Solution. The transpose of these matrices is:

Notice how taking the transpose affects the shape of the matrices:

• a $2\times2$ matrix remains a $2\times2$ matrix.

• a $3\times2$ matrix becomes a $2\times3$ matrix.

• a $3\times3$ matrix remains a $3\times3$ matrix.

In general, if $\boldsymbol{A}$ is an $m\times{n}$ matrix, then $\boldsymbol{A}^T$ will be an $n\times{m}$ matrix. Let's also understand the mathematical definition of the matrix transpose. Observe what happens to the entry $a_{21}=3$ in $\boldsymbol{A}$ after taking the transpose. Since the rows and columns are swapped, this entry is located at $a_{12}$ in $\boldsymbol{A}^T$.

Solution. Since we are swapping the rows and columns, the transpose of $\boldsymbol{A}$ would have $3$ rows and $2$ columns, that is, $\boldsymbol{A}^T\in\mathbb{R}^{3\times2}$.

Properties of matrix transpose

Proof. By definition, $\boldsymbol{A}^T$ is obtained by swapping the rows and columns of $\boldsymbol{A}$. Taking the transpose of $\boldsymbol{A}^T$ will swap back the rows and columns and so we end up with the original matrix $\boldsymbol{A}$.

To be more mathematically precise, suppose $\boldsymbol{A}$ is as follows:

The transpose of $\boldsymbol{A}$ is:

The transpose of $\boldsymbol{A}^T$ is:

This completes the proof.

Proof. Consider the following matrices:

The transpose of $\boldsymbol{A}$ and $\boldsymbol{B}$ is:

The sum $\boldsymbol{A}^T+\boldsymbol{B}^T$ is:

Now, the sum $\boldsymbol{A}+\boldsymbol{B}$ is:

Taking the transpose gives:

Notice that this matrix is equal to \eqref{eq:yJ6VsJMu5sStPhtgG31}. Therefore, we have that:

This completes the proof.

Proof. Suppose $\boldsymbol{A}$ is an $n\times{n}$ matrix. The diagonals of $\boldsymbol{A}$ is $a_{ii}$ for $i=1,2,\cdots,n$. Taking the transpose involves switching the two subscripts, but since they are identical, we also end up with $a_{ii}$ as the diagonal entries of $\boldsymbol{A}^T$. This completes the proof.

Proof. Consider the following matrix:

If $k$ is any scalar, then the product $k\boldsymbol{A}$ is:

The transpose of $k\boldsymbol{A}$ is:

The transpose of $\boldsymbol{A}$ is:

The scalar-matrix product $k\boldsymbol{A}^T$ is:

This is equal to the matrix in \eqref{eq:VMQuvGftyd4iwd5MPJ0}. Therefore, we conclude that:

This completes the proof.

Proof. We will prove the case for $\mathbb{R}^3$ but this easily generalizes to $\mathbb{R}^n$. Let vectors $\boldsymbol{v}$ and $\boldsymbol{w}$ be defined as follows:

Their dot product is:

Similarly, we have that:

Similarly, we have that:

This completes the proof.

Proof. By the previous propertylink and propertylink of dot product, we have that:

This completes the proof.

Proof. From the rules of matrix multiplication, we know that:

Here, the subscript $ij$ represents the value in the $i$-th row $j$-th column. \eqref{eq:hZKaAMSlx63jVMkEVY8} is true for all $i=1,2,\cdots,m$ and $j=1,2,\cdots,p$. For instance, $(\boldsymbol{AB})_{13}$ represents the entry at the $1$st row $3$rd column of matrix $\boldsymbol{AB}$.

We know from the definition of transpose that:

Equating \eqref{eq:aZZgDF30ZHVz6WidSNK} and \eqref{eq:hZKaAMSlx63jVMkEVY8} gives:

Now, consider the following:

Here, we used the fact that the entry $b_{jk}$ in $\boldsymbol{B}^T$ is equal to the entry $b_{kj}$ in $\boldsymbol{B}$.

Equating \eqref{eq:LMNSG63uKXgNU6nkuDe} and \eqref{eq:ZgdDoK2zrOpYmDYO8yN} gives:

Equation \eqref{eq:Bd1Iy1tDn1Ilz2TpbMa} holds true for all $j=1,2,\cdots,p$, and $i=1,2,\cdots,m$. Since the values of two matrices $\boldsymbol{AB}^T$ and $\boldsymbol{B}^T\boldsymbol{A}^T$ are equivalent, these matrices are equivalent, that is:

This completes the proof.

Proof. The proof makes use of our previous theorem $(\boldsymbol{AB})^T=\boldsymbol{B}^T\boldsymbol{A}^T$. We start from the left-hand side of our proposition:

This completes the proof.

We will now generalize this theorem for any number of matrices. The proof uses induction and follows a similar logic.

Proof. We will prove the theorem by induction. Consider the base case when we have $2$ matrices. We have already shown in theoremlink that:

Therefore, the base case holds. We now assume that the theorem holds for $n-1$ matrices:

Our goal is to show that the theorem holds for $n$ matrices:

We now use the inductive assumption \eqref{eq:IHQoYjMe8rq0BGQ1T2f} to get:

By the principle of mathematical induction, the theorem holds for the general case of $n$ matrices. This completes the proof.

The next theorem is useful for certain proofs.

Proof. The matrix-vector product $\boldsymbol{A}\boldsymbol{v}$ results in a vector. We can use theoremlink to convert the dot product of $\boldsymbol{Av}$ and $\boldsymbol{w}$ into a matrix product:

Note the following:

• the third step uses theoremlink, that is, $(\boldsymbol{A}\boldsymbol{B})^T= \boldsymbol{B}^T\boldsymbol{A}^T$.

• the final step uses theoremlink to convert the matrix product into a dot product.

This completes the proof.

Practice problems