On using $A = CDC^{-1}$ and $A=C^{-1}DC$ for diagonalizing a matrix and finding $D$

Question

I am studying matrix diagonalization and I have a few doubts. We know that a matrix $A$ is diagonalizable if there exists an invertible matrix $C$ and a diagonal matrix $D$ such that $A = CDC^{-1}$ is diagonal. Does $A=C^{-1}DC$ also hold true? I've done a few calculations and it seems like so but would like confirmation.

To find $D$, I first find $C$ with the diagonalization method. Then, $D=CAC^{-1}$. After I find $D$, then: $A=CDC^{-1}$. Am I missing something? Any hints would be appreciated.

Answer 1

You have a linear transformation $T$. In one basis (which I will call $a$) this linear transformation is represented by the matrix $A$. In a different basis (which I will call $d$), the same linear transformation is represented by $D$.

The matrix $C$ takes the components of any vector as expressed in one of these bases and translates it to the other base. Sandwitching a matrix (representing some linear transformation) between $C$ and $C^{-1}$ will take the components of that matrix and translate that to the matrix that represents the same linear transformation in the other base.

When you write $A = CDC^{-1}$, then that means that you have chosen $C$ to be the matrix that takes in the components of vectors expressed in $d$ and spits out the components of vectors expressed in $a$. When you write $A = C^{-1}DC$, then that means that you have chosen $C$ to be matrix that takes in vectors expressed in $a$ and spits out vectors expressed in $d$. These are different $C$'s and should never be used in the same context the way I have done in this paragraph. You have to make a single choice and then stick with it.

Usually, when diagonalizing, it is most immediate and therefore most convenient to choose $C$ so that its columns are the eigenvectors of $T$, expressed in the basis $a$ (the so-called "eigenvectors of $A$"). This means that we happen to go with the first option above. This means that $A = CDC^{-1}$. Note that this also means $$ A = CDC^{-1}\\ AC = CDC^{-1}C\\ AC = CD\\ C^{-1}AC = C^{-1}CD = D $$ so that we have $D = C^{-1}AC$, not the other way around.

Answer 2

Absolutely not: the identity $C^2DC^{-2}=D$ for a generic diagonal matrix $D$ and a generic invertible matrix $C$ is not a thing.

Answer 3

If you take $A = \begin{pmatrix}0&1\\-1&0\end{pmatrix}$, then $C=\frac{\sqrt{2}}{2}\begin{pmatrix}1&1\\\mathrm{i}&\mathrm{-i}\end{pmatrix}$ and

$$C^{-1}DC = \begin{pmatrix}0&\mathrm{i}\\\mathrm{i}&0\end{pmatrix} \neq CDC^{-1} = \begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$

Answer 4

Say you have $D=CAC^{-1}$, now how can you find $A$?

Multiply with $C$ to the right: $$DC = CAC^{-1}C\\[0.2cm] DC = CA$$

and $C^{-1}$ to the left:

$$C^{-1}DC = C^{-1}CA\\[0.2cm]\\C^{-1}DC = A$$

You seem confused that since you can choose whether you start with

$$D = CAC^{-1}$$ or $$D = C^{-1}AC$$

it does not mean that you can switch between these once you have chosen one of them.

Answer 5

Let $A=\begin{pmatrix} 0 & i \\-i & 0\end{pmatrix}$ its eigenvalues are $\mp 1$ so $$D=\begin{pmatrix} -1 & 0 \\ 0 & 1 \end {pmatrix}$$ and the doiagonalizinfg matrix is $C=\begin{pmatrix} -i & i \\ 1 & 1 \end{pmatrix}$, then $$C^{-1} D C=\begin{pmatrix} 0 & 1 \\ 1 & 0\\\end{pmatrix}$$ and $$CDC^{-1}=\begin{pmatrix} 0 & i \\ -i & 0 \end{pmatrix}$$