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Chapter 7: Problem 52

Prove that \(\lambda=0\) is an eigenvalue of \(A\) if and only if \(A\) is singular.

Short Answer

Expert verified
\(\lambda=0\) is an eigenvalue of matrix \(A\) if and only if \(A\) is a singular matrix.

Step by step solution

01

Assumption and Proof 'if' Part

Assume that \(\lambda=0\) is an eigenvalue of \(A\). Then by definition of an eigenvalue, there exists a non-zero vector \(x\) such that \(Ax = \lambda x = 0x = 0\). This means that \(A\) has a nontrivial nullspace and that \(A\) is singular.
02

Assumption and Proof 'only if' Part

Assume that \(A\) is singular. As \(A\) is singular, there exists a non-zero vector \(x\) such that \(Ax = 0 = 0x\), meaning \(\lambda = 0\) is an eigenvalue of \(A\).
03

Conclusion

Based on the previous steps, it's shown that \(\lambda=0\) is an eigenvalue of \(A\) if and only if \(A\) is singular. This can be concluded as both directions of the statement have been proven.

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Most popular questions from this chapter

A population has the characteristics listed below. (a) A total of \(75 \%\) of the population survives its first year. Of that \(75 \%, 25 \%\) survives the second year. The maximum life span is 3 years. (b) The average number of offspring for each member of the population is 2 the first year, 4 the second year, and 2 the third year. The population now consists of 120 members in each of the three age classes. How many members will there be in each age class in 1 year? In 2 years?

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