Question

Show that the transpose of a sum of matrices is equal to the sum of the transposes. Also show that${\left({\text{M}}^n\right)}^{\mathbf{\text{T}}}\mathbf{=}{\left({\text{M}}^{\text{T}}\right)}^{\mathbf{n}}$. Hint: Use (9.11)and (9.8).

Short answer

The sum of the transposes is equal to the transpose of the sum of the matrices.

Step-by-step solution

Given information.

The equation, ${\left(M^n\right)}^T={\left(M^T\right)}^n$needs to be proven.

Step 2: Transpose of a matrix.

A matrix's transpose is determined by converting its rows into columns or columns into rows.

Show that the transpose of the sum of the matrix is equal to the sum of their transposes.

Assume A and B are two matrices. Now,

$$\begin{gathered} (A+B{)}_{ij}^T=(A+B{)}_{ij} \\ (A+B{)}_{ij}=A_{ij}+B_{ij} \\ (A+B{)}_{ij}^T=A_{ij}^T+B_{ij}^T \\ {A}_{ij}^T+B_{ij}^T={\left(A^T+B^T\right)}_{ij} \end{gathered}$$

Also,

$$\begin{gathered} (A+B{)}_{ij}^T={\left(A^T+B^T\right)}_{ij} \\ (A+B{)}^T=A^T+B^T \end{gathered}$$

Hence, from the above equation, the sum of the transposes is equal to the transpose of the sum of the matrices. Now,

$$\begin{gathered} {\left(M^n\right)}^T=(M.M\dots M{)}^T\left(\text{for}n\text{times}\right) \\ =\left(M\right(M\dots M){)}^T \end{gathered}$$

Since$A\left(BC\right)=\left(AB\right)C=ABC$,

$$\begin{gathered} (M\dots M{)}^T{M}^T \left(\because {\left(ABCD\right)}^T\left.=D^T{C}^T{B}^T{A}^T\right)\right. \\ =\left(M\right(M\dots .M){)}^T{M}^T \\ =\left({(M\dots M)}^T{M}^T\right)M^T \end{gathered}$$

Solve further.

$$\begin{gathered} =(M\dots M{)}^T{M}^T{M}^T \\ =(MM{)}^T{M}^T\dots M^T \\ =M^T{M}^T{M}^T\dots \dots M^T(n-\text{times}) \\ ={\left(M^T\right)}^n \end{gathered}$$

Hence,

${\left(M^T\right)}^n={\left(M^n\right)}^T$

Hence, proved.