Transpose of Row Matrix is Column Matrix

From ProofWiki
Jump to navigation Jump to search

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 $\mathbf x^\intercal$, 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.

$\blacksquare$


Sources