Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a linear map $f: \mathbb{R^{4}} \rightarrow \mathbb{R^{4}} $ and a base vector a, with (f: a ,a ) matrix to be diagonal?

I've found this theorem:

Let L be the linear transformation

L(v) = Av

then A is diagonalizable with n linearly independent eigenvectors S = {v1, ... >,vn} if and only if the matrix of L with respect to S is diagonal.

Do you know if it is true the other way around?
What I mean is, if A is diagonalizable with n linearly independent eigenvectors $S = {v_1, ... ,v_n}$ then the matrix of L is diagonal. Is that true?

I've found this:

If n characteristic vectors correspond with all different eigenvalues, then these characteristic vectors are linear independent. The characteristic vectors can be used as a basis of V. The matrix of t relative to that basis is diagonal and the eigenvalues are the diagonal elements of the matrix.

Link Here!

Which states that if all the eigenvectors of a matrix A are linearly independent then the matrix of the linear map, is diagonal.

So if we have a matrix A that is diagonalizable, then the linear map, would be f(x) = A*x ?

Thank you for your time!

share|improve this question
    
In what you quoted, $L$ is a linear transformation, not a matrix. $A$ is a matrix, which is probably not diagonal. If there is a basis $S$ consisting of eigenvectors, then $L$ is diagonalizable, which means the matrix of $L$ in that basis is diagonal. It doesn't make sense to say a linear transformation is diagonal, only a matrix. –  GEdgar May 18 '12 at 13:07
    
@GEdgar: It says "A is diagonalizable", and I've said "then the matrix of L is diagonal", not the linear transformation. –  Chris May 18 '12 at 13:10

2 Answers 2

up vote 1 down vote accepted

What you write is essentially true, however you mix up some of the notation.

Given a fixed basis of $\mathbb R^4$, there is a one-to-one correspondence between endomorphisms $\mathbb R^4\to \mathbb R^4$ and $4\times 4$ matrices with entries in $\mathbb R$.

(An "endomorphism" is linear map from a vector space to itself, just in case you didn't know)

Even though one might think of them as being the same thing, they aren't. They are just one-to-one which is somewhat as close to equal as you can get, but still they are not the same thing. In particular you have to fix a basis first and the same endomorphism might have plenty of different matrix representations, always with respect to a different basis.

Hence you can not say that a linear transformation is "diagonal", only a matrix can be diagonal. Moreover even if a matrix is diagonalisable this doesn't necessarily mean that it is diagonal with respect to any basis but just wrt to some particular choices of basis.

However the following are true:

An $n\times n$ matrix $A$ representing the linear transformation $L$ is diagonalisable if and only if there exist $n$ linear independent eigenvectors of the transformation $L$. Moreover the matrix representation of $L$ with respect to this basis is diagonal.

If a linear transformation $L:\mathbb R^n\to \mathbb R^n$ has $n$ different eigenvalues $\lambda_i$ with corresponding eigenvectors $v_i$, then the $v_i$ are linearly independent.

share|improve this answer
    
I think that's ("In particular you have to fix a basis first and the same endomorphism might have plenty of different matrix representations, always with respect to a different basis.") what I did after all! I found some of our teacher, that proved to be exactly what I wanted (and to think I have searched through the notes first :-/) Thank you for your help! –  Chris May 18 '12 at 20:10

First, the wording of those statements in the link you posted looks rather poor to me and, if I may say, even misleading.

Second, I couldn't find anywhere in the link you posted the slightest clue about whose page is that (mathematician, student, a fan...?), it doesn't seem to be linked to some university or stuff.

Now, what you wrote here "Which states that if all the eigenvectors of a matrix A are linearly independent then the matrix of the linear map, is diagonal." is*false*. What is true is: a matrix A is diagonalizable iff there exists a basis (for the space vector we're working with) of eigenvectors(=characteristic vectors) of A.

It also seems to me that your notation (f;a,a) is non-standard and I, for one, have no idea what it means.

Summary: a linear operator $\,L\,$ is said to be diagonalizable iff there exists a basis of eigenvectors of the operator, and this does NOT mean necessarily that if we express the operator in matrix form $\,Lv=Av\,$ (when the RHS must be understood as the action of the matrix A on the corresponding column vector), then A is diagonalizable.

Your last question "So if we have a matrix A that is diagonalizable, then the linear map, would be f(x) = A*x ?" seems to be lacking something. At least I can't understand it.

share|improve this answer
    
I don't know whose is it, I just wanted some information. :/ The notation (f:**a**,a) (a with bold means a vector) and in general it represents a matrix. The matrix of the linear transformation! What does "RHS" mean? As for the last question, sorry, it's something wrong! –  Chris May 18 '12 at 20:02
    
RHS = Right hand side , and LHS is...well, you know. –  DonAntonio May 18 '12 at 22:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.