Transpose, symmetric and skew-symmetric matrices

Turn­ing a matrix on its side, so that its rows become columns, gives its trans­pose. The trans­pose has sim­ple rules, with one sur­prise for prod­ucts, and it leads to two impor­tant fam­i­lies of square matri­ces: sym­met­ric matri­ces, which equal their own trans­pose, and skew-sym­met­ric matri­ces, which equal the neg­a­tive of their trans­pose. Every square matrix turns out to be the sum of one of each.

The trans­pose#

The trans­pose of a matrix $A$, writ­ten $A'$ or $A^T$, is the matrix obtained by writ­ing the rows of $A$ as columns. If $A$ is of order $m \times n$, then $A'$ is of order $n \times m$, and the entry in row $i$, col­umn $j$ of $A'$ is $a_{ji}$.

Exam­ple 1#

Find the trans­poses of $A = \begin{bmatrix} 1 & 4 & -2 \\ 0 & 3 & 5 \end{bmatrix}$ and $B = \begin{bmatrix} 7 \\ -1 \end{bmatrix}$.

The first row of $A$ becomes the first col­umn of $A'$, and the sec­ond row becomes the sec­ond col­umn:

$$A' = \begin{bmatrix} 1 & 0 \\ 4 & 3 \\ -2 & 5 \end{bmatrix}, \qquad B' = \begin{bmatrix} 7 & -1 \end{bmatrix}$$

$A$ is $2 \times 3$ and $A'$ is $3 \times 2$. The trans­pose of a col­umn matrix is a row matrix.

Prop­er­ties of the trans­pose#

For matri­ces of suit­able orders and a scalar $k$:

1. $(A')' = A$
2. $(A + B)' = A' + B'$
3. $(kA)' = kA'$
4. $(AB)' = B'A'$ (the rever­sal law)

The fourth prop­erty is the sur­prise: the trans­pose of a prod­uct is the prod­uct of the trans­poses in the reverse order.

Exam­ple 2#

Ver­ify $(AB)' = B'A'$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$.

From the pre­vi­ous les­son, $AB = \begin{bmatrix} 19 & 22 \\ 43 & 50 \end{bmatrix}$, so $(AB)' = \begin{bmatrix} 19 & 43 \\ 22 & 50 \end{bmatrix}$.

Now $B' = \begin{bmatrix} 5 & 7 \\ 6 & 8 \end{bmatrix}$ and $A' = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$, and

$$B'A' = \begin{bmatrix} 5 + 14 & 15 + 28 \\ 6 + 16 & 18 + 32 \end{bmatrix} = \begin{bmatrix} 19 & 43 \\ 22 & 50 \end{bmatrix}$$

The two agree. (You can check that $A'B'$ gives a dif­fer­ent matrix.)

Sym­met­ric matri­ces#

A square matrix $A$ is sym­met­ric if $A' = A$, that is, $a_{ij} = a_{ji}$ for all $i$ and $j$. The entries are mir­ror images in the prin­ci­pal diag­o­nal.

$$\begin{bmatrix} 2 & 3 & 1 \\ 3 & 0 & 5 \\ 1 & 5 & 7 \end{bmatrix} \text{ is symmetric: } a_{12} = a_{21} = 3, \; a_{13} = a_{31} = 1, \; a_{23} = a_{32} = 5.$$

Skew-sym­met­ric matri­ces#

A square matrix $A$ is skew-sym­met­ric if $A' = -A$, that is, $a_{ij} = -a_{ji}$ for all $i$ and $j$.

On the prin­ci­pal diag­o­nal $i = j$, so $a_{ii} = -a_{ii}$, which gives $2a_{ii} = 0$. Every diag­o­nal entry of a skew-sym­met­ric matrix is $0$.

$$\begin{bmatrix} 0 & 2 & -2 \\ -2 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \text{ is skew-symmetric.}$$

Exam­ple 3#

Find $x$ and $y$ if $\begin{bmatrix} 0 & x & 4 \\ -3 & 0 & y \\ -4 & 6 & 0 \end{bmatrix}$ is skew-sym­met­ric.

We need $a_{12} = -a_{21}$, so $x = -(-3) = 3$; and $a_{23} = -a_{32}$, so $y = -6$.

Split­ting a square matrix#

Result. Every square matrix $A$ can be writ­ten as the sum of a sym­met­ric matrix and a skew-sym­met­ric matrix:

$$\displaystyle A = P + Q, \quad \text{where } P = \frac{1}{2}(A + A') \text{ and } Q = \frac{1}{2}(A - A')$$

Why it works. Adding, $\displaystyle P + Q = \frac{1}{2}(2A) = A$. Using the prop­er­ties of the trans­pose, $\displaystyle P' = \frac{1}{2}(A' + A) = P$, so $P$ is sym­met­ric; and $\displaystyle Q' = \frac{1}{2}(A' - A) = -Q$, so $Q$ is skew-sym­met­ric.

Exam­ple 4#

Express $A = \begin{bmatrix} 2 & 5 & -1 \\ 1 & 0 & 4 \\ 3 & 6 & 7 \end{bmatrix}$ as the sum of a sym­met­ric and a skew-sym­met­ric matrix.

$$A' = \begin{bmatrix} 2 & 1 & 3 \\ 5 & 0 & 6 \\ -1 & 4 & 7 \end{bmatrix}$$

$$\displaystyle P = \frac{1}{2}(A + A') = \frac{1}{2}\begin{bmatrix} 4 & 6 & 2 \\ 6 & 0 & 10 \\ 2 & 10 & 14 \end{bmatrix} = \begin{bmatrix} 2 & 3 & 1 \\ 3 & 0 & 5 \\ 1 & 5 & 7 \end{bmatrix}$$

$$\displaystyle Q = \frac{1}{2}(A - A') = \frac{1}{2}\begin{bmatrix} 0 & 4 & -4 \\ -4 & 0 & -2 \\ 4 & 2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 & -2 \\ -2 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix}$$

$P$ is sym­met­ric, $Q$ is skew-sym­met­ric, and adding them entry by entry gives back $A$.

Prac­tice#

1. Write the trans­pose of $\begin{bmatrix} 3 & -1 & 0 & 2 \end{bmatrix}$ and state its order.
2. Write the trans­pose of $\begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{bmatrix}$.
3. For $A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}$, find $(A + B)'$.
4. With the same $A$ and $B$, ver­ify that $(AB)' = B'A'$.
5. Is $\begin{bmatrix} 1 & 4 \\ 4 & -3 \end{bmatrix}$ sym­met­ric, skew-sym­met­ric or nei­ther?
6. Is $\begin{bmatrix} 0 & -5 \\ 5 & 0 \end{bmatrix}$ sym­met­ric, skew-sym­met­ric or nei­ther?
7. Find $a$, $b$, $c$ if $\begin{bmatrix} 1 & a & 2 \\ 7 & 4 & b \\ c & -3 & 0 \end{bmatrix}$ is sym­met­ric.
8. Why can $\begin{bmatrix} 1 & 2 \\ -2 & 0 \end{bmatrix}$ not be skew-sym­met­ric?
9. Express $\begin{bmatrix} 4 & 7 \\ 1 & 3 \end{bmatrix}$ as the sum of a sym­met­ric and a skew-sym­met­ric matrix.
10. Express $\begin{bmatrix} 3 & 1 & -2 \\ 5 & 0 & 4 \\ 2 & 6 & 1 \end{bmatrix}$ as the sum of a sym­met­ric and a skew-sym­met­ric matrix.

Answers#