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Linear Algebra With Applications
Found in: Page 345
Linear Algebra With Applications

Linear Algebra With Applications

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

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Short Answer

For a given eigenvalue, find a basis of the associated eigenspace. Use the geometric multiplicities of the eigenvalues to determine whether a matrix is diagonalizable. For each of the matrices A in Exercises 1 through 20, find all (real) eigenvalues. Then find a basis of each eigenspace, and diagonalize A, if you can. Do not use technology


The eigenbasis for the 3, so the diagonalization of A in the eigenbasis is 100020003

See the step by step solution

Step by Step Solution

Step 1: Algebraic Versus.

Algebraic versus geometric multiplicity If λ is an eigenvalue of a square matrix A,

then gemu(λ) ≤ almu(λ).

This matrix is upper triangle, so its eigenvalues are the entries units main diagonal, which are


For λ=1, we solve

role="math" localid="1659588503956" A-lx=0010012002x1x2x3=000x2=0,x2+2x3=0,2x3=0x2=0,x3=0

The basic of this eigenspace is


Step 2: To find the values.

First, λ=2,

role="math" localid="1659588665728" A-2λx=0-110002001x1x2x3=000-x1=x2=0,2x3=0 ,x3=0x1-x2=0,x3=0

The basic for this eigenspace is 110=v2

Step 3: To find the values.

Now, λ=3 , we solve


The basic of eigenspace 121=v3

Therefore, V1,V2,V3 is an eigenbasis for 3 , so the diagonalization of A in the eigenbasis is 100020003

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