Matrices and Rank

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)

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?

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.

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.

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

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.

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?.

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.

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.