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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?

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  • $\begingroup$ 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. $\endgroup$ – skyking Sep 20 at 11:15
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You don't need a contradiction.

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.

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    $\begingroup$ +1 You assume $(v_1,...,v_n)$ even orthonormal, $v_i^tv_i=1$? $\endgroup$ – Peter Melech Sep 20 at 11:18
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    $\begingroup$ The proof also works if the vectors are only orthogonal by the property of norm and that a field has no zero divisors. $\endgroup$ – Wuestenfux Sep 20 at 13:28
  • $\begingroup$ That for sure! Just to clarify. $\endgroup$ – Peter Melech Sep 20 at 18:17

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