A diagonalizable matrix $A$ can be written as
$$ A=P \Delta P^{-1} $$where $\Delta$ is a diagonal matrix (the eigenvalues at the diagonal) and $P$ is a matrix which columns are the eigenvectors.
The idea is that we have a basis of the vector space in which the transformation given by $A$ is simply made of scale changes (even negative or null) in the main axes. When the basis change is by mean of an orthogonal matrix then the matrix $A$ is symmetric. That is, we have
It is related to singular value decomposition. Indeed they are equal when the matrix is symmetric and positive-semidefinite. See MSE.
1. Solve $det(A-\lambda I)=0$. The solutions $\lambda_i$ are called eigenvalues.
2. For every $\lambda_i$ we look for a basis of the subspace $V_{\lambda_i}$
$$ (A-\lambda_i I)\begin{pmatrix} x_1\\ x_2\\ \vdots\\ \end{pmatrix}= \begin{pmatrix} 0\\ 0\\ \vdots\\ \end{pmatrix} $$They are the eigenvectors associated to the eigenvalue $\lambda_i$.
3. If we have enough eigenvectors to complete a basis of $\mathbb R^N$ then the matrix $A$ is diagonalizable, and $\Lambda$ is made with the eigenvalues (repeated if necessary, according to the dimension of $V_{\lambda_i}$). The $P$ matrix is made with the eigenvectors. If we don't have enough eigenvectors, the matrix is not diagonalizable and we look for the Jordan canonical form of a matrix. The reasons could be:
a. There are complex solutions in step 1.
b. Any $V_{\lambda_i}$ has not dimension enough "to fill" the multiplicity of $\lambda_i$ in step 1.
If a matrix is diagonalized, its diagonal form is unique, up to a permutation of the diagonal entries. This is because the entries on the diagonal must be all the eigenvalues. For instance,
$$ \left[\begin{matrix} 1 & 0 & 0\\ 0 & 2 & 0 \\ 0 & 0 & 1 \end {matrix}\right] \text { and }\left[\begin{matrix} 1 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 2 \end {matrix}\right] $$are examples of two different ways to diagonalize the same matrix.
Important fact: common eigenvectors.
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Author of the notes: Antonio J. Pan-Collantes
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