But suppose we are still guaranteed that $\big \vert \lambda_1\big \vert \gt \big \vert \lambda_2\big \vert \geq \ldots \geq \big \vert \lambda_n\big \vert$ where the $\lambda_i$'s are all the complex eigenvalues of $P$, then can we say that the given irreducible chain is aperiodic
Yes. This is a basic spectral result, or if you prefer a basic Perron Frobenius theory result. Since we know $\lambda_1 = 1$, i.e. the Perron root, suppose you look at your transition matrix $P$ and 'remove' the perron vectors, i.e. consider the matrix $B := P- \mathbf {1\pi}^{\top}$. The matrix $B$ is what determines the difference between the distribution in your chain and the stationary distribution.
$B$ has maximal modulus eigenvalue $\big \vert \lambda_2 \big \vert \lt 1$. So $\lim_{k \to \infty} B^k \to \mathbf 0$ and $\lim_{k \to \infty} P^k =\mathbf {1\pi}^{\top}$. This limit cannot exist if $P$ is periodic. So $P$ is aperiodic. Finite state chains always have at least one positive recurrent class, with an associated stationary distribution; so the non-existence of a limit is equivalent to a given chain having periodicity.
Equivalently a finite state (time homogenous) markov chain with a single communicating class is periodic iff it only has one eigenvalue on the unit circle.
Addendum
here is a sketch of a different route to the above result, one which relies on some probabilistic and analytic results and very little linear algebra. Everything below assumes there is a single communicating class.
1.) Prove that for any Markov chain with at least one non-zero entry on the diagonal of the n x n transition matrix $A$, you have $A^k \gt 0$ for all $k \geq K$ -- where this inequality is interpreted component-wise -- i.e. all components of $A^k$ are strictly positive. You can tighten the bound but an easy proof would be to select say $K = 5 n^2$ -- all that matters is K is some constant that depends only on $n$.
2.) Since looking at the diagonal gives useful information, consider that
$\text{trace}\big(P^r\big)$
$ = \lambda_1^r + \lambda_2^r + ... +\lambda_n^r $
$= 1 + \sum_{j=2}^n \lambda_j^r $
$= \big \vert 1 + \sum_{j=2}^n \lambda_j^r \big \vert$
$\geq \Big \vert \big \vert 1\big \vert -\big \vert \sum_{j=2}^n \lambda_j^r \big \vert \Big \vert $
$\geq 1 -\big \vert \sum_{j=2}^n \lambda_j^r \big \vert $
$\geq 1- \sum_{j=2}^n \big \vert\lambda_j \big \vert^r $
$\geq 1- (n-1)\cdot \big \vert\lambda_2\big \vert^r$
by application of triangle inequality
and for large enough $r$ we have
$\text{trace}\big(P^r\big) \geq 1- (n-1)\cdot \big \vert\lambda_2\big \vert^r \gt 0$,
i.e. $P^r$ has at least one positive component on its diagonal for all $r\geq R$.
In combination with (1) this implies that $P^m \gt 0$ for all $m \geq M$.
3.) Since for large enough powers of $P$, the transition matrix is strictly positive, we can show that $P^m \to \mathbf {1\pi}^{\top}$. For several different elementary takes consider section 11.4 Fundamental Limit Theorem for Regular Chains in the free book by Grinstead and Snell
-- in particular (i) the main proof that starts the section, (ii) the Doeblin Proof, and (iii) the exercise 9 suggested by Doyle. All three are quite nice.
https://math.dartmouth.edu/~prob/prob/prob.pdf
Some clues: This Markov chain is periodic of period 2. Cyclically alternating subclasses are $C_1=\{1,3\}$ and $C_2=\{2,4\}.$ So the chain is not ergodic.
[if the initial vector, step 1. is $\nu = (1/2,\, 0,\, 1/2,\, 0),$ then it
will visit class $C_1$ only at odd numbered steps.]
However, it is doubly stochastic (columns sum to unity), so the uniform distribution on the four states is a stationary (steady state) distribution. You can easily verify this using $\pi = (1/4,\, 1/4,\, 1/4,\, 1/4)$ in $\pi\mathbf{P}= \pi.$ Can you find others?
The terminology used to discuss Markov chains is not exactly standard from one text to another. I hope I have used terminology you can understand.
Best Answer
Maybe a good starting point would be to look at a simple example. The eigenvalues of the transition matrix $\begin{bmatrix} \frac{1}{2} & 1 \\ \frac{1}{2} & 0\end{bmatrix}$ are:
Of course, the second one cannot be a probability distribution no matter how we renormalize it, because it has negative entries!