Is this endomorphism diagonalizable ?

[God]

f is an endomorphism of R^2.
What I know is that (f-3*identity)o(f-4*identity)=0 (here, o is as in : f(g(x))=f o g (x))
I have to deduce from this whether f is diagonalizable (proving it or disproving it by a counter-example)

I tried to find some counter example at random, but didn't find any, so I thought I would simply try to "solve" this by supposing the matrix of f is
a b
c d

and then multiply
a-3 b
c d-3
with
a-4 b
c d-4
then solve a system so as to make this product equal to zero ((f-3*identity)o(f-4*identity)=0)
and find out if the matrix of f is diagonalizable or not... but it is very very difficult and I got a non-linear system of equations I'm not even sure I can solve by myself, so I take it I am in the wrong track...

Can I have some hint please ? Thanks.

 

[God]

:-'(

 

[daon2]

You have:


\(\displaystyle \text{det}((f-4)(f-3)) = \text{det}(f-4)\text{det}(f-3)=0\)

Which means one of f-3 and f-4 is not invertible.

Now break it into cases.

(1) What if neither are invertible?
(2) What if only f-3 is invertible? Hint: f might not be diagonalizable. To find a counter-example, assume \(\displaystyle f(v)=\lambda v\), what are possible values of lambda?

 

[God]

You have:


\(\displaystyle \text{det}((f-4)(f-3)) = \text{det}(f-4)\text{det}(f-3)=0\)

Which means one of f-3 and f-4 is not invertible.

Now break it into cases.

(1) What if neither are invertible?
(2) What if only f-3 is invertible? Hint: f might not be diagonalizable. To find a counter-example, assume \(\displaystyle f(v)=\lambda v\), what are possible values of lambda?

If neither are invertible, then 3 and 4 are Eigenvalues. In this case, f is diagonalizable, since there are 2 different values, and it is an endomorphism of R^2.
If only f-3 is invertible, then 3 is not an Eigenvalue, but 4 is.
The possible values of lambda are 3 or 4. But I still can't tell whether f is diagonalizable in this case. (There might be another value of lambda, but it can't be told due to the lack of information)

Having \(\displaystyle f(v)=\lambda v\) doesn't imply f is not diagonalizable.
Let M be the matrix of f in the standard basis, f is diagonalizable if \(\displaystyle Kernel(M-\lambda*id)=\)multiplicity of lambda, but in this case I can't do anything with it since I don't know either of them.

 

[God]

Okay I guess I found a counter-example :
After assuming f(v)=3v, I found this matrix :
\(\displaystyle A=\begin{pmatrix} 3 & 3\\ 0 & 3\end{pmatrix}\)
with v=(1,0) eigenvector (Av=3v)

det(A-x*id)=(x-3)² so 3 is an eigenvalue of multiplicity 2
ker(A-3id) only contains vectors such as (x,0), so dim(ker)=1=/=2 so the matrix isn't diagonalizable

However, (A-3id)*(A-4id)=0

So the endomorphism associated to this matrix is not diagonalizable, but (f-3id)o(f-4id)=0

and the counter example would be f(x,y)=(3x+3y,3y)

Is it correct ?

But then why do I get f(x,y)-3id=(3y,0), f(x,y)-4id=(3y-x,-y) and (f-3id) o (f-4id)=(-3y,0)=/=0 ?? This really doesn't make sense, works with matrix but not with functions... help pls




NVM, the matrix product isn't equal to zero.
Well, there is no way I can find it. I have tried everything I could

 

[God]

Please can someone help me so I can finally move on and focus on something else ??
I have tested several matrixes already and they never fulfill all the conditions

 

[daon2]

Please can someone help me so I can finally move on and focus on something else ??
I have tested several matrixes already and they never fulfill all the conditions

My fault. if f-3 is invertible, then multiplying (composing) both sides of (f-3)(f-4) by the inverse of f-3 on the left gives you (f-4)=0.