# Why do we need a Matrix of Eigenvectors in Diagonalization?

I know that if we have a matrix A defined for a space V with dimension n that has n distinct eigenvalues, we can write it as $$X^{-1}DX$$ where D is a diagonal matrix with the eigenvalues of A on its diagonal and X is a matrix of eigenvectors.

It seems that what we are doing is using the fact that we have n distinct eigenvalues to find a basis of eigenvectors of our space. In this basis, the linear transformation which corresponds to A simply rescales each eigenvector so A can be written as a diagonal matrix in this basis.

I understand that we need the matrix X to transform a vector defined in a basis $$f_1,...,f_n$$that was used to define A to the eigenbasis $$e_1,...e_n$$. However, I do not understand why we have to use X as the linear map that does this. Why can we not just define a map $$L:V \to V$$ where $$L(f_i) = e_i$$ $$\forall 1\leq i \leq n$$ and then X is just the identity matrix.

Your question makes little sense: it is not clear what you mean when you say that under your choice of $$L$$, “$$X$$ is just the identity matrix.” But perhaps these comments will clarify the situation.

Diagonalization is really about linear operators, not about the matrices that represent them. Say you have a finite-dimensional space $$V$$ of dimension $$n$$ and an operator $$L:V\to V$$. The question is: does there exist a basis of $$V$$ so that the matrix that represents $$L$$ with respect to that basis is diagonal? The answer (not hard to show, just unraveling what a diagonal representation means) is: such a basis exists if and only if $$L$$ has $$n$$ linearly independent eigenvectors. Such a basis is called an eigenbasis.

Now, let’s translate these basis-free ideas into basis-dependent matrices. Say $$L_e$$ be the matrix that represents $$L$$ with respect to the standard basis $$e$$ (or really any old basis you wish), and let $$L_b$$ be the matrix that represents $$L$$ with respect to the eigenbasis $$b$$. Then $$L_b$$ is diagonal. The question is, how are these two matrices $$L_e$$ and $$L_b$$ related?

The answer is that they are similar, of course; they represent the same transformation with respect to different bases. Say that $$X$$ is the matrix that represents the transformation that sends the basis $$e$$ to the eigenbasis $$b$$, with respect to the basis $$e$$. Then elementary change of basis calculations give the formula

$$L_e=XL_bX^{-1}$$

• What I meant was that it appears as though we are just writing $L_e$ as a composition of $L_b$ and a pair of linear maps that change the basis of the vector we would be inputting into $L_e$ to the basis $b$. I was wondering why we couldn't just choose a map $X$ that does it and has an easy matrix i.e. $X(e_i) = b_i$. Why do we have to use the matrix of eigenvectors specifically? – user446153 Jun 11 at 22:17
• The map that has $X(e_i)=b_i$ is the matrix of eigenvectors, when you represent the map $X$ with respect to the basis $e$. So your question makes no sense. – symplectomorphic Jun 12 at 3:18

Because in that basis, the algebra of matrices becomes really simple and immediate and you can get lot of information just by looking at the matrix itself. Also, the representative matrix of $$L$$ is exactly $$X^{-1}$$.

At the end you will have like: $$A=X^{-1}ID I^{-1} X$$