Mutually orthogonal set of vectors
By Jessica Wood •
Show that the standard basis: $$..$$ $\mathscr{B} = \left\{ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0\\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\}$ of $\mathbb{R}^{4 \times 1}$
is a mutually orthogonal set of vectors.
$\endgroup$ 32 Answers
$\begingroup$Here, the result follows from the definition of "mutually orthogonal". A set of vectors is said to be mutually orthogonal if the dot product of any pair of distinct vectors in the set is 0. This is the case for the set in your question, hence the result.
$\endgroup$ $\begingroup$Hints:
- Two vectors $\bf v$ and $\bf w$ are orthogonal -by definition, in spaces of dimension $>3$ - iff their scalar product is $0$.
- The scalar product of a vector $(x_1,\dots,x_n)^T$ with $(y_1,\dots,y_n)^T$ is defined to be $$x_1y_1+x_2y_2+\dots+x_ny_n\,.$$
