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# Transpose (Matrix)|Definition & Meaning

## Definition

In **matrix** terminology, **transpose** is a function that swaps the **rows** and **columns** of a **matrix**. It is like flipping a **matrix** over its diagonal (in **geometry,** this swaps the x and y axes). It is represented using a capital **superscript** “**T**” after the **matrix** or its name, e.g., M$^T$ where M is any **matrix**. **Transposing** a **transposed** **matrix** gives back the original **matrix**, so (M$^T$)$^T$ = T.

One of the most used methods in linear algebra is the **transpose** of a **matrix**. This method can be achieved by interchanging the **rows** with the **columns** and each column with the **rows** of the given **matrix**. In other words, flipping a **matrix** over its diagonals will carry out the transposition of that **matrix**.

Before going into detail about the **transpose** of a **matrix**, let’s first learn the basic concepts of a **matrix**.

## What Is a Matrix?

**A** **matrix** is a modified form of a table with **rows** and **columns** of numerical data or mathematical functions arranged in rectangular **arrays**. Since the number of **rows** and **columns** can vary, the **matrices** are divided into many types. The simplest form of a **matrix** contains one row and one column:

$\begin{bmatrix} 1 \end{bmatrix}$

In practical mathematics, **matrices** have multiple **rows** and **columns**, like the one given below:

**A** = $\begin{bmatrix} 1 & 2 \\ a & b \\ x & y \end{bmatrix}_{3\times 2}$

In the above **matrix**, there are a total of 3 **rows** and 2 **columns**. This defines the order of a **matrix,** i.e., 3 x 2. Thus the total number of elements is 6.

**A** row is defined by the horizontal array, whereas the vertical array is comprehended as the column.

Let’s take a look at another form of a **matrix**:

**B** = $\begin{bmatrix} 1 & 2 & 3\\ a & b & c \end{bmatrix}_{2\times 3}$

As you can see, there is a visual difference in both matrices **A** and **B**, as there are a greater number of **rows** in matrix **A** than **in matrix** **B**. Such a **matrix** having more **rows** than **columns** is known as a vertical **matrix**, whereas **matrix** **B** has a greater number of **columns** than **A**, which makes it a horizontal **matrix**.

Let’s suppose we have a situation in which every element of **matrix** **A** is present in **matrix** **B,** and both have the same number of elements. Then are these **matrices** equivalent? To answer this question, a formula has been provided, which is stated as:

**A** = $\begin{pmatrix} a_{ij} \end{pmatrix}_{m\times n}$ and **B** = $\begin{pmatrix} b_{ij} \end{pmatrix}_{r\times s}$

For **A** and **B** to be equivalent, the following points have to be considered,

- m = r and n = s such that the order must be the same for the two
**matrices**. - For $a_{ij} = b_{ij}$, every value of i and j must be equal.

## What Is the Transpose of a Matrix?

In **matrix** nomenclature, **transpose** is a procedure that switches the horizontal and vertical arrays of a **matrix**. This feels like reversing a **matrix** over its diagonal. It is characterized using a tick or a capital superscript “T” after the **matrix** or its title, e.g., **A**’ or **A**$^T$ where **A** is any **matrix**.

The general form of a transposition goes by this statement:

If **A** = $\begin{pmatrix} a_{ij} \end{pmatrix}_{m\times n}$ then **A**’ = $\begin{pmatrix} a_{ij} \end{pmatrix}_{n\times m}$

Thus a **transpose** of a **matrix** can also be stated as a method of converting all the **rows** into the corresponding **columns** of a given **matrix** and vice-versa.

## How To Find the Transpose of a Matrix?

Interchange the **rows** and **columns** sound quite straightforward, but that’s not it. Let’s say we have a **matrix** of order 2×3 which means that it contains 2 **rows** and 3 **columns,** and the total number of elements is 6.

The process of the **transpose** of a **matrix** starts by taking the first row and sliding it onto the first column of a new **matrix**. The new **matrix** will thus contain one column with all the row elements of the previous **matrix**.

Similarly, the second row of the given **matrix** will be mapped as the second column of the new **matrix**, with all the elements remaining the same. Thus the new **matrix** now has a total of 3 **rows** and 2 **columns**, defining a new order 3×2 for the new **matrix**.

## Properties of a Matrix Transpose

The properties of **matrix** **transpose** can be better understood by taking two **matrices** of equal order. Let those **matrices** be **A** and **B;** then their **transpose** properties can be defined as follows,

### Double Transpose

Double **transpose** means taking the **transpose** of a **transposed** **matrix**. This results in a **matrix** that is exactly equal to the original one. Such as, the **transpose** of **matrix** **A** is **A**’, and the **transpose** of **A**’ will be (**A**’)’, which is equal to **matrix** **A**.

The concept behind this property is that every element of **A** that is $a_{ij}$ gets flipped to $a_{ji}$ when the first **transpose** is taken. Thus making **A**’. Now, if another **transpose** is taken, the element $a_{ij}$ gets flipped back to $a_{ij}$, which was the element of the original **matrix** **A**.

Let’s say we have a **matrix** Z:

Z = $\begin{bmatrix} 20 & 12 & 6\\ 9 & 7 & 0 \\ 3 & 89 & 20 \end{bmatrix}$

Then:

Z’ = $\begin{bmatrix} 20 & 9 & 3\\ 12 & 7 & 89 \\ 6 & 0 & 20 \end{bmatrix}$

And:

(Z’)’ = $\begin{bmatrix} 20 & 12 & 6\\ 9 & 7 & 0 \\ 3 & 89 & 20 \end{bmatrix}$ = Z

### Addition of Transposed Matrices

Adding two individually **transposed** **matrices** is the same as adding the **matrices** and then taking the **transpose** of their sum. It is more like a distributive property of **matrices**.

To prove that (**A** + **B**)’ = **A**’ + **B**’, let’s say we have two **matrices,** **A** and **B**:

**A** = $\begin{bmatrix} 2 & -3 & 8\\ 21 & 6 & -6 \\ 4 & -33 & 19 \end{bmatrix}$ and **B** = $\begin{bmatrix} 1 & -29 & -8\\ 2 & 0 & 3 \\ 17 & 15 & 4 \end{bmatrix}$

Then:

**A** + **B** = **A** = $\begin{bmatrix} 2 & -3 & 8\\ 21 & 6 & -6 \\ 4 & -33 & 19 \end{bmatrix}$ + **B** = $\begin{bmatrix} 1 & -29 & -8\\ 2 & 0 & 3 \\ 17 & 15 & 4 \end{bmatrix}$ = $\begin{bmatrix} 3 & -32 & 0\\ 23 & 6 & -3 \\ 21 & -18 & 23\end{bmatrix}$

(**A** + **B**)’ = $\begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & -18 \\ 0 & -3 & 23\end{bmatrix}$

Now:

**A**’ + **B**’ = $\begin{bmatrix} 2 & 21 & 4 \\ -3 & 6 & -33 \\ 8 & -6 & 19 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 17 \\ -29 & 0 & 15 \\ -8 & 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & -18 \\ 0 & -3 & 23 \end{bmatrix}$

Thus Proved that (**A** + **B**)’ = **A**’ + **B**’.

### Multiplying a Constant

It is the same as the addition of **transposed** **matrices**, meaning that if a constant is multiplied by a **matrix** and then the **transpose** is taken, this will be equal to multiplying the constant with the **transposed** **matrix** as:

**(kA)’ = kA’**

### Multiplication Property

Multiplying two **matrices** and then taking their **transpose** is equal to multiplying two **transposed** **matrices** but in reverse order such as:

**(AB)’ = B’A’**

## Solved Example of Transpose of a Matrix

For the given **matrices** **A** and **B**, prove that (AB)’ = B’A’.

**A** = $\begin{bmatrix} 9 & 8 \\ 2 & -3 \end{bmatrix}$ and **B** = $\begin{bmatrix} 4 & 2 \\ 1 & 0 \end{bmatrix}$

### Solution

First, we will find the left-hand side:

**A **$\times$ **B** = $\begin{bmatrix} 44 & 18 \\ 5 & 4 \end{bmatrix}$ $\Rightarrow$ (AB)’ = $\begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix}$

**B**’**A**’ = $\begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix}$

= $\begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix}$ = (AB)’

Therefore, (AB)’ = **B**’**A**’.

**A**’**B**’ =$\begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 40 & 9 \\ 26 & 8 \end{bmatrix}$

This shows that (AB)’ $\ne$ **A**’**B**’.

*All images are created using GeoGebra.*