# A Relation between the Dot Product and the Trace

## Problem 638

Let $\mathbf{v}$ and $\mathbf{w}$ be two $n \times 1$ column vectors.

Prove that $\tr ( \mathbf{v} \mathbf{w}^\trans ) = \mathbf{v}^\trans \mathbf{w}$.

## Solution.

Suppose the vectors have components
$\mathbf{v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} \, \mbox{ and } \mathbf{w} = \begin{bmatrix} w_1 \\ w_2 \\ \vdots \\ w_n \end{bmatrix}.$ Then,
\begin{align*}
\mathbf{v} \mathbf{w}^\trans &= \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} \begin{bmatrix} w_1 & w_2 & \cdots & w_n \end{bmatrix}\6pt] &= \begin{bmatrix} v_1 w_1 & v_1 w_2 & \cdots & v_1 w_n \\ v_2 w_1 & v_2 w_2 & \cdots & v_2 w_n \\ \vdots & \vdots & \vdots & \vdots \\ v_n w_1 & v_n w_2 & \cdots & v_n w_n \end{bmatrix}. \end{align*} We can now see that \[\tr( \mathbf{v} \mathbf{w}^\trans) = \sum_{i=1}^n v_i w_i = \mathbf{v}^\trans \mathbf{w}.

## Comment.

Recall that $\mathbf{v}^\trans \mathbf{w}$ is, by definition, the dot product of the vectors $\mathbf{v}$ and $\mathbf{w}$.

So, the dot product of vectors $\mathbf{v}$ and $\mathbf{w}$ is equal to the trace of the matrix $\mathbf{v} \mathbf{w}^\trans$.

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##### Prove that the Dot Product is Commutative: $\mathbf{v}\cdot \mathbf{w}= \mathbf{w} \cdot \mathbf{v}$

Let $\mathbf{v}$ and $\mathbf{w}$ be two $n \times 1$ column vectors. (a) Prove that $\mathbf{v}^\trans \mathbf{w} = \mathbf{w}^\trans \mathbf{v}$. (b)...

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