Transpose of Row Matrix is Column Matrix

Theorem
Let $\mathbf x = \sqbrk x_{1 n} = \begin {bmatrix} x_1 & x_2 & \cdots & x_n \end {bmatrix}$ be a row matrix.

Then $\mathbf x^\intercal$, the transpose of $\mathbf x$, is a column matrix:


 * $\begin {bmatrix} x_1 & x_2 & \cdots & x_n \end{bmatrix}^\intercal = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$

Proof
Self-evident.