# Two vectors can be expressed in a basis where both have all nonzero entries

Suppose I have two vectors $$\vec{v}$$ and $$\vec{w}$$ in $$\mathbb{R}^k$$, each with at least one non-zero entry. Can I find a basis $$\{e_1,...,e_k\}$$ such that when expressed in $$\{e_1,...,e_k\}$$, neither $$\vec{v}$$ or $$\vec{w}$$ have zero entries?

I think the answer is yes and it has to do with Gram-schmidt but I can't quite see how to prove it. There is an obvious trick I miss probably...

• So do you want a basis for which each entry of w and v in that basis is different from zero? Commented Aug 1 at 15:21
• Yes, that is what I want. It has to be the same basis for both vectors. Commented Aug 1 at 15:24

Yes. Rephrasing what you want, you are starting with two nonzero vectors. If you write them as $$k$$-tuples with respect to the natural basis each will have at least one nonzero entry.

If you replace the natural basis by adding the vector $$(c,c,\ldots,c)$$ to each basis element you will still have a basis. Every coordinate of every vector will change when expressed in this new basis. If you choose $$c$$ small enough you will change the zero coordinates of the vectors you started with but won't change the nonzero coordinates by enough to make them $$0$$.

• I see this answer is correct, but wonder if there is a way to make the addition step more rigourous? How can i show that all coordinates of every vector change to be nonzero here? Commented Aug 1 at 15:56
• @MasonTep >How can i show that all coordinates of every vector change to be nonzero? == You show that changing the basis vectors means nothing more than adding or subtracting constants to all the coordinates of the vectors. Then use the fact that if (coordinate) A > B, then A + C > B + C, leaving one of the vectors still greater than the other, and thus, non-zero. Commented Aug 2 at 2:52

The entries of the vector $$v$$ when expressed in basis $$b_1, \ldots, b_k$$ are $$B^{-1} v$$ where $$B$$ is the matrix formed from columns $$b_1, \ldots, b_k$$. Let $$A$$ be a random $$k \times k$$ matrix (a sample from some absolutely continuous probability distribution on $$\mathbb R^{k \times k}$$), and let $$b_1, \ldots, b_k$$ be the columns of $$B = A^{-1}$$ (which exists with probability $$1$$). With probability $$1$$, $$B^{-1} v = A v$$ and $$B^{-1} w = A w$$ both have all nonzero entries.

==
You just remind the audience that changing the basis vectors only means adding or subtracting the same constants to all the coordinates of the vectors.

Then point out that if (coordinate) A ~= B, then A+C ~= B+C, leaving the vectors with different lengths.

Since they both can't be zero, one must be non-zero.

QED [curtsies]