If A is an Idempotent matrix then, is $A^r\forall r\in \Bbb R$ also an idempotent matrix? Question1: If A is an Idempotent matrix then, is $A^r\forall r\in \Bbb R$ also an idempotent matrix?
The purpose of asking this question is actually regarding the following question .
Question 2: Prove that there exist no continuous function such that- $$f:[0,1]\to\{A\in\Bbb M_{2\times 2}(\Bbb R)| A^²=A\}$$ and $f(0)=0,f(1)=I_{2\times2}$ .
My Attempt: (Although I have no idea of what is meant by continuity for such matrix related functions?  I tried this way )
Since , $A^2=A\implies f(x)=A=A^½=A^¼=A^⅛=... =A^\frac{1}{2n}\forall n\in \Bbb N$ . Hence , $f(x)=A=\lim_{n\to \infty} A^\frac{1}{2n}= A^0=I_{2\times2} $. So $f(x)=I_{2\times2}\forall x\in [0,1]$ . Hence $f(0)\neq 0$ . So such function mentioned in Question 2 doesn't exist.
But later I realized can I really write the statement that is ..$``A^2=A\implies f(x)=A=A^½=A^¼=A^⅛=... =A^\frac{1}{2n}\forall n\in \Bbb N"$ . Now it is possible if $\text{A is idempotent} \implies A^½ $is also Idempotent.  That's why and to make it generalized from ½ to any real number r I have asked the question ,if A is an Idempotent matrix then, is $A^r\forall r\in \Bbb R$ also an idempotent matrix?
If my approach is wrong to the Question 2 please suggest an approach.
 A: Defining $A^r$ for any real $r$ is going to be a problem here, and it is also unnecessary.
Note that, for any idempotent matrix $A$ we have $\det(A)=\det(A^2)=\det(A)^2$ so $\det(A)=0$ or $\det(A)=1$. Also, with a reasonable definition of continuity, the map $A\mapsto\det(A)$ is continuous.
This means, if there was a continuous map $f:[0,1]\to M_2(\mathbb R)$ which only takes idempotent matrices as values, and maps $0$ to $0$ and $1$ to $I$, you could compose it with the map $\det$ and you would get a continuous map
$$\det\circ f: [0,1]\to\{0, 1\}$$
which additionally maps $0$ to $\det 0=0$ and $1$ to $\det I=1$. This is impossible due to $[0,1]$ being connected.
A: I'd suggest using the trace, which is a continuous function of the matrix's entries.  This is necessarily a natural number because it gives the rank of an idempotent matrix (I give a very simple proof in the post-script here:  If $PP^*=P^*P$ and $P^2=P$ then show that $P^*=P$)
Thus compose  $\text{trace}\circ f$ and you have have continuous function that is integer valued, hence constant.
Thus $0=\text{rank}\big(f(0\big) = \text{rank}\big(f(1)\big)=n$
which is a contradiction.
This is a finer viewpoint than using the determinant since it also allows you to consider path connectivity claims with respect to singular idempotent matrices.
A: A third answer in the same vein:
Let $\|\cdot\|$ by any submultiplicative norm on $M_{2\times 2}(\Bbb{R})$. Then if $A^2 = A \ne 0$, we have
$$\|A\| = \|A^2\| \le \|A\|^2 \implies \|A\| \ge 1.$$
Therefore if such a function $f$ would exist, the composition $$\|f(\cdot)\| : [0,1] \to \{0\} \cup [1,+\infty\rangle$$ would be a continuous function such that $\|f(0)\| = 0$ and $\|f(1)\| \ge 1$ which contradicts connectedness of $[0,1]$.
