# Diagonalization: Eigenvalues Vs Elementary Row Operations [duplicate]

Using elementary row operations, a matrix A $\in \mathrm{R}^{n \times n}$ can be reduced to a Row-Reduced Echelon (RRE) form. Using the RRE form of A, the bases of Nullspace and Range can be obtained. The RRE form of A is a triangular (almost diagonal) form of A and is row-equivalent to A. If we are able to get a simple form of A, why do we need to get eigenvalues and diagonalize A ? What is the motivation for having an alternative method to simplify(diagonalize) A ? Why would one choose to diagonalize A using eigen values ?

• It seems that the question is equivalent to "what is the eigenstuff good for". See, e.g., here. By ignoring the importance of eigenvalues, bad things can happen. – Algebraic Pavel Apr 14 '15 at 11:13
• In addition, there are many other matrix decompositions or "simplifications", each of them with different properties and theoretical and/or practical applications. – Algebraic Pavel Apr 14 '15 at 11:22
• There are many useful things one can do with matrices without ever performing a row reduction or finding an eigenvalue. There are other things where finding an eigenvalue is very useful. Are you asking for a list of all possible applications of eigenvalues? – David K Apr 14 '15 at 12:36
• No. I want to know what is the advantage gained by using eigenvalues to diagonalize over the method of using EROs, like for e.g, some advantage in computational feasibility, computational complexity or something else. It does not have anything to do with other uses of eigen values or other matrix decompositions. – Spr Apr 14 '15 at 13:38

Some properties of $A$ cannot be deduced from a row decomposition $A = LR$, where $R$ is the RRE-form of $A$. For example, if you want to compute powers of $A$, for example to compute explicit forms of linear recursion equations, you cannot do this easily from a $LR$-decomposition, we have $$A^n = LR\cdot LR \cdots LR$$ but this is in general not simpler as computing $A^n$ by hand. On the other side, in an eigenvalue decomposition $A = P\Lambda P^{-1}$, $\Lambda$ a diagonal matrix, we have $$A^n = P\Lambda P^{-1} \cdot P\Lambda P^{-1} \cdots P\Lambda P^{-1} = P\Lambda^n P^{-1}$$ and $\Lambda^n$ is computed easily (just take the powers of the diagonal). So some things can be computed more easy with the eigenvalues.