Defining real powor of linear operator Let $T$ be invertible linear operator in finite-dimensional vector spaces $V$. How to define $T^a$ for real $a$ such that $T^a T^b = T^{a+b}$ for every $a, b \in \mathbb{R}$?
 A: There is one approach you can use. Write
$$
T^a = (T-I + I)^a = \sum_{n \ge 0} \binom an (T-I)^n.
$$
Note that we are only taking positive integral powers of $T$ in this expression, which means they are all well-defined ; although the series might not be convergent! (Convergent meaning each component of the matrices is convergent.) Suppose you have two exponents $a$ and $b$ such that this is convergent. Then
$$
T^aT^b = \left( \sum_{n \ge 0} \binom an (T-I)^n \right) \left( \sum_{m \ge 0} \binom bm (T-I)^m \right) = \sum_{n,m \ge 0} \binom an \binom bm (T-I)^{n+m}
$$
which you can re-arrange as
$$
\sum_{k \ge 0} \left( \sum_{j=0}^k \binom aj \binom b{k-j} \right) (T-I)^k = \sum_{k \ge 0} \binom{a+b}k (T-I)^k = T^{a+b}.
$$
(To see that last line, just expand the formal power series $(1+x)^{a+b}$ in two different ways, namely either use the binomial theorem directly on $(1+x)^{a+b}$ or use it twice on $(1+x)^a(1+x)^b$, then compare coefficients.)
To determine convergence, since you work with finite dimensional vector spaces, all norms will be equivalent ; just choose a norm and compute
$$
\lim_{n \to \infty} \frac{\left\| \binom a{n+1} (T-I)^{n+1} \right\|}{\left\| \binom an (T-I)^n \right\|}.
$$
If this limit goes smaller than $1$, $T^a$ is defined ; greater than $1$, it won't, $=1$, no idea what happens, but you should expect issues. It's really just a variant of d'Alembert's test. 
Hope that helps,
A: I was wrong because my example is not invertible.  So this is how to do it.  By the Jordon decomposition, we can reduce the problem to $T = \lambda I + N$ where $\lambda \ne 0$ and $N$ is nilpotent.  So $T^a = \lambda^a (I + \lambda^{-1} N)^a$, and to compute $\lambda^a$ you may have to use complex numbers.
Now you can compute $(I + \lambda^{-1} N)^a$ using Taylor's series.  Now I know it is an infinite series if $a$ is not a positive integer.  But because $N$ is nilpotent, only finitely many of the terms in the infinite series will be non-zero.  And so convergence will not be a problem.
There is also some big machinery called the Dunford-Schwartz functional calculus that will allow you to do this, where $T$ can be any bounded operator on a Banach space whose spectrum does not include zero, and such that the connected component of zero in the resolvent set is the same as the unbounded component.  And since the spectrum of a finite dimensional operator is finite, if $T$ is invertible, this will always work.
A: This is impossible. Consider $V=\mathbb R$ and $Tx=-x$. A linear $T^{\frac 1 2}$ would satisfy $T^{\frac 1 2}T^{\frac 1 2} x = -x$. All linear operators on $\mathbb R$ are of the form $Ax = cx$ for some $c\in\mathbb R$. Thus, assuming $T^{\frac 1 2} x = cx$, we get $c^2 x=-x$. For $x=1$ this gives $c^2=-1$ which has no solution in $\mathbb R$.
