# Consider a linear map and a base vector a, is the matrix of the linear map diagonal?

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.

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 ?

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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

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.

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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.

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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