I understand that two matrices $A$ and $B$ (in $\mathbb{R}^{n \times n})$ are similar matrices if there exists an invertible matrix $P \in \mathbb{R}^{n \times n}$ (a change of basis matrix) such that $B = P^{-1}AP$.
My question is whether;
- does a matrix $P$ always exists for any choice of $A$ and $B$;
- and as such doest this imply that the $A$ is always diagonalizable;
- is $B$ always a diagonal matrix;
- in which cases does $P$ not exists?
I would really appreciate detailed answer to help me structure my understanding.
Many thanks
No. For example, if $A$ is the zero matrix (all entries are zero), then $PAP^{-1} = 0$ for any $P$. So for any non-zero choice of $B$, there cannot be a $P$ which satisfies this equation. There are less trivial examples, but this gets the point across.
This is (sort of) answered by question 1. Since not every two matrices are conjugate, it is not necessarily true that all matrices are diagonalizable. For example, $$ \begin{pmatrix} 1&1\\0&1 \end{pmatrix} $$ is not diagonalizable (over $\Bbb{R}$). Its characteristic polynomial is $(1-\lambda)^2$, and so its only eigenvalue is $\lambda=1$. But you can check that it does not have two linearly independent eigenvectors (the only eigen-direction is $(1,0)$). If it were diagonalizable, it would need two linearly independent eigenvectors with eigenvalue $1$.
No. The equation $B = PAP^{-1}$ is often used when talking about diagonalization, but the equation just by itself is just saying two matrices are conjugate.
This was answered in question #1.