# Show that non-zeros rows in REF are linearly independent

1. May 29, 2014

### negation

1. The problem statement, all variables and given/known data

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.

3. 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?

2. May 29, 2014

### 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?

3. May 29, 2014

### negation

For a matrix in REF, the non-zero row with the entry aij is to the left of the non-zero row with the entry a2ij to the left of entry a3ijto the left of a4ij and so on to anij.

4. May 29, 2014

### LCKurtz

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

5. May 29, 2014

### negation

For a matrix in REF, the non-zero row with the entry aaa is to the left of the non-zero row with the entry abb to the left of entry acc to the left of add and so on to ann

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

Edit: I could have been confused by your question.

a3ij is the entry located at the 3rd row 3rd column.

6. May 29, 2014

### HallsofIvy

Staff Emeritus
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 a3ij.

7. May 29, 2014

### LCKurtz

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 $a_{i,j+1}$.

8. May 29, 2014

### negation

Let's stick to " For a matrix in REF, the non-zero row with the entry aaa is to the left of the non-zero row with the entry abb to the left of entry acc to the left of add and so on to ann"

9. May 29, 2014

### LCKurtz

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.

10. May 29, 2014

### Ray Vickson

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.

11. May 29, 2014

### negation

Alright I was sloppy and careless with words.

The matrix has entries aaa, abb ....ann and has zeroes everywhere else.

12. May 29, 2014

### LCKurtz

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 $0$.

13. May 29, 2014

### Ray Vickson

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