Given the basis $\{b_1, b_2,...b_n\}$ and vector $a$, prove that the set $\{a, b_2, b_3,...,b_n\}$ is the basis.

Question

Given the basis $\{b_1, b_2,...b_n\}, n\in\Bbb N$ and vector $a=\sum_{i=1}^{n} \lambda_i b_i$ in vector space V, prove that the set $\{a, b_2, b_3,...,b_n\}$ is the basis for V if and only if $\lambda_1\neq0$.

My attempt: Let's assume the opposite, i.e.

$1)$ $\{a, b_2, b_3,...,b_n\}$ is the basis and $\lambda_1=0$

or

$2)$ $\lambda_1\neq 0$ and $\{a, b_2, b_3,...,b_n\}$ is not the basis for V.

I should get a contradiction in both cases. In the first case $a=\lambda_2b_2+\lambda_3b_3+...+\lambda_nb_n$. How do I now show that $\{a, b_2, b_3,...,b_n\}$ is not the basis? Also, what should I do in the second case?

Answer 1

You have to show that the new set spans $V$ and that the vectors in it are independent.

Span: take $\vec v\in V$. Since the old basis spanned, we know we can write $$\vec v=\sum v_ib_i=v_1b_1+\sum_{i=2}^nv_ib_i$$

But, by construction, we have $$b_1=\frac 1{\lambda_1}a-\sum_{i=2}^n\frac {\lambda_i}{\lambda_1}b_i$$ So $$\vec v = \frac {v_1}{\lambda_1}a+\sum_{i=2}^n\left(v_i-\frac {\lambda_i}{\lambda_1}\right)b_i$$ Which shows that $\vec v$ can be written as a linear combination of the vectors in our set, hence that set spans $V$

Indepedence: Suppose, to the contrary, that we had non-zero coefficients $A_i$ with $$0=A_1a+\sum_{i=2}^nA_ib_i$$ If $A_1=0$ then this would be a dependence between $\{b_2,\cdots,b_n\}$ but that would contradict the independence of the first collection (contradicting the assumption that the first collection was a basis). Hence $A_1\neq 0$. But then, as before, we write $$0=A_1\lambda_1b_1+\sum_{i=2}^n(A_i+A_1\lambda_i)b_i$$ And, again, that would contradict the independence of the first collection (note that $A_1\neq 0$ implies that at least the coefficient of $b_1$ is non-zero).

Note: if you are comfortable with the notion of dimension, then you could skip the Independence argument. Knowing that the first collection is a basis tells us that $\dim V = n$ hence any collection of $n$ vectors which spans $V$ must be independent.

Answer 2

If $\lambda_1\neq 0$, then $b_1=\lambda_1^{-1}\left(a-\displaystyle\sum_{k=2}^n\lambda_k b_k\right)$ so $\{a,b_2,\dots,b_n\}$ spans the same space as $\{b_1,\dots,b_n\}$ which is the entire space $V$, hence it's a basis of this space.

If $\{a,b_2,\dots,b_n\}$ is a basis of $V$, then the linear combination $a-\displaystyle\sum_{k=2}^n\lambda_k b_k=\lambda_1 b_1$ can't be equal to $0$ since the coefficient of $a$ is not $0$, hence $\lambda_1\neq 0$ because $b_1\neq 0$.