I have a linear map:
$$A = \begin{pmatrix} -1 & 3 & -1 \\ -3 & 5 & -1\\ -3 & 3 & 1 \end{pmatrix}$$
Find the basis and its matrix upon transition to which the linear map A becomes the diagonal matrix.
So i understand that i need to use formula of matrix of the transformation $TAT^{-1}$, but what kind of calculations need to be done then?
I calculated eigenvalues and eigenvectors and i got that: $\lambda_{1} = 4, \lambda_{2} = 1$
eigenvectors for $\lambda_{2} = 1$: $x_{2}=x_{3}, x_{1} = -0.4x_{3}$
You can diagonalize a matrix by finding its eigenvalues and eigenvectors.
Set $\det \begin{vmatrix} -1-\lambda & 3 & -1 \\ -3 & 5-\lambda & -1\\ -3 & 3 & 1-\lambda \end{vmatrix}=0.$
I.e., $-\lambda^3+5\lambda^2-8\lambda+4=(\lambda-2)^2(\lambda-1)=0$.
For $\lambda=2$, solutions to $ \begin{pmatrix} -3 & 3 & -1 \\ -3 & 3 & -1\\ -3 & 3 & -1 \end{pmatrix}v=0$
are linear combinations of $\pmatrix{1\\0\\-3}$ and $\pmatrix{0\\1\\3}$.
For $\lambda=1$, solutions to $ \begin{pmatrix} -2 & 3 & -1 \\ -3 & 4 & -1\\ -3 & 3 & 0 \end{pmatrix}v=0$ are multiples of $\pmatrix{1\\1\\1}.$
In a basis of eigenvectors, the matrix is $D=\pmatrix{2&0&0\\0&2&0\\0&0&1}$.
The matrix to transition coordinates from eigenvectors to standard basis has eigenvectors as columns: $T=\begin{pmatrix}1&0&1\\0&1&1\\-3&3&1\end{pmatrix}$.
You can verify that $A=TDT^{-1}$ or $D=T^{-1}AT$ .