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Claim: Let $W$ be a finite dimensional vector space. For any non-zero vector $w\in W$, there exists a basis $w_1,\cdots ,w_n$ such that $w=a_1w_1+\cdots+a_nw_n$ and $a_i\neq 0$ for each $i=1,\dots,n.$

I was considering constructing a valid basis from any basis, say $w_1',\dots ,w_n'$. If $w=a_1w_1'+\cdots+a_nw_n'$, and $a_i=0$, we can replace $w_i$ by any vector not in the span of the remaing $n-1$ vectors, giving us another basis of $W$. But now the values of the coefficients have not changed, so I am not sure if this is leading anywhere.

The context for my question is this solution. If the claim is true then the solution provided would be a very slick approach to the problem.

Problem: Suppose $V$ and $W$ are finite dimensional and $T \in L(V,W)$ . Prove that if $\dim \text{range}~ T = 1$, then there is a basis of $V$ and a basis of $W$ such that with respect to these bases, all entries of $M(T)$ equal $1$.

Solution: Let $\{ v_1 , \ldots ,v_n\}$ be an arbitrary basis of $V$ and $w \in W$ a generator of $\text{range}~ T$. Without loss of generality $T v_i = \lambda_i w$ with $\lambda_i \neq 0$ (as for example you can replace all $v_i$ with $v_i + v$ where $T v = w$).
Write $w = a_1w_1 + \cdots + a_mw_m$ as a linear combination of some basis such that $a_i \neq 0$ for each $i = 1,\ldots , m$.
Then, take $\mathcal{B}_V = \{ \frac{v_1}{\lambda_1} , \cdots , \frac{v_n}{\lambda_n}\}$ and $\mathcal{B}_W = \{a_1w_1 , \cdots, a_mw_m\}$.

Note: My claim is slightly different (it is a more general case) from the fact being used in the solution. If my claim is wrong, I would appreciate any ideas for proving the existence of such a $w$ in the above solution.

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    $\begingroup$ Of course you need $w \neq 0$. Replacing $w_i'$ doesn't change any of the coefficients at all, and $a_i$ is still zero. I'd look for an algorithm that works with $n=2$ then $n=3$, and hopefully it will be easy to generalize from there. $\endgroup$
    – aschepler
    Commented Nov 10, 2023 at 4:10
  • $\begingroup$ @aschepler That was an error from my side. I have also added $w\neq 0$. Thank you. $\endgroup$
    – Sathvik
    Commented Nov 10, 2023 at 4:23
  • $\begingroup$ Choose a basis of the dual space $W^*$ whose vectors are not contained in the hyperplane $\langle w\rangle^{\perp}$, and let $w_1,\ldots,w_n$ be the dual basis in $W$. $\endgroup$
    – blargoner
    Commented Nov 10, 2023 at 4:44

1 Answer 1

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I'll assume you know the fact that you can take any linearly independent set of vectors and extend them to a basis (it's usually proved as a more general version of proving that every vector space does in fact have a basis).

So, starting with your non-zero vector $w$, we set $w_1 = w$, Then the linearly independent set $\{w_1\}$ can be extended to a basis $B = \{w_1, w_2, ... , w_n\}$. Obviously we have

$$ w = 1w_1 + 0w_2 + 0w_3 + ... + 0w_n$$

as the expression of $w$ w.r.t. this basis.

Next, replace $w_1$ with $w_1' = w_1 - w_2 -w_3 ... - w_n$ and consider the set $B' = \{w_1', w_2, ... , w_n\}$. Now we have

$$ w = 1w_1' + 1w_2 + 1w_3 + ... + 1w_n,$$

and it's easy to show that our new set is also a basis: You can either resort to the basic definition of linear independence and play with the coefficients of a linear combination of $B'$ that results in the $0$ vector, or, more elegantly, notice that $B'$ has the same span as $B$ and use the fact that $n$ vectors that span an $n$-dimensional space must be linearly independent.

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    $\begingroup$ Very neat. Thank you! $\endgroup$
    – Sathvik
    Commented Nov 10, 2023 at 6:00

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