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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

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  • $\begingroup$ Do you mean does $P$ always exist for ANY choice of $A$ and $B$? $\endgroup$ – Nick Dec 25 '20 at 19:01
  • $\begingroup$ Yes, that's exactly what I meant. I've updated the question. Thanks $\endgroup$ – dpakasa Dec 25 '20 at 19:10
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    $\begingroup$ 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 $A,B$ $\endgroup$ – Red Phoenix Dec 25 '20 at 20:18
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  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.

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  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).
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