# Matrices and Rank

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#1
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)
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2 years ago
#2
(Original post by 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)
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.
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2 years ago
#3
What do you mean by all scalar multiples of each other ?
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2 years ago
#4
(Original post by 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)
Yes to the first paragraph, no to the second. What if the first column, c1, were zero?
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#5
(Original post by RichE)
Yes to the first paragraph, no to the second. What if the first column, c1, were zero?
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?
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2 years ago
#6
(Original post by 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?
What do you mean by "Oh, I see"? If you do, you should then realize what the difference is.

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 years ago
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#7
(Original post by RichE)
What do you mean by "Oh, I see"? If you do, you should then realize what the difference is.
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.
Last edited by Takeover Season; 2 years ago
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#8
(Original post by 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.
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?
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#9
(Original post by B_9710)
What do you mean by all scalar multiples of each other ?
E.g. You can obtain one column by multiplying the other by a scalar e.g. c2 = b * c1 where b is a acalar.
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2 years ago
#10
(Original post by 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.
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.
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2 years ago
#11
(Original post by 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?.
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.
Last edited by RDKGames; 2 years ago
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#12
(Original post by 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.
No worries and thank you, that clarifies it!
(Original post by 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.
Thank you, that clarifies it!
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