Is induction needed if a process terminates after $\lt \infty$ steps? Is induction needed if a process terminates after finite steps?
I am asking in general. I will provide a specific example to give something specific to talk about.  But I wonder about this all the time, so an answer in general (that merely references my example) would be preferred to an answer that just addresses my example. Thanks so much!
My Example. If $U$ and $T$ are commuting operators on a finite dimensional space and the minimal polynomials of $U$ and $T$ each split completely into linear factors, then, for any $W$ a proper subspace of $V$ that is invariant under both operators, I can show that there exists a vector $\alpha$ such that $\alpha \notin W$ but $T\alpha \in W$ and $U\alpha \in W$. Now I want to use that fact to construct an algorithm that simultaneously triangularizes $U$ and $T$. Start with $W_1 = \{0\}$ and obtain $\alpha_1$, an eigenvector of both $U$ and $T$. Thus $W_2 = \text{span }\alpha_1$ is invariant under $U$ and $T$. So we obtain $\alpha_2$. I can show that $W_3 = \text{span }\alpha_2 + W_1$ is invariant under $U$ and $T$. Using the same steps to show that $W_2$ was invariant, I can show that $W_3$ is invariant, so continue the process.
So, how to conclude the proof from here? Can I claim, "Continue this process. Because it terminates after a finite number of steps, we are done." If so, what features of the above proof allow me to get away with not setting up a full induction? If not, why not? Thanks again!
 A: Yes, induction is needed here.
You can get away without induction if there is a fixed number of steps given to you in the statement you're trying to prove. For example, if $U$ and $T$ are operators on a $10$-dimensional space, you could write down a proof in $10$ steps: one that finds $\alpha_1$, one that finds $\alpha_2$, one that finds $\alpha_3$, and so on through $\alpha_{10}$.
Similarly, if the space were $100$-dimensional, you could write down a very long proof with $100$ very similar steps, and you wouldn't need to use induction.
A proof by induction is just the idea that if all these steps really are identical, except with the index increasing each time, then it's enough to explain what the $i^{\text{th}}$ step looks like, and then you have explained everything that's needed to write a proof for any number of dimensions. This is always what a proof by induction is.
You can write this up in different ways:

*

*Maybe you write "Assume we have constructed $W_1, W_2, \dots, W_i$ invariant under $U$ and $T$. Then to construct $W_{i+1}$, we do such and such. By induction, we can keep going until we get to $W_n$, where $n$ is the dimension of our space."

*Maybe you write "Continue this process. Because it terminates after a finite number of steps, we are done."

These are not different in the underlying math that's happening! The difference is that the first one explains very carefully how the induction works; the second one leaves some of that for the reader to infer. But that's just a matter of proof style, and of knowing your audience so that you know how much detail you can safely omit.
