# Prove that any non-zero orthogonal vectors in $\mathbb R^{n}$ form a basis for $\mathbb R^{n}$

Proposition. Any $$n$$ non-zero orthogonal vectors in $$\mathbb R^{n}$$ form a basis for $$\mathbb R^{n}$$.

My attempt:

Suppose we have non-zero orthogonal vectors $$\bf (v_{1},v_{2},\cdots,v_{n})$$

We know that

$$n$$ linearly independent vectors in $$\mathbb R^{n}$$ span $$\mathbb R^{n}$$.

or in other words

$$n$$ linearly independent vectors in $$\mathbb R^{n}$$ form a basis.

Therefore, if we show that $$\bf (v_{1},v_{2},\cdots,v_{n})$$ are linearly independent, we automatically show that they form a basis.

We will prove linear independence by contradiction.

Suppose $$\bf (v_{1},v_{2},\cdots,v_{n})$$ are linearly dependent.

We have

$$k_1\mathbf{v_{1}} + k_2\mathbf{v_{2}}+ \cdots + k_j\mathbf{v_{j}} + \cdots + k_n\mathbf{v_{n}} = \mathbf O$$

Where at least one scalar, call it $$k_{j}$$, is not zero.

Premultiply both sides by $$\bf v_{j}^{T}$$

$$\mathbf{v_{j}}^{T}\bigl(k_1\mathbf{v_{1}} + k_2\mathbf{v_{2}}+ \cdots + k_j\mathbf{v_{j}} + \cdots + k_n\mathbf{v_{n}}\bigr) = \mathbf {v_{j}}^{T}\mathbf O \implies$$

$$k_1\mathbf{v_{j}}^{T}\mathbf{v_{1}} + k_2\mathbf{v_{j}}^{T}\mathbf{v_{2}}+ \cdots + k_j\mathbf{v_{j}}^{T}\mathbf{v_{j}} + \cdots + k_n\mathbf{v_{j}}^{T}\mathbf{v_{n}} = 0\implies$$

Because our vectors are orthogonal, then $$a≠b \implies \mathbf {v_{a}}^{T}\mathbf {v_{b}} = 0$$

Thus, we have

$$0 + 0 + \cdots k_{j}||\mathbf{v_{j}}||^{2} \cdots 0 = k_{j}||\mathbf{v_{j}}||^{2} = 0$$

We know that $$||\mathbf{v_{j}}||^{2} > 0$$ (it doesn't equal to zero because task specifies our vectors must be non-zero)

We also know that $$k ≠ 0$$

In this case, $$k_{j}||\mathbf{v_{j}}||^{2}$$ cannot equal $$0$$, hence the contradiction. Therefore, $$k_{j} = 0$$.

Since we've considered arbitrary scalar, we can conclude that all scalars must be zero. And therefore linear system in question is independent. And, again, because

$$n$$ linearly independent vectors in $$\mathbb R^{n}$$ form a basis.

We conclude that vectors $$\bf (v_{1},v_{2},\cdots,v_{n})$$ form a basis in $$\mathbb R^{n}$$. $$\Box$$

Is it correct?

• Yes, it looks correct. Maybe the ending of it can be made a little more to the point. As soon you have your contradiction you can conclude that the premise was false and therefore the vector can't be linearly dependent. – skyking Sep 20 at 11:15

Suppose the vectors $$v_1,\ldots,v_n$$ are orthonormal.
Consider the linear combination $$k_1v_1+\ldots+k_nv_n=0.$$ Then for each vector $$v_i$$, $$0 = v_i^t0 = k_1v_i^tv_1 + \ldots + k_iv_i^tv_i +\ldots+k_nv_i^tv_n = k_iv_i^tv_i = k_i.$$ Done.
• +1 You assume $(v_1,...,v_n)$ even orthonormal, $v_i^tv_i=1$? – Peter Melech Sep 20 at 11:18