Proof of Elementary Properties of the Transpose of a Real Matrix

By Transpose of a Real Matrix, $A^{\top}$ is the real $n\times m$ matrix with $(A^{\top})_{ji}=A_{ij}$ for $i\in[m]$ and $j\in[n]$, and by Real Matrix and the Set of Real Matrices two matrices of the same size are equal exactly when all of their entries agree. Sums below are the finite sums of the field of real numbers.

Claim 1. For $i\in[m]$ and $j\in[n]$, $\bigl((A^{\top})^{\top}\bigr)_{ij}=(A^{\top})_{ji}=A_{ij}$.

Claim 2. By Sum of Real Matrices, $\bigl((A+B)^{\top}\bigr)_{ji}=(A+B)_{ij}=A_{ij}+B_{ij}=(A^{\top})_{ji}+(B^{\top})_{ji}=(A^{\top}+B^{\top})_{ji}$. By Scalar Multiple of a Real Matrix, $\bigl((\mu A)^{\top}\bigr)_{ji}=(\mu A)_{ij}=\mu\,A_{ij}=\mu\,(A^{\top})_{ji}=(\mu A^{\top})_{ji}$.

Claim 3. By Identity Matrix the entry $(I_{n})_{ij}$ is $1$ if $i=j$ and $0$ otherwise, a condition unchanged when $i$ and $j$ are interchanged. Hence $(I_{n}^{\top})_{ji}=(I_{n})_{ij}=(I_{n})_{ji}$.

Claim 4. Let $i\in[m]$ and $j\in[p]$. By Product of Real Matrices,

The two families of summands agree by commutativity of multiplication in $\mathbb{R}$, so the two sums agree.

Claim 5. By Difference, Dot Product, and Orthogonality in $\mathbb{R}^n$ and Matrix-Vector Product, and then by claim 3 of Properties of Finite Sums applied to each inner sum,

using commutativity and associativity of multiplication to arrange each summand. In the same way

Interchanging the order of summation in the second double sum by Interchange of a Finite Double Sum shows that the two are equal.

Claim 6. By Inverse Matrix and Invertible Real Square Matrix we have $A\,A^{-1}=I_{n}$ and $A^{-1}A=I_{n}$. Applying the transpose to each identity and using claims 4 and 3,

Hence $A^{\top}$ is invertible with inverse $(A^{-1})^{\top}$, and this inverse is the only one by Uniqueness of the Matrix Inverse.