Let $V$ be an inner product space, and let $S$ be an orthogonal subset of $V$ consisting of nonzero vectors. Then $S$ is linearly independent.

Also, orthogonal set and linearly independent set both generate the same subspace. (Is that right?)

orthogonal $\rightarrow$ linearly independent
orthogonal $\nleftarrow$ linearly independent
is that right?

One more question.
For T/F,
Every orthogonal set is linearly independent (F)
Every orthonormal set is linearly independent (T)

  • 2
    $\begingroup$ Just as a side note, there are several questions here. These might be better off posted as several questions, though they are related. That is just my opinion though. $\endgroup$ Jun 3, 2013 at 2:36

1 Answer 1


For the theorem:

Hint: let $v_{1}, v_{2}, \ldots, v_{k}$ be the vectors in $S$, and suppose there are $c_{1}, \ldots, c_{k}$ such that $v_{1}c_{1} + \cdots + v_{k}c_{k} = 0$. Then take the inner product of both sides with any vector in the set $v_{j}, 1 \leq j \leq k$. Conclude something about the coefficient $c_{j}$ using the fact that $v_{j} \neq 0$ for all vectors $v_{j}$ in the set.

For your next question, orthogonal set implies linearly independent set with the condition that all the vectors in the set are nonzero - we need this in the above proof! (I'll address that in your true false questions).

You're right that linearly independent need not imply orthogonal. To see this, see if you can come up with two vectors which are linearly independent over $\mathbb{R}^{2}$ but have nonzero dot product. (It shouldn't be too hard to do so!)

For your true false question, every orthogonal set need not be linearly independent, as orthogonal sets can certainly include the '$0$' vector, and any set which contains the '$0$' vector is necessarily linearly dependent.

However, every orthonormal set is linearly independent by the above theorem, as every orthonormal set is an orthogonal set consisting of nonzero vectors.

  • $\begingroup$ But we can always use Gram-Schmidt to obtain an orthagonal(even orthonormal basis) from a linearly independent set,right? $\endgroup$
    – user123124
    May 7, 2019 at 8:24
  • $\begingroup$ @AlexWertheim I quite didnt get it. linear independence for two vectors means that there doesnt exist any scalar $c_i$ that you can multiply with one vector to yield another. So is linear independence merely checking for co-linearity? Which is different from checking for orthogonality? $\endgroup$ Jun 5, 2020 at 15:05

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