# What does $A = PDP^{-1}$ really represent?

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

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?

• 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.) Jun 4 at 13:04
• 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. Jun 4 at 13:19

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 $$A$$).