Proving Linear Independence of Non-Zero Rows in Row-Echelon Form

In summary: The matrix is a representation of the set of nonzero row vectors. The position of the vectors is not important.
  • #1
Benny
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0
Hi, can someone help me with the following question?

Q. Show that if [tex]\left\{ {\mathop {v_1 }\limits^ \to ,...,\mathop {v_k }\limits^ \to } \right\}[/tex] is linearly independent and [tex]\mathop {v_{k + 1} }\limits^ \to \notin span\left\{ {\mathop {v_1 }\limits^ \to ,...,\mathop {v_k }\limits^ \to } \right\}[/tex] then [tex]\left\{ {\mathop {v_1 }\limits^ \to ,...,\mathop {v_k }\limits^ \to ,\mathop {v_{k + 1} }\limits^ \to } \right\}[/tex] is linearly independent. Use this to prove that the non-zero rows of a matrix in row-echelon form are linearly independent.

Here is my attempt.

Write [tex]\alpha _1 \mathop {v_1 }\limits^ \to + ... + \alpha _k \mathop {v_k }\limits^ \to + \beta \mathop {v_{k + 1} }\limits^ \to = \mathop 0\limits^ \to ...\left( 1 \right)[/tex]

[tex]
\beta \mathop {v_{k + 1} }\limits^ \to = - \left( {\alpha _1 \mathop {v_1 }\limits^ \to + ... + \alpha _k \mathop {v_k }\limits^ \to } \right)
[/tex]

If [tex]\beta \ne 0[/tex] then [tex]\mathop {v_{k + 1} }\limits^ \to = - \left( {\frac{{\alpha _1 }}{\beta }\mathop {v_1 }\limits^ \to + ...\frac{{\alpha _k }}{\beta }\mathop {v_k }\limits^ \to } \right)[/tex] but this is impossible since [tex]\mathop {v_{k + 1} }\limits^ \to \notin span\left\{ {\mathop {v_1 }\limits^ \to ,...,\mathop {v_k }\limits^ \to } \right\}[/tex]

So beta is equal to zero and equation one reduces to [tex]\alpha _1 \mathop {v_1 }\limits^ \to + ... + \alpha _k \mathop {v_k }\limits^ \to = \mathop 0\limits^ \to [/tex] where all of the a_i are equal to zero by hypothesis. Is that enough to show the given result?

I can't think of a way to tackle the second part with the matrix. Seeing as that's the case I'll just write out whatever I can think of.

I think the key idea is that in row echelon form, each time I 'move up' one row, the vector(represented by a row in the matrix) has at least one additional non-zero component. So let A be the n by k (n columns and k rows) matrix in row echelon form whose rows are the vectors v_i where i = 1,...,k and each of the vectors has at least one non-zero component.

Starting at the bottom of the matrix and moving up to the first non-zero row I a vector which has c non-zero components call it v_1 and {(v_1)} is linearly independent since it consists of a non-zero single vector. Moving up to the next row I get another vector call it v_2 which has at least c + 1 non-zero components. Since v_2 has more non-zero components than v_1 then {v_1, v_2} is linearly independent. From here I'd probably just continue with the same argument. The problem is that what I've said is a pretty clumsy explanation. I wasn't really sure how to do this question either. So can someone please help me with this?

Edit: Ok my attempt for the second part is completely incorrect because I could have something like v_1 = (0,0,1,0,0) and v_2 = (1,0,0,0,0). Help would be appreciated.
 
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  • #2
Do the second part by induction.
n=1 case
the vector is not zero so is linearly independent
n+1 case
n vectors are independent
if (n+1)st vector is in span(first n vectors) the matrix is not in row-echelon form
therfore (n+1)st vector is not in span
thus n+1 vectors are linearly independent
 
  • #3
Thanks for te help lurflurf. However, I am not sure how my answer should be worded. The question refers to a matrix so would I need to make some reference to the matrix? If so how would I do it in a clear and concise manner? For example, do I need to mention the position of the vectors(represented as rows) in the matrix?
 
  • #4
Benny said:
Thanks for te help lurflurf. However, I am not sure how my answer should be worded. The question refers to a matrix so would I need to make some reference to the matrix? If so how would I do it in a clear and concise manner? For example, do I need to mention the position of the vectors(represented as rows) in the matrix?
The matrix is a representation. You can consider the set of the nonzero row vectors. The position of the vectors is not important.
Just say something like
let v1,...,vn be the nonzero row vectors
 

What is the definition of linear independence?

Linear independence refers to a set of vectors in a vector space where no vector can be expressed as a linear combination of the other vectors in the set.

Why is it important to prove linear independence of non-zero rows in row-echelon form?

Proving linear independence of non-zero rows in row-echelon form is important because it shows that the system of equations represented by the rows has a unique solution. This is essential in solving many real-world problems in fields such as physics, engineering, and economics.

How do you prove linear independence of non-zero rows in row-echelon form?

To prove linear independence of non-zero rows in row-echelon form, you can use the reduced row-echelon form of the matrix and check for any rows that are all zero or contain only one non-zero element. If there are no such rows, then the non-zero rows in the original row-echelon form are linearly independent.

What are the implications of a set of vectors being linearly dependent?

If a set of vectors is linearly dependent, it means that at least one vector in the set can be expressed as a linear combination of the other vectors. This can lead to inconsistencies and multiple solutions in a system of equations, making it difficult to find a unique solution.

Is there a quick way to check for linear independence of non-zero rows in row-echelon form?

Yes, there is a quick way to check for linear independence of non-zero rows in row-echelon form. You can use the determinant of the matrix formed by the non-zero rows. If the determinant is non-zero, then the rows are linearly independent. However, if the determinant is zero, then further analysis is needed to determine the linear independence of the rows.

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