Linear operator to zero power

Question

When $T$ is any linear operator acting on a vector space $V$, and $n$ is a natural number, $T^n$ means $T$ applied $n$ times (composition) and that is also a linear operator. That is clear.

When $T$ is a nonzero linear operator acting on a vector space $V$, then $T^0$ is the identity operator $T^0 = I$. But I think that should also be true (true by definition), when $T$ is the zero operator i.e. the operator which sends all vectors to the zero vector.

Why? Because $T^0$ means that we are not applying any operator. So it makes sense to say: OK, all vectors stay unchanged when "applying" $T^0$ even when $T$ is the zero operator. I say "applying" because we're not actually applying anything.

Is that indeed so?

I am asking because this kind of disagrees with what we have for real numbers where $0^0$ is usually left undefined.

EDIT:
What's the context of this question? I was reading a proof for the uniqueness of the Jordan Normal Form. There this expression comes up $2d(\phi^p) - d(\phi^{p-1}) - d(\phi^{p+1})$, where $p$ is a positive integer, and $d$ is the defect of the linear operator in the brackets. The proof is very nice but convoluted and eventually it boils down to proving the uniqueness for a special linear operator which has only $0$ as a characteristic root (as an eigenvalue). So I had some doubts what happens exactly with the expression $\phi^{p-1}$ when $p = 1$, and if we need to put some restrictions on the linear operator $\phi$.

Answer 1

We define $T^0=I$ for any linear operator $T:V \to V$ so that the usual laws of exponents hold for composition. So, you should take this as a definition which is convenient and not anything extraordinarly deep. Here's why.

For a positive integer $n$ we can define $T^n$ to mean $T \circ T \circ \cdots \circ T$ ($n$ factors), and $T^n$ is again a linear operator on $V$. You can then check, thanks to associativity of function composition, that we have $T^n \circ T^m = T^{n+m}$ for positive integers $n, m$. Doesn't that look nice and familiar?

If we extend this definition to include the possibility that $n=0$ by defining $T^0=I$, then the law $T^n \circ T^m = T^{n+m}$ now holds for integers $n,m \geq 0$. After all, $T^n \circ T^0 = T^n \circ I = T^n$, so we have $T^n \circ T^0 = T^{n+0}$ and all is well.

Even better, if $T$ happens to be invertible we can define $T^{-n}$ for an integer $n \geq 0$ as either $(T^n)^{-1}$ or as $(T^{-1})^n$ as they are both the same, which is a great exercise. Then the law of exponents $$ T^n \circ T^m = T^{n+m} $$ holds for all integers $n, m$. Not only is this algebraically reminiscient of real number exponents in a nice way, this makes the applications of polynomial relations to linear operators incredibly powerful.

But, note that none of this has anything to do with whether or not $T$ is the zero transformation. In that case, $T^0=I$ is still the case by definition. (Note also that this is all about composition of functions and not any sort of real number multiplication/exponentiation, so issues related to $0^0$ in the reals are actually not relevant.)

Answer 2

$0^0$ is an indeterminate form. Consider the two limits:

$$\lim_{x \rightarrow 0}\lim_{y \rightarrow 0}\ x^y $$

$$\lim_{y \rightarrow 0}\lim_{x \rightarrow 0}\ x^y $$

For the first limit, we get $x^0$ under the limit as $x$ goes to $0$. For all $x$, except $x=0$, $x^0=1$. Thus $\lim_{x \rightarrow 0}\ x^0 =\lim_{x \rightarrow 0}\ 1=1$. However for the second limit, we have that as $x$ goes to $0$, then we get $0^y$. This is equal to $0$ for all $y$, except $y=0$. This we obtain $\lim_{y \rightarrow 0}\ 0^y= \lim_{y \rightarrow 0}\ 0 = 0$. Therefore there is a discontinuity at $(x,y)=(0,0)$. So we let $0^0$ be undefined and can only talk about $0^0$ when we specify a direction that we are computing that function from. From the x direction, from the y direction, or somewhere in between.

Let $T = a\begin{bmatrix}1,0,0\\0,1,0\\0,0,1\end{bmatrix}=\begin{bmatrix}a,0,0\\0,a,0\\0,0,a\end{bmatrix}$, for an arbitrary constant $a$. Let $\bf{x}$ be a vector in $\mathbb{R}^3$.

Consider two limits:

$$\lim_{a \rightarrow 0}\lim_{n \rightarrow 0}\ T^n {\bf x} $$

$$\lim_{n \rightarrow 0}\lim_{a \rightarrow 0}\ T^n {\bf x} $$

Clearly the first limit, the limit as $n$ goes to 0 gives $(T^0=I)\forall a$, but in the second limit, the limit as $a$ goes to zero gives $(T^n=0)\forall n$. Clearly there is a discontinuity at $(n,a)=(0,0)$.

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I am leaving my original answer intact for the time being, but my answer is correct. I want to drop this link here in case anyone doubts whether or not a matrix can be exponentiated to non-integral powers. I suppose this adds one caveat to my 'proof', that the linear transform $T$ be diagonalizable, a subset of the invertible linear transforms. The question as stated does not specify that the transforms must be invertible and is the exception.

Additional reference