Rank of a matrix

  • Thread starter eg0
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  • #1
eg0
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Hi
I don't understand why only a matrix full of zero has a rank = 0.

"the rank of a matrix A is the number of linearly independent rows or columns of A"

If I have a 3x3 matrix

A = [ 1 1 1
1 1 1
1 1 1 ]

assuming a_i denotes the column or row vector i of A. I can say

a_1 = 1*a_2 + 0*a_3 so a_1 is not linearly independant
a_2 = 1*a_1 + 0*a_3 so a_2 is not linearly independant
a_3 = 1*a_1 + 0*a_2 so a_3 is not linearly independant

So why rank A = 1 and not 0 ?
I know i'm missing something, I don't know what!
 

Answers and Replies

  • #2
lurflurf
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To have rank 0 Ax=0 for all x. The column (or row vectors) are linearly dependent in pair and triples, but linearly dependent in singles. The rank is r if there exist r rows or columns that are linearly independent.
 
  • #3
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rank is the dimension of the subspace composed by the set of points you can reach using constant multiples of the vectors in your matrix.
A 3x3 matrix of ones can reach any point on a line in R3 (which is a subspace) and lines have dimension 1, so rank is 1.
 
  • #4
HallsofIvy
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More abstractly, an n by n matrix represents a linear transformation from an n dimensional vector space to an n dimensional vector space- L: U-> V. The "range" is the dimension of L(U) as a subspace of V. In particular, if you multiply a matrix by the vector having 1 as the ith entry, 0 every where else, you get the ith column of the matrix. But the set of all such vectors form a basis for U and so are mapped into a set that spans L(U). The only subspace with dimension 0 is the set containing only the 0 vector. In other words, to have rank 0, L must map every vector into the 0 vector. That is the "0" linear tranformation which is represented by the 0 matrix.
 
  • #5
Deveno
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a simpler explanation is provided by noting that the set {(0,0,0)} (or any other n-dimensional 0-vector) is a linearly dependent set.

why? because for the 0-vector, we can have c0 = 0, even if c is non-zero.
 
  • #6
eg0
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Thank you. So I was wrong mainly because I had not understood the notion of linearly (in)dependence...
 

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