Space spanned by orthogonal vectors.

Question

If there are $M$ non-zero orthogonal vectors $X_1$,$X_2$,.....$X_M$. The dimension of the vector space spanned by these $M$ vectors is?

Answer given is $M$ but I am not able to understand how is it $M$.

Answer 1

Suppose that those vectors were linearly dependent. Then you can write at least one of them as a linear combination of the others. Without loss of generality suppose that $$x_m = a_1x_1 + \dots + a_{m-1}x_{m-1}$$

and that not all the $a_i$'s are zero.

Since the vectors are orthogonal, $\langle x_i , x_m \rangle = 0$ for all $i = 1, \dots, m-1$. Then using the linearity of the inner product,

\begin{align*} 0 &= \langle x_i, x_m \rangle\\ &= a_1 \langle x_i, x_1 \rangle + \dots a_{m-1} \langle x_i, x_{m-1} \rangle \\ &= a_i \langle x_i , x_i \rangle. \end{align*}

for all $i = 1, \dots, m-1$.But $x_i \neq 0$, so $\langle x_i , x_i \rangle \neq 0$. Therefore $a_i = 0$ for all $i = 1, \dots , m-1$. This is a contradiction, hence the vectors must be linearly independent. Since we have $m$ linearly independent vectors, they span a space of dimension $m$.

Answer 2

Orthogonal vectors are necessarily independent vectors. The span of n independent vectors has those n vectors as basis. Since the space has n vectors as a basis it has dimension n.