# Thread: Inner product space with orthonormal subset

1. ## Inner product space with orthonormal subset

Let V be an inner product space, let $S = \{ v_{1},v_{2},...,v_{n} \}$ be an orthonormal subset of V. Prove that for any $x \in V$, we have $||x||^2 \geq \sum ^{n} _{i=1} ||^2$

proof so far.

by a theorem, there exist unique vectors u in span(S) and v in the orthogonal complement of span(S) such that x = u + v.

Now, since u is orth. to v, meaning $||x||^2 = ||u||^2 + ||v||^2$

Since $u \in \ span{S}$, write $u = \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}$

then $x = u + v = v + \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}$

Since they are all orthogonal to one another, we can write $||x||^2 = ||v||^2 + || \alpha _{1} v_{1} ||^2 + || \alpha _{2} v_{2} ||^2 + . . . + || \alpha _{n} v_{n} ||^2$

Now, I'm a bit stuck... Am I at least doing this right so far?

Let V be an inner product space, let $S = \{ v_{1},v_{2},...,v_{n} \}$ be an orthonormal subset of V. Prove that for any $x \in V$, we have $||x||^2 \geq \sum ^{n} _{i=1} ||^2$

proof so far.

by a theorem, there exist unique vectors u in span(S) and v in the orthogonal complement of span(S) such that x = u + v.

Now, since u is orth. to v, meaning $||x||^2 = ||u||^2 + ||v||^2$

Since $u \in \ span{S}$, write $u = \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}$

then $x = u + v = v + \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}$

Since they are all orthogonal to one another, we can write $||x||^2 = ||v||^2 + || \alpha _{1} v_{1} ||^2 + || \alpha _{2} v_{2} ||^2 + . . . + || \alpha _{n} v_{n} ||^2$

Now, I'm a bit stuck... Am I at least doing this right so far?
Yes, that's the right approach. Now you need to use the fact that since $\{ v_{1},v_{2},...,v_{n} \}$ is an orthonormal basis for S, and u∈S, it follows that $\alpha_j = \langle u,v_j\rangle = \langle x,v_j\rangle$ for each j. Together with the fact that each v_j is a unit vector, that will give you the result.

Why $
\alpha_j = \langle u,v_j\rangle = \langle x,v_j\rangle
$
?

Why $
We know that S is orthonormal, so $\langle v_i,v_j\rangle = 0$ if i≠j, and $\langle v_j,v_j\rangle = 1$.
We also know that $u = \alpha _{1} v_{1} + \alpha _{2} v_{2} + \ldots + \alpha _{n} v_{n}$. Take the inner product with v_j, and we get $\langle u,v_j\rangle = \langle \alpha _{1} v_{1} + \alpha _{2} v_{2} + \ldots + \alpha _{n} v_{n},v_j\rangle = \alpha _{1}\langle v_1,v_j\rangle + \alpha _{2}\langle v_2,v_j\rangle + \ldots + \alpha _{n}\langle v_n,v_j\rangle = \alpha_j$ because of the orthonormality.
Finally, we know that x = u + v, where $v\in S^{\perp}$, and so $\langle x,v_j\rangle = \langle u,v_j\rangle + \langle v,v_j\rangle$. But $\langle v,v_j\rangle = 0$ because v_j∈S.