Suggested languages for you:

Americas

Europe

Q37E

Expert-verified
Found in: Page 338

### Linear Algebra With Applications

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

# consider an eigenvalue ${{\mathbit{\lambda }}}_{{\mathbf{0}}}$of an ${\mathbf{n}}{\mathbf{×}}{\mathbf{n}}$ matrix A. we are told that the algebraic multiplicity of exceeds 1. Show that ${\mathbf{f}}{\mathbf{\text{'}}}\left({\lambda }_{0}\right){\mathbf{=}}{\mathbf{0}}$(i.e.., the derivative of the characteristic polynomial of A vanishes are ${{\mathbf{\lambda }}}_{{\mathbf{0}}}$).

Consider an eigenvalue.

If an eigenvalue ${\lambda }_{0}$of is of algebraic multiplicity greater than ,

it's at least 2.

$\mathrm{f}\text{'}\left(\mathrm{\lambda }\right)=2\mathrm{f}\left(\mathrm{\lambda }\right)\left(\mathrm{\lambda }-{\mathrm{\lambda }}_{0}\right)+\mathrm{f}\text{'}\left(\mathrm{\lambda }\right)\left(\mathrm{\lambda }-{\mathrm{\lambda }}_{0}\right)2\mathrm{f}\text{'}\left({\mathrm{\lambda }}_{0}\right)=0$

See the step by step solution

## Step 1: definition of eigenvalues

The eigenvalue is a number that indicates how much variance exists in the data in that direction; in the example above, the eigenvalue is a number that indicates how spread out the data is on the line.

If an Eigenvalue ${\lambda }_{0}$of is of algebraic multiplicity greater than 1,

it's at least 2.

So, the characteristic polynomial is,

## Step 2: definition function

A function is a relationship between a set of inputs that each have one output.

A function is now, declare the values.

where is a function:

now,

. It is $\mathrm{f}\text{'}\left(\mathrm{\lambda }\right)=2\mathrm{f}\left(\mathrm{\lambda }\right)\left(\mathrm{\lambda }-{\mathrm{\lambda }}_{0}\right)+\mathrm{f}\text{'}\left(\mathrm{\lambda }\right)\left(\mathrm{\lambda }-{\mathrm{\lambda }}_{0}\right)2\mathrm{f}\text{'}\left({\mathrm{\lambda }}_{0}\right)=0$

Hence,

$\mathrm{f}\text{'}\left(\mathrm{\lambda }\right)=2\mathrm{f}\left(\mathrm{\lambda }\right)\left(\mathrm{\lambda }-{\mathrm{\lambda }}_{0}\right)+\mathrm{f}\text{'}\left(\mathrm{\lambda }\right)\left(\mathrm{\lambda }-{\mathrm{\lambda }}_{0}\right)2\mathrm{f}\text{'}\left({\mathrm{\lambda }}_{0}\right)=0$