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A widely used method for estimating eigenvalues of a general matrix $A$ is the $Q R$ algorithm. Under suitable conditions, this algorithm produces a sequence of matrices, all similar to $A$ , that become almost upper triangular, with diagonal entries that approach the eigenvalues of $A$ . The main idea is to factor $A$ (or another matrix similar to $A$ ) in the form $A=Q_{1} R_{1},$ where $Q_{1}^{T}=Q_{1}^{-1}$and $R_{1}$ is upper triangular. The factors are interchanged to form $A_{1}=R_{1} Q_{1},$ which is again factored as $A_{1}=Q_{2} R_{2} ;$ then to form $A_{2}=R_{2} Q_{2}$ and so on. The similarity of $A, A_{1}, \ldots$ follows from the more general result in Exercise $23 .$Show that if $A=Q R$ with $Q$ invertible, then $A$ is similar to $A_{1}=R Q .$

Since $Q$ is invertible we can write$A=Q R=Q R Q Q^{-1}=Q(R Q) Q^{-1}=Q A_{1} Q^{-1}$So, by the definition of similarity $A$ is similar to $A_{1}$

Calculus 3

Chapter 5

Eigenvalues and Eigenvectors

Section 2

The Characteristic Equation

Vectors

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all right, so we're given that A is equal Thio Q R and notes that we can say this is also equal to Q R. I, since multiplied by the identity matrix doesn't really do anything. And we can also express the I as que que in verse because we're given that cute is in vertebral and by the associative property we can groups. Demetris is like this with our Q in the middle. So we're giving que our Q que members, and we are also given that our Q is equal to a one. So we're gonna do that. So basically, we get that A is equal to Q A one Q in verse. And that's basically the whole proof, because this is exactly what the definition of similarity is. So, since a is equal, Thio Q A. One cute in verse by the definition of similarity, A is similar to a one, and that's it.

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