Given the linearly independent set $\{a_1,a_2,...,a_k\}, k\in \Bbb N$ in vector space $V$ and the vector $b\in V\setminus[\{a_1,a_2,...a_k\}]$, prove that the set $\{b,a_1,a_2,...,a_k\}$ is also linearly independent.

I don't know even where to start, should I assume the opposite, that $\{b,a_1,a_2,...,a_k\}$ is linearly dependent and then get the contradiction in the end?

If $\{b,a_1,a_2,...,a_k\}$ is linearly dependent, that means

$\alpha_0b+\alpha_1a_1+...\alpha_ka_k=0 \Rightarrow$ at least one of the scalars is not $0$.

Is this the proper way to start? If so, where do I go from here?

  • $\begingroup$ With $[\dots ]$ you mean the span of $a_1,\dots,a_k$? $\endgroup$ – Alessandro Blasetti Nov 7 '16 at 12:27
  • $\begingroup$ @AlessandroBlasetti Yes, I think that's the span. $\endgroup$ – lmc Nov 7 '16 at 12:29
  • $\begingroup$ If it is the span then $b$ is linearly independent of the set. $\endgroup$ – Masacroso Nov 7 '16 at 12:29
  • $\begingroup$ @Masacroso Can you elaborate, please? $\endgroup$ – lmc Nov 7 '16 at 12:31
  • $\begingroup$ @Now_now_Draco_play_nicely the span of a set, by definition, define all the vectors that are linearly dependent of the set. Then a vector is independent of some set of vectors if and only if it dont belong to it span. $\endgroup$ – Masacroso Nov 7 '16 at 13:23

$$\alpha_0b+\alpha_1a_1+...\alpha_ka_k=0 \\ \alpha_1a_1+...\alpha_ka_k=-\alpha_0 b $$But $b$ is not in the span of the $a_i$, so we must have $\alpha_0=0$. The remaining $\alpha_i$ must then also be $0$ because the $a_i$ are linearly independent.

  • $\begingroup$ Why must $\alpha_0$ be 0? $\endgroup$ – lmc Nov 7 '16 at 12:41
  • 1
    $\begingroup$ Because of all the multiples of $b$, only $0b$ is in the span of the $a_i$. $\endgroup$ – Arthur Nov 7 '16 at 12:42

Suppose that there is no linear independency.

Then $(\lambda,\lambda_1,\dots,\lambda_k)\neq(0,0,\dots,0)$ exists with: $$\lambda b+\lambda_1 a_1+\cdots+\lambda_k a_k=0$$

The assumption $\lambda=0$ leads to the conclusion that $a_1,\dots, a_k$ are not linearly independent, so must be rejected.

But the assumption $\lambda\neq0$ leads to the conclusion that $b\in[\{a_1,\dots,a_k\}]$ so must also be rejected.

A contradiction has been found now.

  • $\begingroup$ There is no reason to make the assumption that $(\lambda,\lambda_1,\dots,\lambda_k)\neq(0,0,\dots,0)$. You can just take $\lambda b+\lambda_1 a_1+\cdots+\lambda_k a_k=0$ and note that $\lambda=0$ leads to all other coefficients being $0$, while $\lambda\neq0$ leads to a contradiction with the definition of $b$. $\endgroup$ – Arthur Nov 7 '16 at 12:48
  • $\begingroup$ @Arthur The real assumption made here is the absence of linear independency. From that we conclude (not assume) that $(\lambda,\lambda_1,\dots,\lambda_k)\neq(0,0,\dots,0)$ exists with...This leads to an absurdity. As you have shown yourself it can also be proved more directly. $\endgroup$ – drhab Nov 7 '16 at 12:54

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