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Q23E

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Found in: Page 336

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# 23: Suppose matrix A is similar to B. What is the relationship between the characteristic polynomials of A and B? What does your answer tell you about the eigenvalues of A and B?

Characteristic polynomials of A and B are the same, therefore their eigenvalues are also same.

See the step by step solution

## Step 1: Eigenvalues

In linear algebra, an eigenvector or characteristic vector of a linear transformation is a nonzero vector that changes at most by a scalar factor when that linear transformation is applied to it. The corresponding eigenvalue, often denoted by \lambda, is the factor by which the eigenvector is scaled.

## Step 2: Relationship between the characteristic polynomials of A and B

If A and B are similar matrices then, ${T}^{-1}-AT=B$, for an invertible T. If det $\left(A-\lambda l\right)=0$, then

$0=de{t}^{-1}det\left(A-\lambda l\right)detT=det\left({T}^{-1}\left(A-\lambda l\right)T\right)=det\left({T}^{-1}AT-\lambda {T}^{-1}lT\right)=det\left({T}^{-1}AT-\lambda l\right)=det\left(B-\lambda l\right)$

This means that the characteristics polynomials of A and B are the same. From this also follows that their eigenvalues are exactly the same.

Hence, characteristic polynomials of A and B are the same, therefore their eigenvalues are also same.