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Let $B = \{v_1,...,v_n\}$ be a basis for a vector space $V$ and let $u_1,..., u_k \in V$. If $\{[u_1]_B,...,[u_k]_B\}$ is linearly independent in $\mathbb{R^n}$, then $\{u_1,...,u_k\}$ is linearly independent in $V$.

For this question, I must prove or disprove this fact. So far, I have no idea where to start. I would like some help to figure out a proof of this question.

Ian Murphy
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  • What is $[u]_B$, where $u$ is a vector ? – servabat Mar 10 '15 at 18:56
  • @servabat: Is $V^k$ o.k.? – Frieder Mar 10 '15 at 18:59
  • @Frieder : I guess $V$ is wrong. There are $k$ vector so you can't be in $V$, you are in $V\times V \times .... \times V = V^k$ – servabat Mar 10 '15 at 19:01
  • @Bye_World : This is wrong. How can a set be in a set. You must use $\subseteq$ (while my version was initially right, I don't get why you edited..) – servabat Mar 10 '15 at 19:03
  • @servabat: Nothing, but a typo! I knew. – Frieder Mar 10 '15 at 19:07
  • @Bye_World : AFAIK, $(u_1, ..., u_k)$ is the common notation for a k-tuple of elements $u_1, ..., u_k$ as well as $V^k$ is the common notation for cartesian product of $V$. I don't get what was wrong :? – servabat Mar 10 '15 at 19:08
  • @Bye_World : I guess that's again a matter of notation, I always seen the notation "Let $(\alpha_1, ..., \alpha_n) \in V^n$" to denote the fact that you pick $n$ elements called respectively $\alpha_1, ... \alpha_n$ in $V$ (but sure your notation looks write and the OP post should be leaved as is, just discussing). – servabat Mar 10 '15 at 19:13
  • @servabat Interesting. I think I have seen that notation before but I certainly don't see it nearly as often as "let $u_1,..., u_k \in V$". Either way, it's perfectly understandable now so I'd just as soon leave it alone. –  Mar 10 '15 at 19:15

2 Answers2

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Hint: Equivalently, we want to prove/test the validity of the following statement: if $\mathbf{u}_{1},\dots, \mathbf{u}_{k} \in V$ are linearly dependent, then $\{[\mathbf{u}_{1}]_B,...,[\mathbf{u}_{k}]_B\}$ are also linearly dependent.

What is the basic definition of linear dependence? $\mathbf{u}_{1},\dots, \mathbf{u}_{k}$ are linearly dependent if and only if there exist coefficients $c_{1}, \dots c_{k}$, with at least one $c_{i} \neq 0$, such that $$ c_{1} \mathbf{u}_{1} + \dots c_{k} \mathbf{u}_{1} = \mathbf{0}. $$

(It may help to first think the case when the basis vectors in $B$ are orthonormal.)

megas
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Well, you should see that if $V$ and $W$ are arbitrary finite-dimensional vector spaces over the same field, and $A:V\rightarrow W$ is a linear isomorphism (invertible linear transformation) then the image of a linearly independent set of vectors will remain independent under the action of $A$.

For example, let $E=\{e_1,...,e_k\}\subset V$ be a set of linearly independent vectors in $V$,and let $Ae_1,...Ae_k$ be their images in $W$.

Assume, that the images $Ae_1,...,Ae_k$ are dependent. Then there exist a linear combination of them with at least some nonzero coefficients, $$ \alpha_1Ae_1+...\alpha_kAe_k, $$ that is the $0_W$ zero vector in $W$. Now, using linearity, $$ 0_W=\alpha_1Ae_1+...\alpha_kAe_k=A(\alpha_1e_1+...+\alpha_ke_k), $$ meaning, that $\alpha_1e_1+...+\alpha_ke_k\in \ker A$. But, since $A$ is invertible, its kernel must be trivial, so $\alpha_1e_1+...+\alpha_ke_k=0_V$. But some of the $\alpha_i$ coefficients are nonzero, which implies that $E$ is linearly dependent, which is a contradiction. Actually, this doesn't require $A$ to be invertible, just that its kernel is trivial.

You can check easily that the map that maps your vectors in $V$ to their components in $\mathbb{R}^n$ is a linear isomorphism, therefore, by what we derived now, it preserves linear independence.

Bence Racskó
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