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not invertible mean that the determinant equals 0.

x(x-2)=0

??

- Thread starter transgalactic
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- #1

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not invertible mean that the determinant equals 0.

x(x-2)=0

??

- #2

Mark44

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Where does x(x -2) come in?

not invertible mean that the determinant equals 0.

x(x-2)=0

??

You have an n x n matrix A that has 0 as an eigenvalue. How do you find the eigenvalues of a matrix? How are eigenvalues

- #3

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x(x -2) its just an example for a polynomial which is not inertible

i thought of proving like this

that if an operator T is not invertible then det T=0 then dim ker differs 0

(we have a row of zeros) so i have eigen value 0

correct

??

i thought of proving like this

that if an operator T is not invertible then det T=0 then dim ker differs 0

(we have a row of zeros) so i have eigen value 0

correct

??

Last edited:

- #4

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"If an equation containing a variable parameter possesses nontrivial solutions only for certain special values of the parameter, these solutions are called eigenfunctions and the special values are called eigenvalues."Where does x(x -2) come in?

You have an n x n matrix A that has 0 as an eigenvalue. How do you find the eigenvalues of a matrix? How are eigenvaluesdefined?

how to use it

??

- #5

Office_Shredder

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[tex]Av = \lambda v[/tex]

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Av=0

how does it show that its not invertible

i know that now v plays the role of kernel A

but i dont know what is the link between the kernel of an operator and its

ability to be invertible

??

- #7

Office_Shredder

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- #8

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i really dont know what is the affect of kernel

i know how to find it in a matrix

i know

dim Im + dim Ker =V

??

- #9

statdad

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- #10

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zero vector is the product of multiplication of the operarator by the kernel vectors

i dont know

??

i dont know

??

- #11

statdad

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[tex]

Ax = b

[/tex]

is

[tex]

x = A^{-1}b

[/tex]

Now, suppose that [tex] v \ne 0 [/tex] is an eigenvector that corresponds to the eigenvalue [tex] \lambda = 0 [/tex]

Now

[tex]

Av = \lamba \, v = 0

[/tex]

tell you about [tex] v [/tex]? What does this contradict, and what assumption led to that contradiction?

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Consider

Ax = 0

A0 = 0

What is A^{-1}(0)?

Ax = 0

A0 = 0

What is A

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what is the answer

??

- #15

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why if dim ker differs 0

then its not invertible

??

then its not invertible

??

- #16

Mark44

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If matrix A has 0 as an eigenvalue, A does not have an inverse.

If A has 0 as an eigenvalue, Ax = 0x = 0, for some nonzero vector x (the eigenvector associated with eigenvalue 0).

So Ax = 0, where x is not the zero vector.

From this equation, is it possible to say either that A definitely has an inverse or that A definitely doesn't have an inverse?

- #17

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but i dont know how it affects the invertiblity of A

??

- #18

Mark44

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There is always at least one vector in ker(A), regardless of whether A has an inverse.

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yes 0,0,0,

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HallsofIvy

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- #21

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i get x=0

what does it mean?

what does it mean?

- #22

HallsofIvy

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That means that "x= 0" is the **only** solution to Ax= 0. What does that tell you about the kernel of A?

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