I know this is a rather simple questions. I've searched for a while and from what i gather about the dominant Eiganvalue.
If i have $\lambda_1$ = 1 $\lambda_2$ = 3 and $\lambda_3$=5 (hypothetically) would 5 be the dominant Eiganvalue? And from that can i assume that given any set of eigenvalues the one with highest weight is always the Dominant?
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$\begingroup$with the highest modulus (magnitude) yes (i.e. if you had $\lambda_1 = -10$ then it would be $\lambda_1$).
This matters for the power iteration, with the rough steps below (see references for a more detailed demonstration): indeed assume that $A$ is diagonalizable so that $A = PDP^{T}$ for an orthogonal matrix $P$ and that you can write a random vector $w$ as a linear combination of the columns of $P$, say $w=\sum \alpha_i p_i$ but $Ap_i = \lambda_i p_i$ so that
$$ Aw = \sum \alpha_i \lambda_i p_i $$
and from there $$ A^k w = \sum \alpha_i \lambda_i^k p_i $$
assuming $\lambda_1$ is the eigenvalue with highest magnitude and that $\alpha_1\neq 0$, you can factor it out:
$$ A^k w = \alpha_1 \lambda_1^k\left[p_1 + \sum {\alpha_j\over \alpha_1}\left({\lambda_j\over \lambda_1}\right)^k p_j \right]$$
and now the fact that $|\lambda_1|>|\lambda_j|$ for $j=2,\dots$ can be used to show that, at the limit, the magnitude of $(\lambda_j/\lambda_1)^k$ vanishes and (provided you normalize the vector along the way) this will lead you at the limit to a multiple times the dominant eigenvector $p_1$.
More details about the power iteration here: (it also clearly states the definition of dominant eigenvalue / eigenvector in case you wanted to see it black on blue) and here:
$\endgroup$ $\begingroup$If the eigenvalues of an $n \times n$ matrix $A$ can be arranged as$$ | \lambda_1 | > | \lambda_2 | \geq | \lambda_3 | \geq \cdots \geq | \lambda_n | $$then $\lambda_1$ is called the dominant eigenvalue of $A$.
The dominant eigenpair $(\lambda_1, \mathbf{v}_1)$ of $A$ is very useful for determining the steady-state (long-term behavior) of linear dynamical systems of the form$$ \dot{\mathbf{x}} = A \mathbf{x} $$or$$ \mathbf{x}_{n + 1} = A \mathbf{x}_n $$
Power method gives a simple numerical procedure for finding the dominant eigenvalue $\lambda_1$ of $A$ (when it exists) and the Inverse Power Method gives a simple numerical procedure for finding the corresponding (dominant) eigenvector of $A$.
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