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First, let us look at some theory regarding diagonalization. A linear transformation on an n-dimensional vector space (or equivalently, an n-by-n matrix), is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to n. Since matrices and linear transformation are equivalent, let us concentrate on an n-by-n matrix, say A. We can assume that it is the matrix representation of some linear transformation (on an n-dimensional vector space, say V) with respect to some basis.
Now, suppose λ1, λ;2, ..., λk are distinct eigenvalues of A. Note that k need not be equal to n, as some eigenvalues may be repeated in the set of all eigenvalues, and therefore, when considering only distinct eigenvalues, the number may be less than n. There is an eigenspace corresponding to each eigenvalue, and each of these eigenspaces can be represented by a set of (one or more) linearly independent vectors. These are the eigenvectors. If x is an eigenvector corresponding to the eigenvalue λ, then:
Ax = ...
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