# Interpretation of Matrix Diagonalization

If $A$ be a non singular matrix then what good does it do by constructing another matrix, say $P$ whose columns are a basis that consists of eigenvectors of $A$? Does it have something to do with thee eigenvectors of $A$ being a set of basis for the transformation. Also, what is the significance of the diagonal matrix, say $\Lambda$?

Why is Diagonalization important?Even more so, what about orthogonal Diagonalization? What do they signify?
[Sorry for asking so many questions at once, I am fairly new to Linear Algebra.]

Any help is Much Appreciated!
Thank You!

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There is no such as thing as the eigenvectors of $A$; you mean, a matrix whose columns are a basis that consists of eigenvectors of $A$. Such a matrix gives you a change-of-basis under the coordinate matrix for the linear transformation determined by $A$ that turns it into a diagonal matrix. Did you see this question on the importance of eigenvectors, and this one on orthogonal diagonalizability? – Arturo Magidin Mar 25 '12 at 3:40
Oh, Okay... edited ! – funktor Mar 25 '12 at 5:09

There are many situations where one needs to calculate the exponential $e^A$ of a matrix $A$. This makes perfect sense as one takes $$e^A=\sum_{k=0}^\infty\frac{A^k}{k!}.$$ Now, good luck calculating all the powers of $A$!
But if we know that $A$ is diagonalizable, then $A=PDP^{-1}$ with $D$ diagonal. Then $$A^2=PDP^{-1}PDP^{-1}=PD^2P^{-1},$$ and similarly $A^k=PD^kP^{-1}$ for all $k$. As $D$ is diagonal, its powers consist simply of the powers of its entries. So $e^D$ is the diagonal matrix with each diagonal entry the exponential of the corresponding entry of $D$, and $$e^A=\sum_{k=0}^\infty\frac{A^k}{k!}=Pe^DP^{-1}.$$