If I have a vector space $V$ ( of dimension $n$ ) over real numbers such that $\{v_1,v_2…v_n\}$ is the basis for the space ( not orthogonal ). Then I can write any vector $l$ in this space as $l=\sum_i\alpha_iv_i$. Here $\alpha_1,\alpha_2…\alpha_n$ are the coefficients that define the vector $l$ according to this basis. Can another set of coefficients $\beta_1,\beta_2…\beta_n$ give the same vector $l$ ? If the basis was orthogonal the answer would be no, but I can't prove for a non orthogonal basis.
[Math] Are the coefficients of a vector according to a basis unique
linear algebravector-spaces
Related Solutions
When you choose a basis $v_{1}, \ldots, v_{n}$ for $\ker \varphi$, you're assuming that it is is finite-dimensional. Since there is no assumption on $V$ being finite-dimensional or any other information about the dimension of $\ker \varphi$, we cannot assure that $\ker \varphi$ is finite-dimensional.
This problem has two parts:
We need to show that for every $v \in V$, there exist $w \in \ker \varphi$ and $\alpha \in F$ such that $v = w + \alpha u$. So you start by considering an arbitrary vector $v \in V$, and then try to find $w$ and $\alpha$ such that $v = w + \alpha u$. When you're trying to prove an existence result, it is sometimes useful to think in reverse: suppose you already have $v = w + \alpha u$ with the desired properties and now try to find out what $w$ and $\alpha$ should be. Let's see how that looks.
Suppose we have have $w \in \ker \varphi$ and $\alpha \in F$ such that $v = w + \alpha u$. Can we get some information about $w$ and $\alpha$ from this?
Well, applying $\varphi$ to both sides, we see that $$ \varphi(v) = \varphi(w) + \alpha\varphi(u) $$
and $w$ is in $\ker \varphi$, so
$$ \varphi(v) = \alpha \varphi(u). $$
We know that $\varphi (u) \neq 0$, so it's time to use that fact:
$$ \alpha = \frac{\varphi(v) }{ \varphi(u) } .$$
So it seems like we've found our candidate for $\alpha$. Up to this point, it must the case that
$$ v = w + \frac{\varphi(v) }{ \varphi(u) } u .$$
Can you now see what $w$ should be?
Of course, what I described above shouldn't be in the final proof. It is only the scratch work. In the final proof you would explicitly say what your candidates for $w$ and $\alpha$ are, and show that they satisfy the required properties.
For the second part, we need to show that the sum is direct. In this case, I think it would be easier to use the fact that the sum of two subspaces $U$ and $W$ is direct if and only if $U \cap W = \{ 0 \} $.
Best Answer
They cannot. Suppose so. Then $\sum \alpha_i v_i = \sum \beta_i v_i$. This implies: $$0 = \sum (\alpha_i - \beta_i) v_i$$ which violates the basis being a set of linearly independent vectors.