Consider the linear combination
\[c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots +c_k \mathbf{v}_k=\mathbf{0}.\]
Our goal is to show that $c_1=c_2=\cdots=c_k=0$.

We compute the dot product of $\mathbf{v}_i$ and the above linear combination for each $i=1, 2, \dots, k$:
\begin{align*}
0&=\mathbf{v}_i\cdot \mathbf{0}\\
&=\mathbf{v}_i \cdot (c_1\mathbf{v}_1+c_2\mathbf{v}_2+\cdots +c_k \mathbf{v}_k)\\
&=c_1\mathbf{v}_i \cdot \mathbf{v}_1+c_2\mathbf{v}_i \cdot \mathbf{v}_2+\cdots +c_k \mathbf{v}_i \cdot\mathbf{v}_k.
\end{align*}

As $S$ is an orthogonal set, we have $\mathbf{v}_i\cdot \mathbf{v}_j=0$ if $i\neq j$.

Hence all terms but the $i$-th one are zero, and thus we have
\[0=c_i\mathbf{v}_i\cdot \mathbf{v}_i=c_i \|\mathbf{v}_i\|^2.\]

Since $\mathbf{v}_i$ is a nonzero vector, its length $\|\mathbf{v}_i\|$ is nonzero.
It follows that $c_i=0$.

As this computation holds for every $i=1, 2, \dots, k$, we conclude that $c_1=c_2=\cdots=c_k=0$.
Hence the set $S$ is linearly independent.

(b) If $k=n$, then prove that $S$ is a basis for $\R^n$.

Suppose that $k=n$. Then by part (a), the set $S$ consists of $n$ linearly independent vectors in the dimension $n$ vector space $\R^n$.

Thus, $S$ is also a spanning set of $\R^n$, and hence $S$ is a basis for $\R^n$.

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