Transpose of Row Matrix is Column Matrix

Theorem
Let $$\mathbf{x} = \left[{x}\right]_{1 n} = \begin{bmatrix}x_1 & x_2 & \cdots & x_n\end{bmatrix}$$ be a row matrix.

Then $$\mathbf{x}^t$$, the transpose of $$\mathbf{x}$$, is a column matrix:


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

Proof
Self-evident.