3
$\begingroup$

Is the Following Proof Correct?

Theorem. If the vectors $\alpha_1,\alpha_2,...,\alpha_n$ constitute a linearly independent list in the Vector space $V$ where as the list of vectors $\alpha_1,\alpha_2,...,\alpha_n,\beta$ is linearly dependent in $V$ then $\beta$ can be uniquely expressed as a linear combination of $\alpha_1,\alpha_2,...,\alpha_n$

Proof. Let $\beta = \alpha_{n+1}$. Now since the list $\alpha_1,\alpha_2,...,\alpha_n,\alpha_{n+1}$ is linearly dependent it follows that there exists some $j\in I = \{1,2,3,...,n,n+1\}$ such that $\alpha_j\in\operatorname{span}(\alpha_1,\alpha_2,...,\alpha_{j-1})$, evidently this $j = n+1$ since assuming that $j\in I\backslash\{n+1\}$ contradicts the fact that the list $\alpha_1,\alpha_2,...,\alpha_n$ is linearly independent in $V$.

We therefore conclude that $\alpha_{n+1} = \beta\in\operatorname{span}(\alpha_1,\alpha_2,...,\alpha_n)$, further more the unique representation of $\beta$ as a linear combination of $\alpha_1,\alpha_2,...,\alpha_n$ follows from the fact that $\alpha_1,\alpha_2,...,\alpha_n$ is linearly independent in $V$.

$\blacksquare$

$\endgroup$
0

2 Answers 2

4
$\begingroup$

I wouldn't say $j$ is evidently $n+1$. Consider the vectors in $\mathbb R^2$:

$$\alpha_{1}=\begin{pmatrix} 1\\0 \end{pmatrix}, \alpha_{2}=\begin{pmatrix} 0\\1 \end{pmatrix}, \alpha_{3}=\begin{pmatrix} 1\\1 \end{pmatrix}.$$ We have $\{\alpha_1,\alpha_2\}$ linearly independent, $\{\alpha_1,\alpha_2,\alpha_3\}$ is linearly dependent, and we have $$\alpha_1=-\alpha_2+\alpha_3.$$ But nonetheless you can show that $j=n+1$ is possible, just with some more detail.

$\endgroup$
2
$\begingroup$

Yes, it is correct. However, you can further explain the assumption $j\in I\backslash \{n+1\}$ and why you don't lose generality.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .