Finding and Proving Invertability of 2x2 Matrices

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Homework Help Overview

The discussion revolves around finding and proving the invertibility of 2x2 matrices, specifically focusing on the conditions under which a matrix satisfies the equation AA = A and the implications of a polynomial equation involving the matrix.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the equation AA = A, with some suggesting that A could be a projection operator. There are attempts to analyze the conditions under which A is invertible based on rearranging polynomial expressions and considering specific cases like A = I.

Discussion Status

Some participants have offered guidance on the approaches to take, such as considering the range of A and the implications of assuming A is regular. Multiple interpretations of the problem are being explored, particularly regarding the nature of the solutions for the equation AA = A.

Contextual Notes

Participants are discussing the need to consider both invertible and non-invertible cases for the matrix A, as well as the implications of the determinant in relation to invertibility. There is an acknowledgment of the complexity involved in finding all solutions to the given conditions.

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Homework Statement


a) find all the 2x2 matrices where AA=A.
b) prove that if
[tex] I + a_{1}A + a_{2}A^2 + ... + a_{k}A^k = 0[/tex]
then A is invertable


Homework Equations


1)det(A) = 0 iff A isn't invertable


The Attempt at a Solution


a) I'm not sure how to approch this. I found that if A is invertable then the only solution is A=I but how do i cover the other cases?

b) by rearanging:
[tex] A(a_{1}I + a_{2}A^1 + ... + a_{k}A^{k-1}) = -I [/tex]
and if i take the determinant on each side i see that |A| <> 0 so it's invertable. Is that correct?
Thanks.
 
Last edited:
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b) looks correct.

Regarding a), assume A is regular and see what follows from that. Further on, consider A = I as a special case.
 
AA=A says that A is a projection operator. So consider what the range of A is. The 'other' case you are after is where it is a one-dimensional subspace of R^2. BTW, you don't have to show det(A) is non-zero, you've constructed an explicit inverse.
 
You could also just set A=[[a,b],[c,d]] and write the condition A^2=A. You can then eliminate two of the variables (except for singular cases). There are a two parameter family of these babies.
 

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