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Chapter 7: Q23E (page 345)

Find all eigenvalues and eigenvectors ofA=(1101). Is there eigenbasis? Intercept your result geometrically.

Short Answer

Expert verified

The matrix is of gemu(1)<almu(1). There is no eigenbasis.

Step by step solution

01

Algebraic Versus.

Algebraic versus geometric multiplicity If 位 is an eigenvalues of a square matrix A,

then gemu(1)<almu(1)

This matrix is upper triangle, so its eigenvalues are the entries on its main diagonal, which are 位1,2=1,位=1.

02

Solve the matrix

A-I=00100x1x2=00x2=0

E1=span10

Therefore,gemu(1)<almu(1) , there is no eigenbasis.

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