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Why is a matrix singular if the determinant is zero?

by brownman
Tags: determinant, singular, singular matrix
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brownman
#1
Feb3-13, 05:09 PM
P: 13
I'm looking for the deeper meaning behind this law/theorem/statement (I don't know what it is, please correct me). My textbook just told us a matrix is not invertible if the determinant is zero.
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tiny-tim
#2
Feb3-13, 05:22 PM
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hi brownman!

an nxn matrix is a linear transformation on ℝn to itself

its determinant is the ratio of the n-dimensional volume of the image of a box to the volume of the box itself

if the determinant is 0, the image of the box is flattened by at least one dimension (because its volume is 0), so the transformation obviously isn't invertible (it isn't one-to-one)

if the determinant isn't 0, the box isn't flattened, so the transformation can always be reversed
HallsofIvy
#3
Feb3-13, 07:11 PM
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It is also true that det(AB)= det(A)det(B). so if det(A)= 0, it is impossible to have AB= I for any matrix B.

Ferramentarius
#4
Feb4-13, 03:00 AM
P: 22
Why is a matrix singular if the determinant is zero?

As the determinant is the product of the eigenvalues of a matrix it being zero means at least one of the eigenvalues is zero as well. By definition it follows that Ax = 0x = 0 for some vector x ≠ 0. In case A was invertible we would have (A^-1)Ax = 0 meaning x = 0 which contradicts that x ≠ 0 and therefore A is not invertible.
brownman
#5
Feb5-13, 11:50 PM
P: 13
Okay the combined definitions from all of you seem to make a general sort of sense, thank you for the help guys :)
fortissimo
#6
Feb12-13, 09:36 AM
P: 24
A definition of the determinant of an n*n matrix as the n-volume spanned by its column vectors gives this result easily. It can also be proved using other definitions with somewhat more hassle.
lavinia
#7
Feb14-13, 12:54 PM
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Quote Quote by brownman View Post
Okay the combined definitions from all of you seem to make a general sort of sense, thank you for the help guys :)
A way to "see" that a determinant is a volume measure is to work it out for two vectors in the plane. It is a simple exercise in Euclidean geometry.

For instance suppose the two vectors are (5,0) and (3,2)

The area of the parallelogram that they span is the height times the base which is 2(the height) x 5(the base). The determinant of the matrix

5 0
3 2

is 5 x 2 - 3x0.

Can you generalize this example?
fortissimo
#8
Feb14-13, 02:57 PM
P: 24
Quote Quote by fortissimo View Post
A definition of the determinant of an n*n matrix as the n-volume spanned by its column vectors gives this result easily. It can also be proved using other definitions with somewhat more hassle.
Also, it is the signed volume, which depends on the order of the vectors.
lavinia
#9
Feb15-13, 10:03 AM
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Another way to think about determinates is to think of them as multilinear functions of the rows (or columns) and investigate how this algebraic property is related to computing volumes. The approach leads to the idea of the exterior algebra of a vector space.

Question: Is every alternating multilinear map a determinant?


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