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I am following Gilbert Strang's Linear Algebra course (18.06, link). In lecture 30 on "Linear transformations and their matrices", he mentions that the eigenvector basis leads to a diagonal transformation matrix $\Lambda$, using the example of a projection matrix.

I am not able to understand this statement. To determine the transformation matrix $A$, we need to determine first what the input and output bases are. But if we have already chosen the bases, then how can we choose the bases that are the eigenvectors of $A$? I am thinking of it like the chicken and the egg problem.

Suppose I want a diagonal transformation matrix $\Lambda$. For determining $\Lambda$, we need the bases, but for determining the bases, we need the matrix $\Lambda$ because we have to compute its eigenvectors.

Can someone please help me to understand what I am not understanding correctly here?

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  • $\begingroup$ The existence of a basis doesn't preclude the existence of another. A linear transformation depends on a vector space or two; it is not bound to a specific basis, and thus can be represented by (different) matrices with respect to different bases. $\endgroup$ Jun 27, 2021 at 13:45
  • $\begingroup$ @darijgrinberg I understand that, but if I want to choose "good" bases such that the transformation matrix I get is diagonal, then how do I do it? I don't know the matrix for which I need the eigenvectors $\endgroup$
    – temp_user
    Jun 27, 2021 at 13:49
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    $\begingroup$ The eigenvectors, too, are defined in terms of the linear transformation alone (not of the matrix). Thus, you can first choose some arbitrary basis to compute the eigenvectors, then switch to a basis made of these eigenvectors (and throw away the first basis). $\endgroup$ Jun 27, 2021 at 13:50

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Purely from the definitions, you don’t have a problem here: You have some linear transformation $\phi : V \to V$ on some vector space. If you can choose a basis $v_1, \dots, v_n$ of $V$ such that every $v_i$ is an eigenvector of $\phi$, the transformation matrix will be a diagonal matrix.

Now, you’re right that often, we like to compute a transformation matrix if we want to do computations with the linear transformation and we don’t know the eigenbasis yet. But that is not a problem: Choose any basis $w_1, \dots, w_n$ first and represent $\phi$ in that basis (use the same basis for the domain and codomain!), giving a matrix $A$. Then do your computations with $A$. The algorithm for diagonalizing the matrix will give you a change-of-basis matrix $T$ in addition to the diagonal matrix $\Lambda$. This change of basis matrix then allows you to compute a basis $v_1, \dots, v_n$ such that the transformation matrix of $\phi$ with respect to $v_1, \dots, v_n$ is the diagonal matrix $\Lambda$.

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  • $\begingroup$ So, will all transformation matrices (corresponding to different choices of input basis = output basis) have the same eigenvectors? $\endgroup$
    – temp_user
    Jun 27, 2021 at 14:09
  • $\begingroup$ They will have the same eigenvalues but not the same eigenvectors. This is because the eigenvectors of the matrix are really the coordinate vectors of the eigenvectors in the chosen basis. If you change the basis, these coordinate vectors will change. $\endgroup$ Jun 27, 2021 at 17:40

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