$$A =PDP^{-1}$$

I'm kind of new to the topic, we find $P$ which is columns representing a basis for $A$. But step by step, what does $PD$, then $PDP^{-1}$ say about the process? Are we changing the basis, why $D$ becomes an eigenvalued diagonal matrix-> how does it work?

  • 3
    $\begingroup$ I downvoted this question because it never bothers to say what A, P, or D are, or under what circumstances the given equation is supposed to hold. (Being "kind of new" to a topic is no justification for this.) $\endgroup$
    – Dan Asimov
    Jun 4 at 13:04
  • 1
    $\begingroup$ Multiplying by $P^{-1}$ is a change of basis operation. We change basis to the basis consisting of the columns of $P$. So, this equation tells us that multiplying by $A$ is equivalent to first changing basis, multiplying by a diagonal matrix (simple), then changing back to the original basis. $\endgroup$
    – littleO
    Jun 4 at 13:19

1 Answer 1


Lets say that $\mathbf{e}_1$,$\mathbf{e}_2$ (in the simplest case) are eigenvectors (not basis vectors) for $A$, so $A\mathbf{e}_1= \lambda_1\mathbf{e}_1$ and $A\mathbf{e}_2= \lambda_2\mathbf{e}_2$. Because $P$ has columns which are these eigenvectors, it maps $\mathbf{i}$ to $\mathbf{e}_1$ and $\mathbf{j}$ to $\mathbf{e}_2$ (where $\mathbf{i}$ and $\mathbf{j}$ are the standard basis of $\mathbb{R}^2$). Then $P^{-1}$ must map $\mathbf{e}_1$ to $\mathbf{i}$ and $\mathbf{e}_2$ to $\mathbf{j}$.

Then the composition $PDP^{-1}$ maps $\mathbf{e}_1$ to $\mathbf{i}$ to $\lambda_1 \mathbf{i}$ to $\lambda_1\mathbf{e}_1$, and similarly $\mathbf{e}_2$ to $\lambda_2\mathbf{e}_2$: it maps the two eigenvectors of $A$ to the same things as $A$ does - so it $\it is \rm$ $A$, provided the eigenvectors of $A$ form a basis of $\mathbb{R}^2$ (if this isn't the case, you can't diagonalise $A$).


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