Show that non-zeros rows in REF are linearly independent

[negation]

Homework Statement

Given

2 1 1 0
0 0 1 1
0 0 0 3(i) Show that the rows of A are linearly independent.
(ii) Show that the nonzero rows of any matrix in row echelon form are linearly independent.

The Attempt at a Solution

i)
REF gives

1 0 0 | 0
0 1 0 | 0
0 0 1 | 0
0 0 0 | 0
x₁ = x₂ = x₃ =x₄ = 0

The solution set is non-trivial due to the row of zeroes and the system is linearly dependent. The rank is 3 and the basis is {(1,0,0),(0,1,0),(0,0,1)}

ii) Can someone give me a step by step guidance on this part?

 

[jbunniii]

Suppose a matrix is in row echelon form. Consider any nonzero row of the matrix. What can you say about that row which is not true of any row below it?

 

[negation]

Suppose a matrix is in row echelon form. Consider any nonzero row of the matrix. What can you say about that row which is not true of any row below it?

For a matrix in REF, the non-zero row with the entry a_ij is to the left of the non-zero row with the entry a_2ij to the left of entry a_3ijto the left of a_4ij and so on to a_nij.

 

[LCKurtz]

$a_{ij}$ refers to the entry in the $i$th row and $j$th column of the matrix. Where is the element $a_{3ij}$?

 

[negation]

$a_{ij}$ refers to the entry in the $i$th row and $j$th column of the matrix. Where is the element $a_{3ij}$?

For a matrix in REF, the non-zero row with the entry a_aa is to the left of the non-zero row with the entry a_bb to the left of entry a_cc to the left of a_dd and so on to a_nn

Tell me if there's a better way of sub-scripting the entries.

Edit: I could have been confused by your question.

a_3ij is the entry located at the 3rd row 3rd column.

 

[HallsofIvy]

For a matrix in REF, the non-zero row with the entry a_aa is to the left of the non-zero row with the entry a_bb to the left of entry a_cc to the left of a_dd and so on to a_nn

Tell me if there's a better way of sub-scripting the entries.

No, that's perfectly good. But you are not using it below.

Edit: I could have been confused by your question.

a_3ij is the entry located at the 3rd row 3rd column.

According to what you said above, the "entry located at the 3rd row 3rd column" would be a₃₃, NOT a_3ij.

 

[LCKurtz]

Edit: I could have been confused by your question.

a_3ij is the entry located at the 3rd row 3rd column.

It's the notation that is confusing you. Three subscripts makes no sense. The entry in the third row and third column is $a_{33}$. The element to its right is $a_{34}$. The element to the right of $a_{ij}$ is $a_{i,j+1}$.

 

[negation]

No, that's perfectly good. But you are not using it below.


According to what you said above, the "entry located at the 3rd row 3rd column" would be a₃₃, NOT a_3ij.

It's the notation that is confusing you. Three subscripts makes no sense. The entry in the third row and third column is $a_{33}$. The element to its right is $a_{34}$. The element to the right of $a_{ij}$ is $a_{i,j+1}$.

Let's stick to " For a matrix in REF, the non-zero row with the entry a_aa is to the left of the non-zero row with the entry a_bb to the left of entry a_cc to the left of a_dd and so on to a_nn"

 

[LCKurtz]

Let's stick to " For a matrix in REF, the non-zero row with the entry a_aa is to the left of the non-zero row with the entry a_bb to the left of entry a_cc to the left of a_dd and so on to a_nn"

You can stick to that if you wish, but I don't have any idea what it means for one row to be to the left of another row.

 

[Ray Vickson]

Let's stick to " For a matrix in REF, the non-zero row with the entry a_aa is to the left of the non-zero row with the entry a_bb to the left of entry a_cc to the left of a_dd and so on to a_nn"

How can one row be to the right of another?

Anyway, would it not be much easier just to say that all the elements in column $i$, lying below $a_{ii}$, are zero. REF is like an upper-triangular matrix but possibly with some zero rows inserted.

 

[negation]

You can stick to that if you wish, but I don't have any idea what it means for one row to be to the left of another row.

How can one row be to the right of another?

Anyway, would it not be much easier just to say that all the elements in column $i$, lying below $a_{ii}$, are zero. REF is like an upper-triangular matrix but possibly with some zero rows inserted.

Alright I was sloppy and careless with words.

The matrix has entries a_aa, a_bb ...a_nn and has zeroes everywhere else.

 

[LCKurtz]

Alright I was sloppy and careless with words.

The matrix has entries a_aa, a_bb ...a_nn and has zeroes everywhere else.

I suppose you mean those are nonzero, which you should state if that's what you mean. (That "sloppy and careless with words" thing again). Even so, does that describe this matrix?$$
\begin{bmatrix}
1&0&0&0\\
0&0&1&0\\
0&0&0&1
\end{bmatrix}$$Here both $a_{22}$ and $a_{33}$ are $0$.

 

[Ray Vickson]

Alright I was sloppy and careless with words.

The matrix has entries a_aa, a_bb ...a_nn and has zeroes everywhere else.

The USUAL form of REF does not conform to what you say. A matrix like
A = \begin{bmatrix}<br /> 1 &amp; 2 &amp; 0 &amp; -1 \\<br /> 0 &amp; 3 &amp; 1 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 &amp; 2<br /> \end{bmatrix} <br />
is a perfectly good REF. As I had said already, you can think of $A$ as arising from an upper-triangular matrix
B = \begin{bmatrix}<br /> 1 &amp; 2 &amp; 0 &amp; -1 \\<br /> 0 &amp; 3 &amp; 1 &amp; 1 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 2<br /> \end{bmatrix}<br />
by removing the zero row.