# How do I show that the vectors of an invertible MX are indepedent?

1. Mar 23, 2013

### Minhtran1092

Suppose we have an nxn matrix A with column vectors v1,...,vn. A is invertible. With rank(A)=n. How do I prove that v1,...,vn are linearly independent?

I think I can prove this by using the fact that rank(A)=n, which tells me that there is a pivot in each of the n columns of the rref(A) matrix (because rref(invertible mx) gives an identity matrix). I'm not sure how to interpret this result to show that each column vector are linearly independent though.

Should I look at the linear combination of an identity matrix to establish independence?

2. Mar 23, 2013

### mathman

I don't know how to go about the proof. However, if the v's are linearly dependent, then det(A) = 0 and the matrix is not invertible.

3. Mar 23, 2013

### lavinia

Not sure what you mean by pivot and rref so I can't help you directly.

But if the vectors were linearly dependent the some linear combination of them would be zero - by definition. The coefficients of this linear combination form another vector. What is the matrix multiplied by this vector?

4. Mar 23, 2013

### Minhtran1092

rref stands for the Reduced Row Echelon Form operation on a calculator (which operates on a matrix to give the reduced row echelon form of some given matrix). rref(A) gives reduced row echelon form of A.

A pivot of a row refers to the leading 1 in the respective row of the rref of some MX.

I don't follow where you mentioned that the linear comb. of some dependent vectors would be zero.

5. Mar 23, 2013

### AlephZero

That is just the definition of "linearly dependent".

If the $v_i$ are linearly dependent, then $\sum c_iv_i = 0$ where the scalars $c_i$ are not all zero.

The $v_i$ are column vectors of A. So think how to write $\sum c_iv_i = 0$ as $Ax = 0$, for a non-zero vector $x$.