# Linear Independence and Inner Product Spaces

Let $S=\{v_1,v_2,\dots,v_n\}$ be a linearly independent subset of an inner product space $V$, and $w \in V$ where $w$ is orthogonal to each vector in $S$. Prove, using only the definition of linear independence, orthogonal vectors and the inner product space axioms that $S\cup\{w\}$ is also linearly independent.

I'm just not quite sure how to combine the definition of linear independence, orthogonal vectors and the inner product space axioms to show that $S\cup\{w\}$ is also Linearly Independent.

• What is $\langle u, b w + \sum_{j=1}^n a_j v_j \rangle$, where $u \in \{w\} \cup \{v_i\}_{i=1}^n$? What does this tell you about $b$ and the $a_j$? – Eric Towers May 6 '16 at 16:30

Suppose that $c_1v_1+\dotsb+c_nv_n+c_{n+1}w=0$ for some scalars $c_i$. Take the inner product of the left hand side and right hand side with $w$ and use the fact $w$ is orthogonal to each vector in $S$ to deduce that $c_{n+1}=0$. Finally use the fact that the $\{v_i\}$ form a linearly independent set to deduce that $c_i=0$ for $1\leq i\leq n$.

• I assume $w$ has to be non-zero? – snulty May 6 '16 at 16:36
• Yes. Otherwise there is no way that $S\cup\{w\}$ is a linearly independent set. – Foobaz John May 6 '16 at 16:37
• But $0$ is orthogonal to every vector and $S\cup\{0\}$ is linearly dependent. I feel this should've been part of the question? – snulty May 6 '16 at 16:40
• Yes it should be. – Foobaz John May 6 '16 at 16:40

Suppose

$$w=\sum_{k=1}^n a_kv_k\implies\,\forall\,j\;,\;\;1\le j\le n\;:$$

$$0\stackrel{\text{given}}=\left\langle\,w,v_j\,\right\rangle=\sum_{k=1}^n a_k\langle v_k,v_j\rangle=a_k\langle v_j,v_j\rangle\implies a_k=0$$

since $\;v_1,...,v_n\;$ l.i.

Fill in details, in particular: why is the above enough?

• why is it enough ? – Peter B. May 6 '16 at 16:46
• Because a set of vectors is linearly dependent if and only if there is one of them linearly dependent in the ones preceeding it. If you look at $\;\{v_1,...,v_n,w\}\;$ , just as you wrote, this means it must be $\;w\;$ a linear combination of the preceeding ones. And as noted in other comments, it must be $\;w\neq0\;$ , otherwise the claim is false. – DonAntonio May 6 '16 at 16:48