# Matrices and Rank

Watch
Announcements

Page 1 of 1

Go to first unread

Skip to page:

Hi, I was just wondering if I've got a nxn matrix and it's got a rank of 1, does this

E.g., in a 3x3 matrix, if I have a rank of 1, does this

**necessarily**mean that the columns and/or rows are all scalar multiples of each other?E.g., in a 3x3 matrix, if I have a rank of 1, does this

**DEFINITELY**mean that I can write c2 as c2 = ac1 and c3 = bc1 for some scalars**a**and**b**, and the same for the rows? (Where c2 = column 2 of the matrix)
0

reply

Report

#2

(Original post by

Hi, I was just wondering if I've got a nxn matrix and it's got a rank of 1, does this

E.g., in a 3x3 matrix, if I have a rank of 1, does this

**Takeover Season**)Hi, I was just wondering if I've got a nxn matrix and it's got a rank of 1, does this

**necessarily**mean that the columns and/or rows are all scalar multiples of each other?E.g., in a 3x3 matrix, if I have a rank of 1, does this

**DEFINITELY**mean that I can write c2 as c2 = ac1 and c3 = bc1 for some scalars**a**and**b**, and the same for the rows? (Where c2 = column 2 of the matrix)Rank of a matrix is the maximal number of of linearly independent columns of the matrix.

If the rank is 1 then there is a singleton linearly independent vector whose span includes all columns of the matrix.

0

reply

Report

#4

**Takeover Season**)

Hi, I was just wondering if I've got a nxn matrix and it's got a rank of 1, does this

**necessarily**mean that the columns and/or rows are all scalar multiples of each other?

E.g., in a 3x3 matrix, if I have a rank of 1, does this

**DEFINITELY**mean that I can write c2 as c2 = ac1 and c3 = bc1 for some scalars

**a**and

**b**, and the same for the rows? (Where c2 = column 2 of the matrix)

1

reply

(Original post by

Yes to the first paragraph, no to the second. What if the first column, c1, were zero?

**RichE**)Yes to the first paragraph, no to the second. What if the first column, c1, were zero?

I suppose my second paragraph then holds if every column is a non zero vector in R3 right?

0

reply

Report

#6

(Original post by

Oh, I see. Yes, but what is the difference between my first and second paragraph? I used the second paragraph to demonstrate my first paragraph's meaning.

I suppose my second paragraph then holds if every column is a non zero vector in R3 right?

**Takeover Season**)Oh, I see. Yes, but what is the difference between my first and second paragraph? I used the second paragraph to demonstrate my first paragraph's meaning.

I suppose my second paragraph then holds if every column is a non zero vector in R3 right?

EDIT: sorry I was abrupt here. I should have gone and re-read your original post. I had read into the first paragraph that the rows had to be scalar multiples of some row - which is correct. But you clearly say of each other. So my answer should have been no to both in the case that one or two of the rows are zero.

Last edited by RichE; 2 months ago

0

reply

(Original post by

What do you mean by "Oh, I see"? If you do, you should then realize what the difference is.

**RichE**)What do you mean by "Oh, I see"? If you do, you should then realize what the difference is.

But, if all of the columns were non zero vectors, then does a rank 1 matrix imply you can write each column as a scalar multiple of each other column e.g c1 = a c2, c1 = b c3, and vice versa c2 = 1/a c1 etc, c3 = 1/b c1 etc.

Last edited by Takeover Season; 2 months ago

0

reply

(Original post by

Isn't it in the definition?

Rank of a matrix is the maximal number of of linearly independent columns of the matrix.

If the rank is 1 then there is a singleton linearly independent vector whose span includes all columns of the matrix.

**RDKGames**)Isn't it in the definition?

Rank of a matrix is the maximal number of of linearly independent columns of the matrix.

If the rank is 1 then there is a singleton linearly independent vector whose span includes all columns of the matrix.

0

reply

(Original post by

What do you mean by all scalar multiples of each other ?

**B_9710**)What do you mean by all scalar multiples of each other ?

0

reply

Report

#10

(Original post by

I mean I can see why it will not work if c1 is the zero vector in R3, then of course if c2 and c3 are linearly dependent but non-zero, then we will have a rank 1 and c2 and c3 are scalar multiples of each other and we can write c1 as a scalar multiple of c2 and c3 i.e. 0c2, 0c3, but not the same for c1 as a scalar multiple of c2 and c3 because the latter are non-zero.

But, if all of the columns were non zero vectors, then does a rank 1 matrix imply you can write each column as a scalar multiple of each other column e.g c1 = a c2, c1 = b c3, and vice versa c2 = 1/a c1 etc, c3 = 1/b c1 etc.

**Takeover Season**)I mean I can see why it will not work if c1 is the zero vector in R3, then of course if c2 and c3 are linearly dependent but non-zero, then we will have a rank 1 and c2 and c3 are scalar multiples of each other and we can write c1 as a scalar multiple of c2 and c3 i.e. 0c2, 0c3, but not the same for c1 as a scalar multiple of c2 and c3 because the latter are non-zero.

But, if all of the columns were non zero vectors, then does a rank 1 matrix imply you can write each column as a scalar multiple of each other column e.g c1 = a c2, c1 = b c3, and vice versa c2 = 1/a c1 etc, c3 = 1/b c1 etc.

0

reply

Report

#11

(Original post by

Thank you, yes I think that is what it is defined as. So, it spans all of the columns individually right i.e. each column can be written as a scalar multiple of that basis vector where the basis vector is one of the columns, so it means each column can be written as a scalar multiple of that basis vector column?.

**Takeover Season**)Thank you, yes I think that is what it is defined as. So, it spans all of the columns individually right i.e. each column can be written as a scalar multiple of that basis vector where the basis vector is one of the columns, so it means each column can be written as a scalar multiple of that basis vector column?.

FYI the issue with your original statement is that you say "scalar multiples of

**each other**" but this is not true because if one vector is zero and another non-zero, then the non-zero vector is clearly not a scalar multiple of the zero vector.

You can refine the way you phrase this by either imposing that the matrix has no zero vectors, or allow them and say that every vector is a scalar multiple of any single non-zero vector.

Last edited by RDKGames; 2 months ago

0

reply

(Original post by

Sorry I was abrupt earlier. I should have gone and re-read your original post. I had read into the first paragraph that the rows had to be scalar multiples of some row - which would have been correct. But you clearly say of each other. So my answer should have been no to both in the case that one or two of the rows are zero.

**RichE**)Sorry I was abrupt earlier. I should have gone and re-read your original post. I had read into the first paragraph that the rows had to be scalar multiples of some row - which would have been correct. But you clearly say of each other. So my answer should have been no to both in the case that one or two of the rows are zero.

(Original post by

More or less.

FYI the issue with your original statement is that you say "scalar multiples of

You can refine the way you phrase this by either imposing that the matrix has no zero vectors, or allow them and say that every vector is a scalar multiple of any single non-zero vector.

**RDKGames**)More or less.

FYI the issue with your original statement is that you say "scalar multiples of

**each other**" but this is not true because if one vector is zero and another non-zero, then the non-zero vector is clearly not a scalar multiple of the zero vector.You can refine the way you phrase this by either imposing that the matrix has no zero vectors, or allow them and say that every vector is a scalar multiple of any single non-zero vector.

0

reply

X

Page 1 of 1

Go to first unread

Skip to page:

### Quick Reply

Back

to top

to top