# How to prove basis that is based on an orthonormal basis in linearly independent

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?

• vectors of a basis are by default linearly independent Sep 28, 2018 at 18:26
• You may use better " the set of vectors" instead of base. Sep 28, 2018 at 18:30
• Yes, you are right about the determinant. Note that the orthonormal base is irrelevant for this case. Sep 28, 2018 at 18:31
• 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.
– amd
Sep 28, 2018 at 22:55

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.