# Matrix similarity and diagonalizable matrices

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;

1. does a matrix $$P$$ always exists for any choice of $$A$$ and $$B$$;
2. and as such doest this imply that the $$A$$ is always diagonalizable;
3. is $$B$$ always a diagonal matrix;
4. in which cases does $$P$$ not exists?

I would really appreciate detailed answer to help me structure my understanding.

Many thanks

• Do you mean does $P$ always exist for ANY choice of $A$ and $B$? – Nick Dec 25 '20 at 19:01
• Yes, that's exactly what I meant. I've updated the question. Thanks – dpakasa Dec 25 '20 at 19:10
• Let me give an intuition. Two matrices are similar iff they represent same linear transformation. Therefore they must satisfy some properties which are connected to that transformation. Let $A,B$ two matrices and they represent $T:V\ to W$. If there is a $x \in V$ s.t. $Tx=8x$, then $8$ must be an eigenvalue of $A,B$ – Red Phoenix Dec 25 '20 at 20:18

1. does a matrix $$P$$ always exists for any choice of $$A$$ and $$B$$?

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.

1. and as such doest this imply that the A is always diagonalizable?

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 $$1$$.

1. is B always a diagonal matrix?

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.

1. In which cases does $$P$$ not exist?

This was answered in question #1.

1. No. For instance, if $$A$$ is the identity matrix and if $$B$$ is the null matrix, no such matrix $$P$$ exists.
2. There are non-diagonalizable matrices, such as $$\left[\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right]$$.
3. No.
4. When $$A$$ and $$B$$ are not similar (by definition).