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Chapter 7: Q40E (page 346)

For which values of constants a, b, and c are the matrices in Exercises 40 through 50 diagonalizable?

A=[1a01]

Short Answer

Expert verified

The matrix a=0 is diagonalizable.

Step by step solution

01

Characteristic equation 

Consider an n 脳n matrix A and a scalar 位. Then 位 is an eigenvalue5 of A.

det(A-位濒)=0

This is called the characteristic equation (or the secular equation) of matrix A.

02

Solution of the problem

We solve:

detA-位濒=01-位a01-位=01-位2=0位1,2=1

03

Substitute λ=1

For 位=1, we solve:

A-lx=00a00x1x2=00ax2=0Fora鈮�0,wehaveE1=span10

So A is not diagonalizable. For a = 0, though, we get A-l=0, thus E1=鈩�2, and therefore A is diagonalizable.

Here, the final result is a = 0.

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Most popular questions from this chapter

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

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