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$S=(v_1, \cdots v_n)$ of vectors in $\mathbb{F^n}$ is a basis iff the matrix obtained by forming a matrix (call it A) of the co-ordinate vectors of $v_i$ is invertible

My Idea: I was able to prove the reverse direction wherein we can show that $AX =0$ has trivial solutions so linearly independent and also spans the given vector space so it is a basis.

I am not sure about the forward direction that is to prove that A is invertible. I am just confused about the coordinate vector for $v_i$. That is,we assume $S$ is a basis then how can each element (namely $v_i'$'s) in a basis have a coordinate vector? Wouldn't it just be a single column for different vector in the vector space?

James Bond
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2 Answers2

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HINT: The determinant of a matrix is zero iff the vectors which form it are linearly independent.

See also:

Using the Determinant to verify Linear Independence, Span and Basis

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Leo Azevedo
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  • I was thinking in a way to visualize this geometrically and what I got was the following. Solving $Ax=0$ is the same as asking "how many vectors x are perpendicular to all the vectors (rows) in my matrix A?". If the vectors in A form a basis of the space, there can be no vector x, except the null one, which is perpendicular to every vector in A. – Leo Azevedo Jan 29 '14 at 07:14
  • Ah! Great intuitive explanation. Thanks a lot – James Bond Feb 03 '14 at 18:06
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You can rewrite $A$ as a row of vectors $A_1,\dots,A_n$. Then $$ Ax = x_1 A_1 + \dots + x_n A_n $$ Think about

$Ax=b$ always solvable means $A_i$ generate $V$

$Ax=b$ uniquely solvable (if at all) means $A_i$ independent

If these two conditions are met, you can find $n$ solutions for $Ax_i=e_i$ (put them in a matrix). This matrix will be the inverse.

If $A$ is invertible, then for every $b$ the equation $Ax=b$ has the unique solution $x=A^{-1}b$. So the $A_i$ are a basis.

Blah
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