Transpose of the Matrix
The transpose of a matrix is an important phenomenon in the matrix theory. Let us understand the concept of the transpose of a matrix. If I say roughly then the process of taking transpose of a matrix is something equivalent to changing the rows of the matrix into columns and columns of the matrix into rows. If $A$ is a matrix then its transpose is denoted by $A^T$ or $A^t$ or $A'$.
Consider the following example of matrix and its transpose.
$$ \begin{aligned} A&=\left[\begin{matrix}1&7&2&9\\2&0&0&5\\1&9&8&3\end{matrix}\right]_{3\times4} \end{aligned} $$
The transpose of matrix $A$ is
$$ \begin{aligned} A^T&=\left[\begin{matrix}1&2&1\\7&0&9\\2&0&8\\9&5&3\end{matrix}\right]_{4\times 3} \end{aligned} $$
Here we can observe that first row of the matrix $A$ is first column of the matrix $A^T$; second row of the matrix $A$ is second column of the matrix $A^T$; and third row of the matrix $A$ is third column of the matrix $A^T$; second thing order of the matrix $A$ is $3\times 4$ while the order of the matrix $A^T$ get reversed and it is $4\times 3$.
Now let us give formal definition of the transpose of the matrix.
Definition of Transpose of the Matrix
Let $A$ be a matrix then the transpose of the matrix $A$ is the matrix obtained on interchanging all the rows of $A$ into columns and vice versa.
More precisely, if
$$ \begin{aligned} A&=\left[\begin{matrix} a_{11}&a_{12}&\ldots&a_{1i}&\ldots&a_{1j}&\ldots&a_{1n}\\ a_{21}&a_{22}&\ldots&a_{2i}&\ldots&a_{2j}&\ldots&a_{2n}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{i1}&a_{i2}&\ldots&a_{ii}&\ldots&a_{ij}&\ldots&a_{in}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{j1}&a_{j2}&\ldots&a_{ji}&\ldots&a_{jj}&\ldots&a_{jn}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{m1}&a_{m2}&\ldots&a_{mi}&\ldots&a_{mj}&\ldots&a_{mn}\\ \end{matrix}\right]_{m\times n} \end{aligned} $$
then the transpose of matrix $A$ is
$$ \begin{aligned} A^T&=\left[\begin{matrix} a_{11}&a_{21}&\ldots&a_{i1}&\ldots&a_{j1}&\ldots&a_{n1}\\ a_{12}&a_{22}&\ldots&a_{i2}&\ldots&a_{j1}&\ldots&a_{n2}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{1i}&a_{2i}&\ldots&a_{ii}&\ldots&a_{ji}&\ldots&a_{ni}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{1j}&a_{2j}&\ldots&a_{ij}&\ldots&a_{jj}&\ldots&a_{nj}\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ a_{1m}&a_{2m}&\ldots&a_{im}&\ldots&a_{jm}&\ldots&a_{nm}\\ \end{matrix}\right]_{n\times m} \end{aligned} $$
In other words if $A=[a_{ij}]_{m\times n}$, then the transpose of the matrix $A$ is $A^T=[a_{ji}]_{n\times m}$.
Thus on taking transpose of the matrix, we find two major changes; first one is $(i,j)^{th}$ entry of the matrix $A$ becomes $(j,i)^{th}$ entry of the matrix $A^T$ and second one is order of the transposed matrix get reversed than the order of the original matrix.
Examples of Transpose of Matrix.
Transpose of a square matrix is a square matrix
$$ \begin{aligned} A&=\left[\begin{matrix}1&2&9\\2&0&5\\1&9&8\end{matrix}\right]_{3\times 3} \end{aligned} $$
Transpose of matrix $A$ is
$$ \begin{aligned} A^T&=\left[\begin{matrix}1&2&1\\2&0&9\\9&5&8\end{matrix}\right]_{3\times 3} \end{aligned} $$
Transpose of a row matrix is a column matrix
$$ \begin{aligned} B&=\left[\begin{matrix}3&2&7&8\end{matrix}\right]_{1\times 4} \end{aligned} $$
Transpose of matrix $B$ is
$$ \begin{aligned} B^T=\left[\begin{matrix}3\\2\\7\\8\end{matrix}\right]_{4\times 1} \end{aligned} $$
Transpose of a column matrix is a row matrix
$$ \begin{aligned} C&=\left[\begin{matrix}1\\0\\3\\8\end{matrix}\right]_{4\times 1} \end{aligned} $$
Transpose of matrix $C$ is
$$ \begin{aligned} C^T&=\left[\begin{matrix}1&0&3&8\end{matrix}\right]_{1\times 4} \end{aligned} $$
Transpose of a matrix is it self
$$ \begin{aligned} D&=\left[\begin{matrix}1&0&3\\0&1&2\\3&2&9\end{matrix}\right]_{3\times 3} \end{aligned} $$
Transpose of matrix $D$ is
$$ \begin{aligned} D^T&=\left[\begin{matrix}1&0&3\\0&1&2\\3&2&9\end{matrix}\right]_{3\times 3}\\ &=D \end{aligned} $$
Transpose of a matrix is negative of the original matrix
$$ \begin{aligned} E&=\left[\begin{matrix}0&2&-5\\-2&0&3\\5&-3&0\end{matrix}\right] \end{aligned} $$
Transpose of matrix $E$ is
$$ \begin{aligned} E^T&=\left[\begin{matrix}0&-2&5\\2&0&-3\\-5&3&0\end{matrix}\right]\\ &=-\left[\begin{matrix}0&2&-5\\-2&0&3\\5&-3&0\end{matrix}\right]\\ &=-E \end{aligned} $$
The process of taking transpose produce a new matrix from the given matrix. Then there are some natural questions like; How the process of taking transpose behaves it applies more than one time? How the process of taking transpose behaves with sum and product of matrices? The following properties of transpose of matrices answer these questions.
Properties of Transpose of Matrices
Property 1. $(A^T)^T=A$
This property says that the transpose of the transpose is original. Let us verify this property by examples.
Verification of Property 1 by Examples
- Example 1
$$ \begin{aligned} A&=\left[\begin{matrix}2&-7&0&6\\1&8&5&-2\\7&8&-1&6\end{matrix}\right] \end{aligned} $$
The transpose of matrix $A$ is
$$ \begin{aligned} A^T&=\left[\begin{matrix}2&1&7\\-7&8&8\\0&5&-1\\6&-2&6\end{matrix}\right]\\ \end{aligned} $$
Now, the transpose of $A^T$ is
$$ \begin{aligned} (A^T)^T&=\left[\begin{matrix}2&-7&0&6\\1&8&5&-2\\7&8&-1&6\end{matrix}\right]\\ &=A \end{aligned} $$
- Example 2
Consider a matrix $B$ as follows:
$$ \begin{aligned} B&=\left[\begin{matrix}1&0&8\\3&1&7\\7&8&-1\end{matrix}\right] \end{aligned} $$
The transpose of matrix $B$ is
$$ \begin{aligned} B^T&=\left[\begin{matrix}1&3&7\\0&1&8\\8&7&-1\end{matrix}\right] \end{aligned} $$
Now, transpose of the $B^T$ is
$$ \begin{aligned} (B^T)^T&=\left[\begin{matrix}1&0&8\\3&1&7\\7&8&-1\end{matrix}\right]\\ &=B \end{aligned} $$
Property 2. $(k\;A)^T=k\;A^T$
This property says that the process of taking the transpose does not affect the scalar which is multiplied to the matrix. Let us verify this property by examples.
Verification of Property 2 by Examples
Consider a $2\times 2$ matrix $A$ as
$$ \begin{aligned} A&=\left[\begin{matrix}2&-7\\1&8\end{matrix}\right]\\ \Longrightarrow\;2A&=2\left[\begin{matrix}2&-7\\1&8\end{matrix}\right]\\ &=\left[\begin{matrix}4&-14\\2&16\end{matrix}\right] \end{aligned} $$
Now
$$ \begin{aligned} (2A)^T&=\left[\begin{matrix}4&-14\\2&16\end{matrix}\right]^T\\ &=\left[\begin{matrix}4&2\\-14&16\end{matrix}\right]\\ &=2\left[\begin{matrix}2&1\\-7&8\end{matrix}\right]\\ &=2A^T\end{aligned} $$
- Example 2
$$ \begin{aligned} B &=\left[\begin{matrix}1&5&-2\\7&-1&6\end{matrix}\right]\\ \end{aligned} $$
$$ \begin{aligned} 5B &=5\left[\begin{matrix}1&5&-2\\7&-1&6\end{matrix}\right]\\ &=\left[\begin{matrix}5&25&-10\\35&-5&30\end{matrix}\right] \end{aligned} $$
Now
$$ \begin{aligned} (5B)^T&=\left[\begin{matrix}5&25&-10\\35&-5&30\end{matrix}\right]^T\\ &=\left[\begin{matrix}5&35\\25&-5\\-10&30\end{matrix}\right]\\ &=5\left[\begin{matrix}1&7\\5&-1\\-2&6\end{matrix}\right]\\ &=5B^T \end{aligned} $$
Property 3. $(A+B)^T=A^T+B^T$
This property says that the first you add two matrices and then take the transpose of its addition or first you take the transpose of the matrices and then take its addition these two things are same. i.e. the process of addition and taking transpose are commutative. Let us verify this property by an example.
Verification of Property 3 by an Example :
Let
$$ \begin{aligned} A&=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right],\\ B&=\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right] \end{aligned} $$
$$ \begin{aligned} A^T&=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]\\ B^T&=\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right] \end{aligned} $$
Then
$$ \begin{aligned} A^T+B^T&=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]+\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right]\\ &=\left[\begin{matrix}5&3\\0&9\\7&5\end{matrix}\right] \end{aligned} $$
Also
$$ \begin{aligned} A+B&=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right]+\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right]\\ &=\left[\begin{matrix}5&0&7\\3&9&5\end{matrix}\right] \end{aligned} $$
Thus
$$ \begin{aligned} (A+B)^T&=\left[\begin{matrix}5&3\\0&9\\7&5\end{matrix}\right]\\ &=A^T+B^T \end{aligned} $$
Property 3 can be generalized for finitely many matrices and it can be put mathematically as
$$(A_1+A_2+\cdots+A_n)^T=A_1^T+A_2^T+\cdots+A_n^T$$
$$i.e. \;\;\left(\sum\limits_{i=1}^nA_i\right)^T=\sum\limits_{i=1}^nA_i^T$$
Property 4 $(A-B)^T=A^T-B^T$
This property is analogue to the property 3, it says that you can deal the process of taking transpose of the subtraction of two matrices is same as the case of addition. Let us verify this property by an example.
Verification of Property 4 by an Example :
Let
$$ \begin{aligned} & A=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right], B=\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right]\;\\ \Longrightarrow & A^T=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]\text{ and } B^T=\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right]\end{aligned} $$
Then
$$ \begin{aligned} A^T-B^T&=\left[\begin{matrix}2&1\\-7&8\\3&2\end{matrix}\right]-\left[\begin{matrix}3&2\\7&1\\4&3\end{matrix}\right]\\ &=\left[\begin{matrix}-1&-1\\-14&7\\1&-1\end{matrix}\right] \end{aligned} $$
Also
$$ \begin{aligned} A-B&=\left[\begin{matrix}2&-7&3\\1&8&2\end{matrix}\right]-\left[\begin{matrix}3&7&4\\2&1&3\end{matrix}\right]\\ &=\left[\begin{matrix}-1&-14&1\\-1&7&-1\end{matrix}\right] \end{aligned} $$
Thus
$$ \begin{aligned} (A-B)^T &=\left[\begin{matrix}-1&-1\\-14&7\\1&-1\end{matrix}\right]\\ &=A^T-B^T. \end{aligned} $$
Property 5. (Shock Shoes Property) $(A\cdot B)^T=B^T\cdot A^T$
This property says that the transpose of product of two matrices is product of transposed matrices in the reverse order.
This property is also know as shock shoes property. Why? Let us try to justify why. You think LHS i.e. $(A\cdot B)^T$ be a process of put on the shock and shoes and the RHS i.e. $B^T\cdot A^T$ be a process of take off shock and shoes. Then during the put on first we put on the shock and then shoes while during taking off it first we took off shoes and then shock. That is the process get reversed and exactly this happened in the case of takeing transpose of the product of two matrices.
Verification of Property 5 by an Example
Let
$$ \begin{aligned} A&=\left[\begin{matrix}2& 5& 3\\-1& 3& 6\end{matrix}\right] \end{aligned} $$
$$ \begin{aligned} B&=\left[\begin{matrix}2 &7& 9\\3& -2& 0\\ 1& 0 &8\end{matrix}\right] \end{aligned} $$
Now,
$$ \begin{aligned} A^T&=\left[\begin{matrix}2&-1\\5&3\\3&6 \end{matrix}\right] \end{aligned} $$
$$ \begin{aligned} B^T&=\left[\begin{matrix}2&3&1\\7&-2&0\\9&0&8\end{matrix}\right] \end{aligned} $$
Then
$$ \begin{aligned} B^T\cdot A^T&=\left[\begin{matrix}2&3&1\\7&-2&0\\9&0&8\end{matrix}\right]\left[\begin{matrix}2&-1\\5&3\\3&6 \end{matrix}\right]\\ &=\left[\begin{matrix}22&13\\4&-13\\42&39\end{matrix}\right] \end{aligned}. $$
Also
$$ \begin{aligned} A\cdot B &=\left[\begin{matrix}2& 5& 3\\-1& 3& 6\end{matrix}\right] \left[\begin{matrix}2 &7& 9\\3& -2& 0\\ 1& 0 &8\end{matrix}\right]\\ &=\left[\begin{matrix}22&4&42\\13&-13&39\end{matrix}\right] \end{aligned} $$
Thus
$$ \begin{aligned} (A\cdot B)^T &=\left[\begin{matrix}22&13\\4&-13\\42&39\end{matrix}\right]\\ &=B^T\cdot A^T \end{aligned} $$
Property 5 can be generalized for finitely many matrices and it can be put mathematically as
$$ (A_1\cdot A_2\cdot\cdots\cdot A_n)^T=A_1^T\cdot A_2^T\cdot\cdots\cdot A_n^T $$
$$ i.e.\;\;\left(\prod\limits_{i=1}^nA_i\right)^T=\prod\limits_{i=1}^nA_i^T $$
Summary of Properties of Transpose of Matrices
$(A^T)^T=A$$(k\;A)^T=k\;A^T$$(A+B)^T=A^T+B^T$$(A_1+A_2+\cdots+A_n)^T=A_1^T+A_2^T+\cdots+A_n^T$i.e.$\left(\sum\limits_{i=1}^nA_i\right)^T=\sum\limits_{i=1}^nA_i^T$.$(A-B)^T=A^T-B^T$$(A\cdot B)^T=B^T\cdot A^T$$(A_1\cdot A_2\cdot\cdots\cdot A_n)^T=A_1^T\cdot A_2^T\cdot\cdots\cdot A_n^T$i.e.$\left(\prod\limits_{i=1}^nA_i\right)^T=\prod\limits_{i=1}^nA_i^T$.
