Let $B$ be a basis for $\mathbb{R}^n$. Prove that the vectors $v_1, v_2, \dots, v_k $ form a linearly independent set if and only if the vectors $[v_1]_B, [v_2]_B, \dots, [v_k]_B$ form a linearly independent set.

I know that the coordinate vectors of $v_1, \dots v_k $ relative to $B$ will be produce zero vectors since only the trivial solution exists if they are linearly independent which implies that a certain vector $v \in B$ cannot be written as a combination of the other vectors implying linear independence of the basis.

My issue is writing this out concretely rather than using intuition (if its correct).

A secondary question is the assertion that is the coordinate vectors span $\mathbb{R}^n$ implies that the basis spans $\mathbb{R}^n$.

I am not sure how to go about proving this assertion


Suppose the vectors $v_1,\dots,v_k$ are linearly dependent. Then there exist non-trivial (not all zero) coefficients $a_1,\dots,a_k$ so that $$ a_1v_1+\dots+a_kv_k=0. $$ Now express this in the basis $B$: $$ 0 = [a_1v_1+\dots+a_kv_k]_B = a_1[v_1]_B+\dots+a_k[v_k]_B. $$ Therefore the vectors $[v_1]_B,\dots,[v_k]_B$ are linearly dependent.

If you assume that the vectors $[v_1]_B,\dots,[v_k]_B$ are linearly dependent, you can follow the same steps backwards to show that $v_1,\dots,v_k$ are linearly dependent.

We have shown that one set of vectors is linearly dependent if and only if the other one is. Therefore one set of vectors is linearly independent if and only if the other one is.

  • $\begingroup$ So you are basically proving the contrapositive ? $\endgroup$ – Potato Apr 6 '17 at 20:28
  • $\begingroup$ @Potato Yes. It is sometimes more convenient to prove $\neg A\iff\neg B$ than $A\iff B$, although they are equivalent. $\endgroup$ – Joonas Ilmavirta Apr 6 '17 at 20:32

First, recall this facts about coordinate vectors: $$[v_1 + v_2]_B = [v_1]_B + [v_2]_B$$ $$[\alpha v]_B = \alpha[v]_B$$ $$v = 0 \iff [v]_B = 0$$

Now we can prove your claim. Let $\alpha_1, \alpha_2, ..., \alpha_k$ be scalars in the field.

We know that:

$\alpha_1v_1 + \alpha_2v_2 + ... + \alpha_kv_k = 0 \iff [\alpha_1v_1 + \alpha_2v_2 + ... + \alpha_kv_k]_B = 0 \iff [\alpha_1v_1]_B + [\alpha_2v_2]_B + ... + [\alpha_kv_k]_B = 0 \iff \alpha_1[v_1]_B + \alpha_2[v_2]_B + ... + \alpha_k[v_k]_B = 0$

Therefore we can see that if $\{v_1, v_2, ..., v_k\}$ are linearly independent, it means that all the $\alpha$'s in the phrase above are $0$ and we can conclude that $\{[v_1]_B, [v_2]_B, ...,[v_k]_B\}$ are also linearly independent: We took a linear combination that equals to 0 and it forced all of the coefficients to be 0. The other way is similar.


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