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I'm confronted with this question:

Let $V$ be an inner product space and $B=\{u_{1}, ..., u_{n}\}$ a basis of $V$.

Suppose there exists $\lambda_{1},...,\lambda_{n} \in F (=R \text{ or } C)$ such that: $$||\sum_{k=1}^{n}\lambda_{k}u_{k}||^2=\sum_{k=1}^{n}|\lambda_{k}|^2$$

Prove or disprove: $B$ is an orthonormal basis.

Not sure where to begin.

I tried finding a counter example in the case of $V=R^2$ using the identity: $$||u+v||^2=||u||^2+2Re\langle u, v\rangle+||v||^2$$ which gave me this result: $$||\lambda_{1}u_{1}+\lambda_{2}u_{2}||^2=|\lambda_{1}|||u_{1}||^2+2Re\langle\lambda_{1}u_{1},\lambda_{2}u_{2}\rangle+|\lambda_{2}|||u_{2}||^2=|\lambda_{1}|^2+|\lambda_{2}|^2$$

But I can't see how this helps me.


Here's a little follow up question if anyone is interested: suppose the equality above is true for all $\lambda_{1},...,\lambda_{n}$, does that imply the basis is orthonormal?

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It helps you a lot. You want the two terms to be equal, right? One way to do this is to set $||u_1|| = ||u_2|| = 1$ and then figure out how you can get the middle term to vanish without forcing $u_1, u_2$ to be orthogonal. – Qiaochu Yuan Mar 13 '11 at 21:04
Inn the way the question is stated, this is not true (take $\lambda_1\neq 0$ and $\lambda_i=0$ for $i>1$). Are there any other conditions on the $\lambda s$ (maybe they cannot be zero)? – Ronaldo Mar 13 '11 at 21:08
Using < and > for the inner product produces bad spacing. Use \langle and \rangle instead. – Arturo Magidin Mar 13 '11 at 21:49
If the equality holds for all choices, then the conclusion follows. First, selecting $\lambda_j=\delta_{ij}$ shows that $||u_i||=1$, so the set is normal. Then take $\lambda_1=\lambda_2=1$ to show the real part of $\langle u_1,u_2\rangle$ is $0$; and $\lambda_1=i$, $\lambda_2=1$ to show the imaginary part is $0$, so $u_1\perp u_2$. Similar arguments show it for other pairs. – Arturo Magidin Mar 14 '11 at 13:23
up vote 1 down vote accepted

It is false. A simple example would be to take $\lambda_i = 0$, $\forall i$ or for $i$'s which are not orthogonal to the rest.

If you assume $\lambda_i \neq 0$, $\forall i$, then here is a counter example in $2D$.

Let $u_1 = 2e_1 + e_2$ and $u_2 = e_1 + 2e_2$, where $e_1, e_2$ form the conventional orthonormal basis. Clearly this forms a basis. Let $\lambda_i \in \mathbb{R}$

Then $||\lambda_1 u_1 + \lambda_2 u_2 ||_2^2 = |\lambda_1|^2 + |\lambda_2|^2$ implies $(2 \lambda_1 + \lambda_2)^2 + (\lambda_1 + 2 \lambda_2)^2 = \lambda_1^2 + \lambda_2^2$.

Hence, $(4 \lambda_1^2 + 4 \lambda_1 \lambda_2 + \lambda_2^2) + (4 \lambda_2^2 + 4 \lambda_1 \lambda_2 + \lambda_1^2) = \lambda_1^2 + \lambda_2^2$.

Hence, $(4 \lambda_1^2 + 4 \lambda_1 \lambda_2) + (4 \lambda_2^2 + 4 \lambda_1 \lambda_2) = 0 \Rightarrow (\lambda_1 + \lambda_2)^2 = 0$

Hence, if we choose $\lambda_1 = - \lambda_2$ we are done.

Hence, if $u_1 = 2e_1 + e_2$ and $u_2 = e_1 + 2e_2$ and if $\lambda_1 = - \lambda_2 \in \mathbb{R}$, then we have $$||\lambda_1 u_1 + \lambda_2 u_2||_2^2 = \lambda_1^2 + \lambda_2^2$$

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First of all, $\lambda_1$ and $\lambda_2$ should be squared in the middle expression of the last equality in your question. Next, for the question to be nontrivial, an asumption like $\lambda_k\ne0$ for all $k$ must be added. Otherwise, a counterexample is given by taking all $\lambda_k=0$.

The answer to the question is that the system need not be an orthonormal basis. In $\mathbb{R}^2$ take $u_1=(1,0)$, $\lambda_1=1$. Then, for any $u_2\in\mathbb{R}^2$ with $\langle u_1,u_2\rangle\ne0$, let $\lambda_2=2\langle u_1,u_2\rangle/(1-\|u_2\|^2)$. Then $\|\lambda_1u_1+\lambda_2u_2\|^2=|\lambda_1|^2+|\lambda_2|^2$ but $u_1$ and $u_2$ are not orthogonal. They may be even linearly dependent. The example may be generalized to $\mathbb{R}^n$.

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