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I dont have much of a background in linear algebra and am trying to solve a problem for a vector analysis class.

I am trying to prove that a certain basis is linearly independent.

Lets say I have an orthonormal basis

$\hat{e_1}$, $\hat{e_2}$, $\hat{e_3}$

then change to another basis where the new basis vectors are combinations of the original (change of basis?)

How do i show that the new basis vectors are linearly independent?

Is it as simple as checking for a non zero determinant of the matrix where the rows are the new basis vectors?

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    $\begingroup$ vectors of a basis are by default linearly independent $\endgroup$
    – user 1987
    Sep 28, 2018 at 18:26
  • $\begingroup$ You may use better " the set of vectors" instead of base. $\endgroup$
    – dmtri
    Sep 28, 2018 at 18:30
  • $\begingroup$ Yes, you are right about the determinant. Note that the orthonormal base is irrelevant for this case. $\endgroup$
    – dmtri
    Sep 28, 2018 at 18:31
  • $\begingroup$ The elements of the second basis can’t help but be linear combinations of the vectors in the first basis—that’s a consequence of the very definition of basis—so that bit of information tells you nothing useful. If you’re trying to show that a specific set of linear combinations of the first basis vectors also form a basis, that’s very different. If the latter is indeed the case, then update your question with those specifics. $\endgroup$
    – amd
    Sep 28, 2018 at 22:55

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Let say you have three new vectors $v_1,v_2,v_3$. Then they are linear combination of $\hat{e_1},\hat{e_2},\hat{e_3}$, ie $$ v_i = \sum\limits_{j=1}^3 a_{ij} \hat{e_j} $$ In other words, the coordinates of the $v_i$ with respect to the basis $\{\hat{e_1},\hat{e_2},\hat{e_3}\}$ form the rows of the matrix $$ A:=\begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} $$ Hence $\det A =0 $ if and only if its rows are linearly dependent if and only if the $v_i$ are linearly dependent.

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