Choosing a basis such that all coefficients in linear combination are non-zero.

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.

• 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. Commented Nov 10, 2023 at 4:10
• @aschepler That was an error from my side. I have also added $w\neq 0$. Thank you. Commented Nov 10, 2023 at 4:23
• 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 $W$. Commented Nov 10, 2023 at 4:44

1 Answer

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.

• Very neat. Thank you! Commented Nov 10, 2023 at 6:00