# Isomorphic transformation question

1. Jun 21, 2011

### nhrock3

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 couldnt 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 dont have any matrix here to do

|A-labdaI|=0

i dont have any matrices only some transformation

which i dont 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

?

2. Jun 21, 2011

### micromass

Hi nhrock3!

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?

3. Jun 21, 2011

### nhrock3

i dont 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 cant 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: Jun 21, 2011
4. Jun 21, 2011

### nhrock3

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)

5. Jun 21, 2011

### micromass

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...