Are the eigenvectors calculated by Mathematica miscalculated?

Questions
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1
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The eigenbase matrix (similar matrix) calculated by Mathematica is not a diagonal matrix.
The eigenvectors of the K matrix are {{1, 5}, {-1, 2}} calculated by me.
The same is true for the b matrix, and the ideal eigenbase can be obtained only by using the w matrix. It is a diagonal matrix.

2

I don’t know what you mean, but for Matrix-Products you should use Dot : a.b

a = {{0, 1}, {1, 1}}
MatrixForm[a]
a.a   (*   your a^2 or a*a   * )
v = {{2, 2}, {1 + Sqrt[5], 1 - Sqrt[5]}}
Inverse[v].a.v
(Inverse[v].a.v). (Inverse[v].a.v)
v. (Inverse[v].a.v).(Inverse[v].a.v). Inverse[v]//FullSimplify

I hope this helps

3

Take care of the notation. The Matrix of the eigenvectors is a list of vectors, which must be transposed to get the known results:

eva = Eigenvectors[a] // Transpose
Inverse[eva].a.eva // FullSimplify // MatrixForm

K = {{4, -5}, {2, -3}}
evK = Eigenvectors[K] // Transpose
Inverse[evK].K.evK // FullSimplify // MatrixForm
4

In the change-of-basis matrix, the eigenvectors go in the columns:

kk = {{4, -5}, {2, -3}};
Inverse[Transpose@Eigenvectors[kk]] . kk . Transpose@Eigenvectors[kk]
(*  {{2, 0}, {0, -1}}  *)