Please correct my thinking, if anything not make sense to you.

A vector in $R^n$ is nothing but an assemblage of its co-ordinate w.r.t. some basis in the form of $n \times 1$ matrix.

Statement: If set of vectors(w.r.t. some basis) is linearly independent, then it is also a linearly independent set w.r.t. any other basis.

Proof : Since coordinate of a vector w.r.t. some basis B1 can be transformed to coordinate of same vector w.r.t. some other basis B2 by a transformation T that takes basis B2 to basis B1, and which has the full rank.

Thus M1(coordinates of the set of vectors w.r.t. B1 arranged in matrix) $=$ T(matrix obtained by transformation) . M2 (coordinates of the set of vectors w.r.t. B2 arranged in matrix), from this we conclude rank M1 = rank M2.

Is my proof make sense ? Any suggestion will be appreciated.

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    $\begingroup$ I take issue with your first sentence. Vectors can be polynomials, functions on some domain that isn’t a set of numbers, and so on. $\endgroup$ – amd Jun 26 '19 at 3:57
  • $\begingroup$ There is isomorphism between n- dimensional vector space over a field F and coordinate space $F^n$ $\endgroup$ – user602672 Jun 26 '19 at 4:19
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    $\begingroup$ Indeed there is. That’s not at all the same as saying that a vector is an assemblage of coordinates. $\endgroup$ – amd Jun 26 '19 at 5:02
  • $\begingroup$ Ok, I have restricted to Euclidean space then. Same is edited. $\endgroup$ – user602672 Jun 26 '19 at 6:51
  • $\begingroup$ You don’t need to restrict the space: just include these isomorphisms explicitly. $\endgroup$ – amd Jun 26 '19 at 18:57

You needn't constrict your question to $\mathbb{R}^n$, this works in every finite-dimensional vector space $V$. Like said in the comments, every $n$-dimensional vector space $V$ is isomorphic to $\mathbb{R}^n$ by the coordinate mapping.

It might also help to make things more clear for both yourself and others to formulate your claim as "linear independence of a set of vectors is basis-independent".

Your method seems sound to me if you can also prove that $\text{rank}(TM_2)=\text{rank}(M_2)$.

Another nice way to prove that linear (in)dependence is basis-independent is by using the determinant. Remember that the determinant of an $n\times n$-matrix is non-zero iff the rows (and collumns) are linear-independent in $\mathbb{R}^n$. Also, remember the product rule for determinants and ask yourself what the determinant of $T$ will be (or rather - won't be).

Another way is the following:

Let $\{v_1,\dots,v_n \}$ be a linearly independent set of vectors in vector space $V$. Let $T:V\to V$ be a basis transformation. Then the question is if this implies $$\sum_{i=1}^n\lambda_i T(v_i)=0 \iff \forall i\leq n: \lambda_i=0.$$

Since $T$ is linear we have

$$\sum_{i=1}^n\lambda_i T(v_i)=T\Big(\sum_{i=1}^n\lambda_i v_i\Big)=0,$$ so $\sum_{i=1}^n\lambda_i v_i\in \text{ker}(T)$. Now, you should figure out what $\text{ker}(T)$ is and how this implies that $\forall i \leq n:\lambda_i=0$. This then proves that $\{T(v_1),\dots,T(v_n)\}$ is also linearly independent.

You might also want to change the title of your question to something less vague. I would have said this in a comment but I don't have enough reputation.

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  • $\begingroup$ Thanks for the clarification and answer. I am making the title as you suggested. So by way your method- can I conclude that changing basis means the unique transformation? I am taking your hint to prove $\rank(TM_2)=\rank(M2)$; since $\det(TM_2)=\det(T).\det(M_2)$ and $\det(T)\neq 0$ $\endgroup$ – user602672 Jun 26 '19 at 13:16
  • $\begingroup$ @user602672, a change of basis is always unique and completely determined by how the basis vectors from one basis are mapped to the other. See Change of basis. With regards to the hint to prove the equality of the ranks, I think I am misunderstood. I didn't mean that you could prove the equality of the ranks by using the determinants, I meant that your original question can be answered by using determinants - without referring to ranks. You should think of such a proof. I will edit my answer. $\endgroup$ – PaleBlueDot Jun 26 '19 at 13:28
  • $\begingroup$ Do let me know if you eventually have found a proof or the proofs I suggested. All three ways (by rank, by determinants, by characterisation of linear independence) are good exercises if you are new to linear algebra. If you haven't managed to find them, or if you want to see me write it out, just say so and I will add them in my answer tomorrow. $\endgroup$ – PaleBlueDot Jun 26 '19 at 13:35
  • $\begingroup$ Ok I will work on your suggestions shortly..your suggestions really helpful to me. Thanks again $\endgroup$ – user602672 Jun 26 '19 at 13:57
  • $\begingroup$ For rank I found the proof here math.stackexchange.com/a/1200571/602672 . By using determinants, I guess (if I understood correctly) we require full set of n vectors (which is not required in other two methods) and only make sense if we talk about determinant of $[v_1, \cdots, v_n]$ and $[Tv_1, \cdots,T v_n] but then I don't how to prove determinant of the latter one is non-zero. The third method 'by the characterization of linear independence', I can follow your answer..there is no doubt. $\endgroup$ – user602672 Jun 26 '19 at 17:40

I believe you're looking from a wrong point of view.

A vector is independent from bases. It's an abstract object belonging to some vector space. As soon as you fix a basis, you can consider the coordinates of this vector with respect to the basis. If the vector space has dimension $n$, then the coordinate vector belongs to $\mathbb{R}^n$.

Your question can be reformulated as follows.

Suppose we have $\{v_1,v_2,\dots,v_k\}$, a set of vectors in the $n$-dimensional vector space $V$. If $\mathscr{B}$ is a basis of $V$, then we can consider the set of coordinate vectors $\{C_{\mathscr{B}}(v_1),C_{\mathscr{B}}(v_2),\dots,C_{\mathscr{B}}(v_k)\}$, which is a set of vectors in $\mathbb{R}^n$.

Then $\{v_1,v_2,\dots,v_k\}$ is linearly independent (in $V$) if and only if $\{C_{\mathscr{B}}(v_1),C_{\mathscr{B}}(v_2),\dots,C_{\mathscr{B}}(v_k)\}$ is linearly independent in $\mathbb{R}^n$.

The proof is simple: the map $C_{\mathscr{B}}\colon V\to\mathbb{R}^n$ that associates to each vector $v\in V$ its coordinate vector is linear and bijective.

If $\{v_1,v_2,\dots,v_k\}$ is linearly independent and $$ \alpha_1C_{\mathscr{B}}(v_1)+\alpha_2C_{\mathscr{B}}(v_2)+\dots+\alpha_kC_{\mathscr{B}}(v_k)=0 $$ then, by linearity, $$ C_{\mathscr{B}}(\alpha_1v_1+\alpha_2v_2+\dots+\alpha_kv_k)=0 $$ and, by injectivity, $\alpha_1v_1+\alpha_2v_2+\dots+\alpha_kv_k=0$. Since the set is linearly independent, we obtain $\alpha_1=\alpha_2=\dots=\alpha_k=0$.

The converse is similar.

How's the map $C_{\mathscr{B}}$ defined? If $\mathscr{B}=\{e_1,e_2,\dots,e_n\}$, then for $v\in V$ we have $$ C_{\mathscr{B}}(v)=(x_1,x_2,\dots,x_n) \text{ if and only if } v=x_1e_1+x_2e_2+\dots+x_ne_n $$ The proof that $C_{\mathscr{B}}$ is linear and bijective consists in applying the definitions.

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  • $\begingroup$ I was thinking geometrically that if $\{C_{\mathscr{B_1}}(v_1),C_{\mathscr{B_1}}(v_2),\dots,C_{\mathscr{B_1}}(v_n)\}$ is linearly independent set then so is $\{C_{\mathscr{B_2}}(v_1),C_{\mathscr{B_2}}(v_2),\dots,C_{\mathscr{B_2}}(v_n)\}$. $\endgroup$ – user602672 Jun 26 '19 at 15:18
  • $\begingroup$ @user602672 That's obvious, isn't it? Just apply the statement I proved. $\endgroup$ – egreg Jun 26 '19 at 15:21
  • $\begingroup$ Yes, but after getting answers from you people :) $\endgroup$ – user602672 Jun 26 '19 at 15:37

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