# Theorectical diagonalazation question

1. Mar 2, 2009

### transgalactic

there is
$$A\epsilon M_{2x2}(Q)$$
I am given that A is diagonazable
prove that
A)
$$A^{10}+12A$$
is diagonizable too

B)give an example for a matrix$$B\epsilon M_{2x2}(Q)$$
that is not diagonizable,but b^2 is diagonisable
??

i know that the eigenvalues of a matrix are the same as for every matrix
like A^10 or A^3+2A+3I etc..

but i dont now how to show what they ask
??

Last edited: Mar 3, 2009
2. Mar 2, 2009

### HallsofIvy

Staff Emeritus
What do you mean by $A^{10}+ 12A$?

3. Mar 3, 2009

### transgalactic

i mean operator of A n the power 10 plus 12 A
that matrix is diagonizable too
?

4. Mar 3, 2009

### yyat

A) If A is diagonalizable, then $$D=T^{-1}AT$$, where D is diagonal. Now, note that sums and powers of diagonalizable matrices are also diagonal.

B) Hint: a certain rotation in the 2-D plane...
To show that a matrix is not diagonalizable over Q, it is sufficient to show that the characteristic polynomial has no rational roots.

5. Mar 3, 2009

### transgalactic

regarding A:
i tried like this:
A=p^-1*D*P
P(A^10+12A)P^-1=P*A^10*P^-1 +12 P*A*P^-1
next..
why
P*A^10*P^-1=(P*A*P^-1)^10
??

regarding b:
i took matrix which polinomial is b=x^2+x+1 =[1,1,1,0]

b^2=[2,1,1,1]=x^3+x^2+x+2
which has only one eigen value of -2
how to deside ith square polinomial that i got is diagonizable??

6. Mar 3, 2009

### Office_Shredder

Staff Emeritus
If you only get one eigenvalue, you need to check if there are two linearly independent eigenvectors

7. Mar 3, 2009

### transgalactic

actually -2 is not an eigen value
so it crashes the theory proposed by YYAT

8. Mar 3, 2009

### Staff: Mentor

You can't possibly get a 3rd degree characteristic polynomial from a 2x2 matrix. Furthermore, a matrix is not equal to its characteristic polynomial, as you show above.

9. Mar 3, 2009

### transgalactic

"yes i can"

we got 4 coordinated each one represents a member of a polinomial
so there is no much room for desitions
{1,x,x^2,x^3}

10. Mar 3, 2009

### Staff: Mentor

Then it is apparent that you don't know how to calculate the characteristic polynomial of a matrix. The characteristic polynomial of an n x n matrix A comes from the equation
$det(A - \lambda I) = 0$.

11. Mar 3, 2009

### transgalactic

i know this thing 100%
give me a way to find a polinomial which doesnt have eigen values
but its square does have
??

12. Mar 3, 2009

### transgalactic

regarding A:
i tried like this:
A=p^-1*D*P
P(A^10+12A)P^-1=P*A^10*P^-1 +12 P*A*P^-1
next..
why
P*A^10*P^-1=(P*A*P^-1)^10
??

13. Mar 3, 2009

### HallsofIvy

Staff Emeritus
You mean "matrix" that doesn't have eigenvalues, not "polynomial". Every square matrix has eigenvalues over the complex numbers so I assume here you are talking about real numbers. Try
$$\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$$

14. Mar 3, 2009

### yyat

$$(PAP^{-1})^2=PAP^{-1}PAP^{-1}=PA^2P^{-1}$$

since $$P^{-1}P=I$$. I think you can see the general pattern.

15. Mar 3, 2009

### transgalactic

the polinomial that comes from it is
x-x^2
it has an eigen values
x(1-x) x=0 x=1
i need a matrix which polinomial doesnt have eigen values but the square of this matrix has
??

16. Mar 3, 2009

### yyat

The characteristic polynomial is

det(A-xI)=x^2+1

so the matrix has no real eigenvalues.

17. Mar 3, 2009

### transgalactic

how to find the matrix of
x^2+x+3

??

18. Mar 3, 2009

### Staff: Mentor

I'm just being curious, but why is this important to find? I don't see anything in this thread that seems related to this particular polynomial.

19. Mar 4, 2009

### transgalactic

i dont know how to think of a matrix which polynomial doesnt have eigenvalues.
on the other hand i can think of such polynomial but how to get its matrix.
??

20. Mar 4, 2009

### Staff: Mentor

Well, you're dealing with 2 x 2 matrices, with entries [a b; c d] (reading row by row). Maybe you can fiddle with these values to make (a - lambda)(d - lambda) - bc come out the way you want. The preceding quadratic in lambda is det(A - lambda*I).