\left[\begin{array}{cc}4& -2\\ 8& 6\end{array}\right]-\left[\begin{array}{cc}3& 15\\ -6& 0\end{array}\right]=\left[\begin{array}{cc}1& -17\\ 14& 6\end{array}\right]$$

An everyday example. Two shops' monthly sales of rice, wheat and dal in two pack sizes can be stored as $2\times 3$ matrices; adding them gives combined sales, and multiplying by 3 estimates sales over three months.

The substance. Equal numbers of entries are not enough — a $2\times 3$ and a $3\times 2$ matrix cannot be added, even though both have six entries.

**The transpose $A^T$ is formed by turning the rows of A into columns, so an $m\times n$ matrix becomes $n\times m$, and it satisfies $(A^T{)}^T=A$, $(A\pm B{)}^T=A^T\pm B^T$, $(kA{)}^T=k{A}^T$ and $(AB{)}^T=B^T{A}^T$.

Worked example.** Let $A=\left[\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\end{array}\right]$, of order $2\times 3$. Then

$$A^T=\left[\begin{array}{cc}1& 4\\ 2& 5\\ 3& 6\end{array}\right]$$

of order $3\times 2$, and transposing again returns A.

Worked example 2. Verify $(A+B{)}^T=A^T+B^T$ for $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}0& -1\\ 5& 2\end{array}\right]$. Here $A+B=\left[\begin{array}{cc}1& 1\\ 8& 6\end{array}\right]$, whose transpose is $\left[\begin{array}{cc}1& 8\\ 1& 6\end{array}\right]$, and

$$A^T+B^T=\left[\begin{array}{cc}1& 3\\ 2& 4\end{array}\right]+\left[\begin{array}{cc}0& 5\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}1& 8\\ 1& 6\end{array}\right]$$

An everyday example. Swapping the rows and columns of a class marks spreadsheet, so that students run across the top instead of down the side, is taking a transpose.

The substance. **$(AB{)}^T$ is $B^T{A}^T$, not $A^T{B}^T$** — for non-square matrices, $A^T{B}^T$ often cannot even be formed.