# Let $V$ be an inner product space. suppose $S={v_1,v_2, … v_n}$ is an orthogonal set of nonzero vectors [closed]

Let $$V$$ be an inner product space. suppose $$S=\{v_1,v_2, ... v_n\}$$ is an orthogonal set of nonzero vectors in $$V$$ such that $$V = \text{Span}(S)$$. Prove that $$S$$ is a basis for $$V$$.

Problem

## closed as off-topic by Adrian Keister, Chris Custer, Brahadeesh, KReiser, CesareoDec 13 '18 at 8:53

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• I edited your question to clean up the $\LaTeX$ a little bit. Cheers! – Robert Lewis Dec 2 '18 at 19:29
• It really boils down to showing that orthogonality implies linear independence; see Anurag A's answer. – Dave Dec 2 '18 at 20:09
• Thanks Dave got it right :) – Anas Dec 3 '18 at 4:24
• @RobertLewis Haha thanks man! – Anas Dec 3 '18 at 4:25
• My pleasure my friend! 😁😁😁 – Robert Lewis Dec 3 '18 at 4:31

Consider, $$c_1v_1+c_2v_2+ \dotsb +c_nv_n =\mathbf{0}.$$ Now take the inner product with vector $$v_k$$, to get $$c_1 \langle v_1, v_k \rangle +c_2\langle v_2, v_k \rangle+ \dotsb +c_k\langle v_k, v_k \rangle + \dotsb +c_n\langle v_n, v_k \rangle=0.$$ From orthogonality it follows that all terms on the left side are zero, except one term. So we get $$c_k \langle v_k, v_k \rangle =c_k\|v_k\|^2=0.$$ But $$v_k$$ is a nonzero vector. Thus $$c_k=0$$. Likewise we can have all the coefficients as $$0$$. Thus the set is linearly independent and since it already spans, therefore a basis.

I am going to assume that $$V$$ is an inner product space over the real field $$\Bbb R$$; where the inner product is a bilinear mapping

$$\langle \cdot, \cdot \rangle: V \times V \to \Bbb R \tag 1$$

such that, for all $$v \in V$$,

$$\Vert v \Vert^2 = \langle v, v \rangle \ge 0, \tag 2$$

where equality holds if and only if

$$v = 0; \tag 3$$

then if a linear dependence existed between the elements of $$S$$, we would have

$$\alpha_i \in \Bbb R, \; 1 \le i \le n, \tag 4$$

such that

$$\exists i, \; 1 \le i \le n, \alpha_i \ne 0, \tag 5$$

that is, not all the $$\alpha_i$$ vanish, such that

$$\displaystyle \sum_1^n \alpha_i v_i = 0. \tag 6$$

If we take the inner product of each side of this equation with any $$v_j$$, we find that

$$\alpha_j \langle v_j, v_j \rangle = \displaystyle \sum_1^n \alpha_i \langle v_j, v_i \rangle = \left \langle v_j, \displaystyle \sum_1^n \alpha_i v_i \right \rangle = \langle v_j, 0 \rangle = 0, \tag 7$$

since $$i \ne j$$ implies

$$\langle v_j, v_i \rangle = 0 \tag 8$$

by the orthogonality of the members of $$S$$. Since

$$v_j \ne 0, \; 1 \le j \le n, \tag 9$$

we have

$$\langle v_j, v_j \rangle \ne 0, \tag{10}$$

whence (7) yields

$$\alpha_j = 0, \; 1 \le j \le n; \tag{11}$$

it follows that the set $$S$$ is linearly independent over $$\Bbb R$$, and hence since

$$V = \text{Span}(S), \tag{12}$$

that $$S$$ is a basis for $$V$$.

Since the vectors $$v_1,\ldots,v_n$$ are orthogonal to each other, they are all linear independent. So we have $$n$$ linear independent vectors of the right dimension, thus $$S$$ is a basis for $$V$$.

Edit: Why are the vectors linear independent? First, checkout the definition here.

For any linear combination $$v = \sum_{j=1}^n a_j v_j$$ of the non-zero vectors $$v_j$$ we know that if $$v = 0$$ holds, then all $$a_j$$ must be zero to fulfill the equality (since all $$v_j \neq 0$$), thus fulfilling the definition of linear independence.