Isomorphic transformation question

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there is a transformation T:R^4 ->R^4

so

dim Im(T+I)=dim Ker(3I-T)=2

prove that T-I is isomorphic



first of all i couldn't understand the first equation

because T is a transformation which is basicly a function

but I is the identity matrices

so its like adding kilograms to tempreture.



then my prof told me that here I is a transformation too

that its not a matrix

its the identity transformation.



and i was told that i need to get the eigenvectors from there

so i told him

"how i don't have any matrix here to do

|A-labdaI|=0

i don't have any matrices only some transformation

which i don't have even the formula to T in order to get the representing matrices out if it





how to find eigen vectors from here

?

and how to proceed in order to prove that

T-I

is isomorphic

?
 
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Hi nhrock3! :smile:

Matrices and transformations are equivalent. Given a matrix, there is a corresponding transformation and vice versa. Thus we can define the notion of eigenvalue for a transformation by

T(x)=\lambda x

thus the eigenvectors of T are the elements of

Ker(T-\lambda I)

and lambda is the corresponding eigenvalue.

Now, can you use the fact that

dim Ker(T+I)=2

to find an eigenvalue of T? And what multiplicity does the eigenvalue have?
 
micromass said:
Hi nhrock3! :smile:

Matrices and transformations are equivalent. Given a matrix, there is a corresponding transformation and vice versa. Thus we can define the notion of eigenvalue for a transformation by

T(x)=\lambda x

thus the eigenvectors of T are the elements of

Ker(T-\lambda I)

and lambda is the corresponding eigenvalue.

Now, can you use the fact that

dim Ker(T+I)=2

to find an eigenvalue of T? And what multiplicity does the eigenvalue have?

i don't understand the transition
i know that
T(v)=Av=\lambda v
its the definition of the link between eigen vectors and eigen values

Ker(T+I) means (T+I)(v)=0

but T and I are both transformation
so i can't use it like here
because
in here
Ker(T-\lambda I)
I is a mtrix
but my I is the identity transformation
its a function not a matrix
 
Last edited:
ok i understand your idea
that
Ker(T+I) meand the eigen vectors of eigenvalue -1
but still matheticly
i need to replace T with some matrices and I (the transformation) needs to be I(the identety matrix)
 
Why do you change everything to matrices?? You can do that, of course, but you can leave everything in transformation form too.

Saying that the vectors in

Ker(T+I)

are the eigenvectors of T with eigenvalue -1, is perfectly fine for transformations. There's no need to change everything to matrices. But you can change to matrices if it's easier for you...
 
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