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I've been going over my notes and I've found that I'm missing the proof of the following. I might have missed it in class or something but it's not in there(my work book) so can someone help me out with showing the following?

Let A be the matrix whose columns are the vectors v_1, v_2,...,v_k(which are elements of R^n). Reduce to row-echelon form. The k vectors are linearly independent in R^n iff Rank(A) = k.

Hopefully there are no uncertainties as to what the definition of row-echelon form and rank are. The definition I have of rank is the number of non-zero rows of a matrix in row-echelon form.

Ok here's how I'd start off. Write down a linear dependence equation.

[tex]

\alpha _1 \mathop {v_1 }\limits_{} + \alpha _2 \mathop {v_2 }\limits_{} + ... + \alpha _k \mathop {v_k }\limits_{} = \mathop 0\limits_{}

[/tex]

The a_i are scalars and the v_i are vectors and the RHS is the zero vector.

By definition of A, the augmented matrix of the equation is:

[tex]

\left( {\left. {\begin{array}{*{20}c}

{} \\

A \\

{} \\

\end{array}} \right|\begin{array}{*{20}c}

0 \\

\vdots \\

0 \\

\end{array}} \right) \approx \left( {\begin{array}{*{20}c}

{} \\

B \\

{} \\

\end{array}\left| {\begin{array}{*{20}c}

\cdots \\

\cdots \\

\cdots \\

\end{array}} \right.} \right)

[/tex]

The above is supposed to mean that B is the row reduced form of A and that the two matrices are row equivalent.

The two matrices have n rows(since the v_i are in R^n) and from how A has been defined. The matrix B has n >= k zeros in the column to the right of the augmentation line.

(1) If rank(A) = k then there are k non-zero rows in B.(the column to the right of the augmentation line contains all zeros)

There are k columns in B(to the left of the augmentation line) corresponding to the k non-zero rows in B.

=> system has unique solution for the a_i.

=> all the a_i = 0 since the column to the right of the augmentation line in B consists of all zeros. So if rank(A) = k then the v_i are linearly indepedent.

(2) If the v_i are linearly independent vectors then the system has the unique solution a_i = 0 for 1 <= i <= k. This implies rank(A) = k.

That's all I've been able to come up with. Any help would be good thanks.

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# Homework Help: Linear indepdence

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