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Chapter 5: Problem 7

Let \(T\) be a linear operator on a finite-dimensional vector space \(V\), and let \(\beta\) be an ordered basis for \(\mathrm{V}\). Prove that \(\lambda\) is an eigenvalue of \(\mathrm{T}\) if and only if \(\lambda\) is an eigenvalue of \([\mathrm{T}]_{\beta}\).

Short Answer

Expert verified
To prove that 位 is an eigenvalue of T if and only if 位 is an eigenvalue of [T]_尾, we first assume 位 is an eigenvalue of T and show that 位 is an eigenvalue of [T]_尾 by expressing eigenvector v and its image T(v) in terms of the ordered basis 尾 and matrix representation [T]_尾. Conversely, we assume 位 is an eigenvalue of [T]_尾 and show that 位 is an eigenvalue of T by reversing the process and expressing the eigenvector [v]_尾 and its image [T]_尾[v]_尾 in terms of the ordered basis 尾 and the linear operator T. Thus, we conclude that 位 is an eigenvalue of T if and only if 位 is an eigenvalue of [T]_尾.

Step by step solution

01

Writing Down Given Information

We are given a linear operator T on a finite-dimensional vector space V, and an ordered basis 尾 for V. We want to prove that 位 is an eigenvalue of T if and only if 位 is an eigenvalue of [T]_尾.
02

Define Eigenvalue and Matrix Representation

Recall the definition of an eigenvalue: 位 is an eigenvalue of T if there exists a non-zero vector v in V such that T(v) = 位v. The matrix representation [T]_尾 is the matrix A such that [Tv]_尾 = A[v]_尾 for all v in V.
03

Assume 位 is an Eigenvalue of T

Let's assume 位 is an eigenvalue of T, so there exists a non-zero vector v in V such that T(v) = 位v.
04

Express Eigenvector in Terms of the Matrix Representation

We want to express the eigenvector v and its image T(v) in terms of the ordered basis 尾 and matrix representation [T]_尾. Using the given connection between a linear operator and its matrix representation, we have [T(v)]_尾 = [位v]_尾 = [T]_尾[v]_尾.
05

Prove that 位 is an Eigenvalue of [T]_尾

From Step 4, since [T(v)]_尾 = [T]_尾[v]_尾, we have [位v]_尾 = [T]_尾[v]_尾. Thus 位 is an eigenvalue of the matrix [T]_尾 with corresponding eigenvector [v]_尾.
06

Conversely, Assume 位 is an Eigenvalue of [T]_尾

Now, let's assume 位 is an eigenvalue of [T]_尾, so there exists a non-zero vector [v]_尾 such that [T]_尾[v]_尾 = [位v]_尾.
07

Express Eigenvector in Terms of the Linear Operator

Reversing the process from Step 4, we want to express the eigenvector [v]_尾 and its image [T]_尾[v]_尾 in terms of the ordered basis 尾 and the linear operator T. We have [T]_尾[v]_尾 = [位v]_尾 = [T(v)]_尾.
08

Prove that 位 is an Eigenvalue of T

Since [T(v)]_尾 = [位v]_尾, it follows that T(v) = 位v. Thus, 位 is an eigenvalue of T with corresponding eigenvector v.
09

Conclusion

We have shown that if 位 is an eigenvalue of T, then it is an eigenvalue of [T]_尾, and if 位 is an eigenvalue of [T]_尾, then it is an eigenvalue of T. Hence, 位 is an eigenvalue of T if and only if 位 is an eigenvalue of [T]_尾.

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Most popular questions from this chapter

Let \(A\) be an \(n \times n\) matrix that is similar to an upper triangular matrix and has the distinct eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{k}\( with corresponding multiplicities \)m_{1}, m_{2}, \ldots, m_{k}$. Prove the following statements. (a) \(\operatorname{tr}(A)=\sum^{k} m_{i} \lambda_{i}\) (b) $\operatorname{det}(A)=\left(\lambda_{1}\right)^{m_{1}}\left(\lambda_{2}\right)^{m_{2}} \ldots\left(\lambda_{k}\right)^{m_{k}}$.

Determine whether or not each of the following linear maps is nonsingular. If not, find a nonzero vector \(v\) whose image is 0 (a) \(F: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(F(x, y)=(x-y, x-2 y)\) (b) \(G: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) defined by \(G(x, y)=(2 x-4 y, 3 x-6 y)\)

Prove: (a) The zero mapping \(0,\) defined by \(\mathbf{0}(v)=0 \in U\) for every \(v \in V,\) is the zero element of \(\operatorname{Hom}(V, U) .\) (b) The negative of \(F \in \operatorname{Hom}(V, U)\) is the mapping \((-1) F,\) that is, \(-F=(-1) F\)

Prove that if both \(T_{w}\) and \(\bar{T}\) are diagonalizable and have no common eigenvalues, then \(T\) is diagonalizable. The results of Theorem \(5.21\) and Exercise 28 are useful in devising methods for computing characteristic polynomials without the use of determinants. This is illustrated in the next exercise.

Show that each of the following linear operators \(T\) on \(\mathbf{R}^{3}\) is nonsingular and find a formula for \(T^{-1}\), where (a) \(T(x, y, z)=(x-3 y-2 z, y-4 z, z)\) (b) \(T(x, y, z)=(x+z, x-y, y)\)
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