# Theorectical diagonalazation question

#### 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:
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#### HallsofIvy

Science Advisor
Homework Helper
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

Staff Emeritus
Science Advisor
Gold Member
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

Mentor
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

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
$det(A - \lambda I) = 0$.

#### transgalactic

i know this thing 100%
give me a way to find a polinomial which doesnt 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

Science Advisor
Homework Helper
i know this thing 100%
give me a way to find a polinomial which doesnt 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
$$\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$$

#### 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
$$\begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}$$

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 doesnt 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 doesnt 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

Mentor
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 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.
??

#### Mark44

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).

#### transgalactic

regarding A:
$$A^{10}+12A=pD^{10}p^{-1}+12pDp^{-1}$$

what now
how to prove that its diagoniazable??

#### transgalactic

regarding A:
$$A^{10}+12A=pD^{10}p^{-1}+12pDp^{-1}$$

what now
how to prove that its diagoniazable??
how to show that its diagonizable?

#### transgalactic

is it a correct solution

$$P(A^{10}+12A)P^{-1}=PA^{10}P^{-1}+12PAP^{-1}=(PAP^{-1})^{10}+12D=D^{10}+12D$$

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

#### Mark44

Mentor
Yes, because matrix multiplication is distributive.

thanks

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