A question in finding invertion

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Homework Help Overview

The discussion revolves around determining the invertibility of an operator defined on polynomials, specifically focusing on the polynomial t^3 - 4t and its eigenvalues. Participants are exploring the implications of eigenvalues on the operator's properties.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between eigenvalues and invertibility, particularly questioning how the presence of a zero eigenvalue affects the operator's invertibility. There is also mention of the determinant's role in this context.

Discussion Status

Some participants have provided insights regarding the implications of having a zero eigenvalue and its connection to the determinant being zero, suggesting that the operator is not invertible. The discussion reflects a progression towards understanding these concepts, although no explicit consensus is reached.

Contextual Notes

There is an emphasis on the relationship between eigenvalues, diagonalization, and the determinant, with participants referencing laws related to invertibility. The original poster expresses uncertainty about applying these concepts to the problem.

transgalactic
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i am given an operator
S:R3[x]->R3[x]

and we have the polinomial from which we take the eigenvalues from
t^3 - 4t
find wether S invertable or not?i tried to think about that and i got that the aigenvalues are 2 , 0, -2
but that only prooves that it diagonizable

i know a law that if the determinant differs zero then its invertable
but i don't know how to apply it here
?
 
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If it has a zero eigenvalue then it has a nontrivial kernel. It's not invertible.
 
transgalactic said:
i know a law that if the determinant differs zero then its invertable
Hopefully, you know that works both ways because it is the other way you need here: If the determinant of a matrix is 0, then it is not invertible.

When you diagonalize a matrix, its determinant stays the same. The determinant of a diagonal matrix is the product of the numbers on its diagonal. Since the eigenvalues are 2, 0, and -2, the "diagonalized" matrix would have those numbers on its diagonal and so its determinant is 0.

Of course, Dick's method works just as well.
 
ahhh because its determinant is the multiplication of the diagonal
and we got a zero in there
then the whole thing is zero

thanks
 

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