Is there an isomorphism between projections?

Question

Let $V$ a finite dimension vector space and let $T$ be a linear operator such that $T^2 = T$. Suppose there's another linear operator $R$ such that $R^2 = R$. Is there an isomorphism $\rho$ such that $R= \rho^{-1} \circ T \circ \rho$ ?

I was thinking the answer is yes, due to the fact that $T$ and $R$ are diagonalizable, and so, I can find $\beta \text{ and } \beta^{\prime}$ basis of $V$ such that $A = [T]_{\beta}$ and $B=[R]_{\beta^{\prime}}$. Then I can find a matrix $Q$ such that $Q$ is a matrix change coordinates from $\beta$ to $\beta^{\prime}$, and then $$B = Q^{-1}AQ$$

I don't know if my reasoning is correct and, regardless of being right or wrong, I'm wondering if there's another proof, a more formal proof, for this.

Answer 1

This is not true is general as consider two transformation $$A=\begin{pmatrix}1 &0\\ 0&1\end{pmatrix}$$ and $$A=\begin{pmatrix}1 &0\\ 0&0\end{pmatrix}$$

They are similar if and only if they have same rank!!

Proof: Two idempotent matrices are similar iff they have the same rank

Answer 2

It's not true. For a counterexample, look at the matrices

$T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \tag 1$

and

$R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}; \tag 2$

Both

$R^2 = R, \; T^2 = T \tag 3$

hold, but $R$ and $T$ are not similar; their characteristic polynomials are different; we have

$p_T(x) = (x - 1)x^2 = x^3 - x^2, \tag 4$

but

$p_R(x) = x(x - 1)^2 = x^3 - 2x^2 + x. \tag t$