Theorectical diagonalazation question

[transgalactic]

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

B)give an example for a matrixB\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 don't now how to show what they ask
??

 

[HallsofIvy]

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

 

[transgalactic]

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

 

[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.

 

[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??

 

[Office_Shredder]

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

 

[transgalactic]

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

 

[Mark44]

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

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.

which has only one eigen value of -2
how to deside ith square polinomial that i got is diagonizable??

 

[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}

 

[Mark44]

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.

 

[transgalactic]

i know this thing 100%
give me a way to find a polinomial which doesn't have eigen values
but its square does have
??

 

[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
??

 

[HallsofIvy]

i know this thing 100%
give me a way to find a polinomial which doesn't have eigen values
but its square does have
??

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 &amp; 1 \\ -1 &amp; 0\end{bmatrix}

 

[yyat]

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
??

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

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

 

[transgalactic]

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 &amp; 1 \\ -1 &amp; 0\end{bmatrix}

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 doesn't have eigen values but the square of this matrix has
??

 

[yyat]

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 doesn't have eigen values but the square of this matrix has
??

The characteristic polynomial is

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

so the matrix has no real eigenvalues.

 

[transgalactic]

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

??

 

[Mark44]

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

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.

 

[transgalactic]

i don't know how to think of a matrix which polynomial doesn't have eigenvalues.
on the other hand i can think of such polynomial but how to get its matrix.
??

 

[Mark44]

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).

 

[transgalactic]

regarding A:
<br /> A^{10}+12A=pD^{10}p^{-1}+12pDp^{-1}<br />

what now
how to prove that its diagoniazable??

 

[transgalactic]

regarding A:
<br /> A^{10}+12A=pD^{10}p^{-1}+12pDp^{-1}<br />

what now
how to prove that its diagoniazable??

how to show that its diagonizable?

 

[transgalactic]

is it a correct solution

<br /> P(A^{10}+12A)P^{-1}=PA^{10}P^{-1}+12PAP^{-1}=(PAP^{-1})^{10}+12D=D^{10}+12D<br />

is it legal to do the split like
<br /> P(A^{10}+12A)P^{-1}=PA^{10}P^{-1}+12PAP^{-1}<br />

 

[Mark44]

Yes, because matrix multiplication is distributive.

 

[transgalactic]

thanks