- #1

Benny

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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.

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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