Rectangular matrices question

  • Thread starter roadworx
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  • #1
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hi,

I have a question on determining whether rectangular matrices are singular.

[tex]\left[1 0 0 1 0\right][/tex]
[tex]\left[1 0 0 0 1\right][/tex]
[tex]\left[0 1 0 1 0\right][/tex]
[tex]\left[0 1 0 0 1\right][/tex]
[tex]\left[0 0 1 1 0\right][/tex]
[tex]\left[0 0 1 0 1\right][/tex]

The book says it's singular. But the explanation isn't very clear. It says something about the first 3 columns treated as vectors give a column of 1's, the final two columns also give a column of 1's. Any better explanation?
 

Answers and Replies

  • #2
HallsofIvy
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What do you mean by "singular"? The most general definition I know is that a linear transformation (representable by a matrix) is not invertible. That is, T:U->V is "singular" if there is no linear transformation S:V-> U such that ST(u)= u for all u in U and TS(v)= v for all v in U. If T is represented by an non-rectangular matrix (of dimension "n by m" with n not equal to m), then U and V do not have the same dimension and T is either not "1-to-1" or not "onto". In either case it has no inverse. That is every non-square matrix is "singular".

If you are using a different definition of "singular", please tell us what it is.
 

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