POTW A Modified Basis in an Inner Product Space

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Given an orthonormal basis $\{e_1,\ldots, e_n\}$ in a complex inner product space $V$ of dimension $n$, show that if $v_1,\ldots, v_n\in V$ such that $\sum_{j = 1}^n \|v_j\|^2 < 1$, then $\{v_1 + e_1,\ldots, v_n + e_n\}$ is a basis for $V$.

 

[julian]

Made a mistake

 

[julian]

Spoiler

We prove linear independence. First, we have $\| v_i + e_i \| \geq \| e_i \| - \| v_i \| > 0$, so none of the $\{ v_1 + e_1 , \dots , v_n + e_n \}$ is the zero vector. We wish to prove that if

$$
\sum_{i=1}^n \alpha_i (v_i + e_i) = 0
$$

then $\alpha_i = 0$ for $i = 1 , \dots , n$. We assume that all the $\alpha$'s being zero is not the only solution and arrive at a contradiction. The above condition implies

$$
\| \sum_{i=1}^n \alpha_i e_i \|^2 = \| \sum_{i=1}^n \alpha_i v_i \|^2 .
$$

We have

\begin{align*}
\| \sum_{i=1}^n \alpha_i v_i \|^2 & \leq (\sum_{i=1}^n | \alpha_i | \cdot \| v_i \| )^2
\nonumber \\
& \leq ( \sum_{i=1}^n |\alpha_i|^2) ( \sum_{i=1}^n \| v_i \|^2 )
\nonumber \\
& < \sum_{i=1}^n | \alpha_i |^2 .
\end{align*}

Note strict inequality. However,

$$
\| \sum_{i=1}^n \alpha_i e_i \|^2 = \sum_{i=1}^n | \alpha_i |^2
$$

and so we have a contradiction. The only way out of the contradiction is for all the $\alpha$'s to be zero.


A standard result is that any set of $n$ linearly independent vectors in an $n-$dimensional vector space, $V$, forms a basis for that space. Proof: Consider

$$
c_0 v+\sum_{i=1}^n c_i (v_i + e_i)
$$

with arbitrary $c_i$ $(i= 0, 1, \dots , n)$ and $v \in V$. The equation

$$
c_0 v+\sum_{i=1}^n c_i (v_i + e_i) = 0 \qquad (*)
$$

cannot imply

$$
c_i = 0 \qquad i= 0, 1, \dots , n
$$

since that would mean that there are $n+1$ independent vectors in $V$,

$$
\{ v , v_1 + e_1 , \dots , v_n + e_n \}
$$

and the dimension of $V$ would not be $n$, but would be at least $n+1$. Therefore, a set $\{ c_i \}$ exists, with at least two non-zero members, such that $(*)$ is satisfied. One cannot have $c_0 = 0$, since that would lead to the conclusion that all the $c_i$'s are zero. Therefore, we can write

$$
v = - \sum_{i=1}^n \frac{c_i}{c_0} (v_i +e_i) .
$$

Therefore, an arbitrary vector $v$ has been expressed as a linear combination of the vectors $\{ v_1 +e_1 , \dots , v_n + e_n \}$. This proves, in addition to being linearly independent, that they span the space and hence are a basis.