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Chapter 13: Problem 13

Let \(T\) be a symmetric operator. Show that (a) The characteristic polynomial \(\Delta(t)\) of \(T\) is a product of linear polynomials (over \(\mathbf{R})\); (b) \(T\) has a nonzero eigenvector. (a) Let \(A\) be a matrix representing \(T\) relative to an orthonormal basis of \(V\); then \(A=A^{T}\). Let \(\Delta(t)\) be the characteristic polynomial of \(A\). Viewing \(A\) as a complex self-adjoint operator, \(A\) has only real eigenvalues by Theorem 13.4. Thus, $$ \Delta(t)=\left(t-\lambda_{1}\right)\left(t-\lambda_{2}\right) \cdots\left(t-\lambda_{n}\right) $$ where the \(\lambda_{i}\) are all real. In other words, \(\Delta(t)\) is a product of linear polynomials over \(\mathbf{R}\). (b) By (a), \(T\) has at least one (real) eigenvalue. Hence, \(T\) has a nonzero eigenvector.

Short Answer

Expert verified
(a) The characteristic polynomial of a symmetric operator $T$ is given by \[ \Delta(t) = (t-\lambda_1)(t-\lambda_2) \cdots (t-\lambda_n), \] where the $\lambda_i$ are real eigenvalues of $T$. (b) Since $T$ has at least one real eigenvalue, it must also have a corresponding nonzero eigenvector.

Step by step solution

01

(a) Characteristic Polynomial

Let \(A\) be a matrix representing operator \(T\) relative to an orthonormal basis of \(V\), such that \(A= A^T\). We need to find the characteristic polynomial \(\Delta(t)\) of \(A\). Since \(A\) is symmetric, it is also a complex self-adjoint operator, and thus its eigenvalues are real by Theorem 13.4. Therefore, we can write the characteristic polynomial \(\Delta(t)\) as a product of linear polynomials with real coefficients: \[ \Delta(t) = (t-\lambda_1)(t-\lambda_2) \cdots (t-\lambda_n), \] where \(\lambda_i\) are real eigenvalues of \(A\).
02

(b) Nonzero Eigenvector

From part (a), we established that the operator \(T\) has at least one real eigenvalue, since its characteristic polynomial is a product of linear polynomials with real coefficients. By definition of eigenvalues, if an operator has an eigenvalue, then it must also have a corresponding eigenvector, which will be nonzero in this case. Thus, the symmetric operator \(T\) has at least one nonzero eigenvector.

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For any operator \(T,\) show that \(T+T^{* *}\) is self-adjoint and \(T-T^{*}\) is skew-adjoint

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