Diagnonalization

Let $A$ be some matrix we are trying to diagonalize. Then $P$ is the matrix formed by compiling a matrix where the columns are $v_1,v_2,\dots ,v_n$ which are the vectors of $A$, and $D$ is the matrix where its diagonals are the eigenvalues of $A$. Then…

$PA{P}^{-1}=D$ Here, $A$ and $D$ are similar matrices. They have the same:

• determinant
• eigenvalues
• rank
• sum of diagonal entries (trace)

Not all matrices are diagonalizable. An $n\text{ x }n$ matrix is diagonalizable iff there exists a basis for ${\mathbb{R}}^n$ of eigenvectors of A. From this, the following theorem can be deduced.

Note

A matrix can be diagonalized iff every eigenvalue of A has a geometric multiplicity equal to its algebraic multiplicity.

That is to say, if an $n\text{ x }n$ matrix has $n$ distinct eigenvalues.