# Orthonormal basis for a subspace of a Hilbert space

On a Hilbert space $H$ with an orthonormal basis $(e_j)_{j \ge 1}$, define $E$ to be spanned by $e_1 + e_2, e_3 + e_4,$ and $e_2 + e_3$.

What's an orthonormal basis for $E$?

I have no idea where to begin. Usually we are given some vectors, and then I'd apply Gram-Schmidt, but what now?

• You are given three vectors that span $E$. How is this any different from usual? – Paul Sinclair Oct 28 '15 at 19:23
• How is this any different than usual? Well, for starters, the vectors aren't explicitly given which rules out any calculations with them? – Seth Oct 28 '15 at 20:16
• You're terminology is off. By definition, any element of a vector space is a "vector". What you would like to work with is a vertical array of numbers, which is generally called a "column vector" or an "element of $\Bbb R^n$". – Omnomnomnom Oct 28 '15 at 20:18
• Why would I like to work with a vertical array or a "column vector"? If $E$ was spanned by $1, x$ and $x^2$, I could deduce that these are linearly independent, and then I would apply Gram-Schmidt to $1$, $x$ and $x^2$, giving an ONB of $1, x - 1$ and $x^2 - 2x - 2/3$. This is what I am used to, and the lack of these explicit definitions is what's confusing me. – Seth Oct 28 '15 at 20:23
• Oh, I see where you're coming from. The key here is that orthonormality is all you need. – Omnomnomnom Oct 28 '15 at 20:30

Hint: We don't lose any information if we represent these vectors as column vectors. In particular, set $$e_1 = \pmatrix{1\\0\\0\\0}, \dots, e_4 = \pmatrix{0\\0\\0\\1}$$ Now, apply Gram Schmidt.
Note, however, that we don't necessarily need coordinate vectors to apply Gram-Schmidt. In particular, we begin the the vectors $$v_1 = e_1 + e_2\\ v_2 = e_3 + e_4\\ v_3 = e_2 + e_3$$ We want to find an orthonormal basis $u_1,u_2,u_3$ of the span. We begin by saying $$u_1 = \frac{1}{\|v_1\|}v_1 = \frac 1{\sqrt{\langle e_1 + e_2, e_1 + e_2 \rangle}}(e_1 + e_2)$$ Note, however, that $$\langle e_1 + e_2, e_1 + e_2 \rangle = \langle e_1, e_1 \rangle + \langle e_2, e_1 \rangle + \langle e_1, e_2 \rangle + \langle e_2, e_2 \rangle =\\ 1 + 0 + 0 + 1 = 2$$ so, we have $$v_1 = \frac 1{\sqrt 2} e_1 + \frac 1{\sqrt 2} e_2$$ perhaps you can take it from there.
• They're coordinate vectors for elements of the span of $\{e_1,e_2,e_3,e_4\}$. – Omnomnomnom Oct 28 '15 at 20:16