I read the theorem on Apostol Calculus II :
Theorem $4.10$
Let $T : V \to V$ be a linear transformation, where $V$ has scalars in
$F$, and $\dim V = n$. Assume that the characteristic polynomial of
$T$ has $n$ distinct roots $\lambda_1,\dots, \lambda_n$ in $F$. Then
we have:
The corresponding eigenvectors $u_1,\dots,u_n$ form a basis for $V$.
The matrix of $T$ relative to the ordered basis $U = [u_1,\dots,u_n]$ is the diagonal matrix $\Lambda$ having the
eigenvalues as diagonal entries:$$\Lambda = \text{diag}(\lambda_1,\dots,\lambda_n) $$
- If $A$ is the matrix of $T$ relative to another basis $E = [e_1,\dots, e_n]$ then $$\Lambda = C^{-1}AC$$ where $C$ is the
nonsingular matrix relating the two bases by the equation $U = EC$.
But something went wrong when I applied it:
I took the matrix $A= \begin{bmatrix}1 & 2 \\5 & 4\end{bmatrix}$ and I tried to find the matrix $C$ in $\Lambda =C^{-1}AC$ :
First, I found the eigenvalues which are $6$ and $-1$, then I chose the eigenvectors $\begin{pmatrix}2 \\5 \end{pmatrix}$ and $\begin{pmatrix}1 \\-1 \end{pmatrix}$ which then give me the matrix $C = U =\begin{bmatrix}2 & 1 \\5 & -1\end{bmatrix}$ if I choose for the $E$ mentioned in the theorem above the basis composed by the unit coordinate vectors. Then $C^{-1} = \frac{1}{7} \begin{bmatrix}1 & 1 \\ 5 & -2\end{bmatrix}$ which indeed gives me
$$\Lambda = C^{-1} A C = \begin{bmatrix}
6 & 0 \\
0 & -1
\end{bmatrix}$$
which is the correct result.
But then I tried to use a different basis instead of the one composed by the unit coordinate vectors, I chose $E = \begin{bmatrix}1 & 0 \\1 & 1\end{bmatrix}$ and I used for $U$ the same matrix as before as I think it should be done, so $U = \begin{bmatrix}2 & 1 \\5 & -1\end{bmatrix}$, then, since $C = E^{-1}U$ and $C^{-1} = U^{-1}E$, we have that $\Lambda = U^{-1}EAE^{-1}U$.
However, calculating it on Wolfram unfortunately doesn't return a diagonal matrix.
Where did I make a mistake? Am I misapplying the theorem?
Best Answer
Hint:
The matrix $A$ represents the linear transformation $L$ (this is not a matrix) in the standard basis, and in this standard basis the linear transformation $L$ is characterized by the eigenvectors and the eigenvalues of the matrix $A$. But, if you change the basis to a new basis $E$, te same linear transformation is represented by another matrix $B=EAE^{-1}$ that have different eigenvectors (that represent the same eigenspace but in the new basis), and in this new representation of the same eigenspace, the eigenvalues are, in general, different.