# Show that if $S$ is an orthogonal set of nonzero vectors in $\mathbb{R}^n$ then $S$ is a linearly independent subset of $\mathbb{R}^n$.

Use the dot product to show that if $S = \{\vec{v_1}, \vec{v_2}, ..., > \vec{v_n}\}$ is an orthogonal set of nonzero vectors in $\mathbb{R}^n$ then $S$ is a linearly independent subset of $\mathbb{R}^n$.

I believe that since the vectors are orthogonal to one another, this implies they are linearly independent from one another, and since they are nonzero, this further affirms that the set is not dependent, and since the vectors are a linearly independent set of $n$ vectors, this implies $\text{dim(span}{(S)}) = n$, which should show that $S$ is linearly independent, and a subset of $\mathbb{R}^n$.

However (and this is the problem I have most often in my linear algebra course) I am uncertain of how to put all of this together in a coherent and logically sound proof.

• If my answer doesn't suit you, let me now. – C. Falcon Nov 15 '15 at 20:49
• @Sheol It's pretty messy, symbolically, and it really lacks descriptiveness--It didn't address anything I put forward. My question is tagged proof-writing and proof-verification, because I want to have feed-back on what I've said, and have a well worded explanation of how I should go forward with what I've written. I appreciate your trying, but as of now I'm still waiting for an answer which addresses my concerns. – alxmke Nov 15 '15 at 21:09

Let $(\lambda_k)_{k\in\{1,\ldots,n\}}\in k^n$ such that: $$\sum_{k=1}^n\lambda_kv_k=0.$$ Let $i\in\{1,\ldots,n\}$, since $\langle\cdot,v_i\rangle$ is linear, one has :$$(1)\;\left\langle\sum_{k=1}^n\lambda_kv_k,v_i\right\rangle=0.$$ Besides, once more using linearity of $\langle\cdot,v_i\rangle$, one has: $$(2)\;\left\langle\sum_{k=1}^n\lambda_kv_k,v_i\right\rangle=\sum_{k=1}^n\lambda_k\langle v_k,v_i\rangle.$$ Since $\{v_1,\ldots,v_n\}$ is a set of orthogonal vectors, one has: $$\forall k\in\{1,\ldots,n\},\langle v_k,v_i\rangle=\delta_{i,k}.$$ Therefore, with $(2)$ it follows: $$\left\langle\sum_{k=1}^n\lambda_kv_k,v_i\right\rangle=\lambda_ie_i.$$ Now, using $(1)$ one has $\lambda_ie_i=0$ and since $e_i\neq 0$, $\lambda_i=0$. Hence, one has: $$\forall i\in\{1,\ldots,n\},\lambda_i=0.$$ Finally, $\{v_1,\ldots,v_n\}$ is linearly independent.