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.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 30$. Linear Equations
• 1980: A.J.M. Spencer: Continuum Mechanics ... (previous) ... (next): $2.1$: Matrices