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Chapter 6: Problem 18

Let \(A\) be a diagonalizable \(n \times n\) matrix. Prove that if \(B\) is any matrix that is similar to \(A,\) then \(B\) is diagonalizable.

Short Answer

Expert verified
To prove that a matrix B is diagonalizable if it is similar to a diagonalizable matrix A, we followed these steps: 1. Recall the definitions of matrix similarity and diagonalizable matrices. 2. Write the characteristic polynomial of A, given that it is diagonalizable with an invertible matrix P and a diagonal matrix D such that \(A = PDP^{-1}\). 3. Show that A and B have the same characteristic polynomials, which implies they have the same eigenvalues. 4. Prove that there is an invertible matrix Q that relates the eigenvectors of A and B. 5. Conclude that B is diagonalizable since it can be expressed as a product of an invertible matrix and a diagonal matrix. Therefore, if matrix B is similar to a diagonalizable matrix A, it is also diagonalizable.

Step by step solution

01

Definition of matrix similarity and diagonalizable matrices

Recall that two matrices A and B are similar if there exists an invertible matrix P such that \(A = P^{-1}BP\). A matrix is said to be diagonalizable if it can be written as a product of a diagonal matrix D and an invertible matrix P.
02

Define the polynomial for the matrix A

Since A is diagonalizable, there exists an invertible matrix P and a diagonal matrix D such that \(A = PDP^{-1}\). Then, the characteristic polynomial of A is given by \(p_A(\lambda) = \det (A - \lambda I_n)\), where \(I_n\) is the identity matrix of size n.
03

Relate the polynomials for matrices A and B

Since B is similar to A, there exists an invertible matrix Q such that \(B = Q^{-1}AQ\). Then, we can substitute A in the definition of the characteristic polynomial of A with the equation involving B. We get \(p_A(\lambda) = \det (Q^{-1}AQ - \lambda I_n) = \det (Q^{-1}(A - \lambda I_n)Q) \). Using the properties of determinants, we obtain \(p_A(\lambda) = \det (Q^{-1})\det (A - \lambda I_n)\det (Q) = p_B(\lambda)\), where \(p_B(\lambda)\) is the characteristic polynomial of B.
04

Eigenvalues of matrices A and B

From step 3, it is clear that matrices A and B have the same characteristic polynomials. Since the eigenvalues of a matrix are the roots of its characteristic polynomial, it follows that A and B share the same eigenvalues.
05

Eigenvectors of matrices A and B

Let \(\lambda\) be an eigenvalue of A, and let x be an eigenvector associated with \(\lambda\) (i.e., \(Ax = \lambda x\)). Since B is similar to A, we have \(B = Q^{-1}AQ\), and we can write \(Q^{-1}Ax = Q^{-1}\lambda x \Rightarrow B(Qx) = \lambda (Qx)\). Because Q is invertible, \(Qx \neq 0\), so (Qx) is an eigenvector of B corresponding to the eigenvalue \(\lambda\).
06

Conclude the proof by showing B is diagonalizable

We have shown that matrices A and B have the same eigenvalues and that there exists an invertible matrix Q such that the eigenvectors of A can be transformed into the eigenvectors of B using Q. This implies that B can be expressed as a product of an invertible matrix and a diagonal matrix, and thus B is diagonalizable. Therefore, if matrix B is similar to a diagonalizable matrix A, it is also diagonalizable.

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Most popular questions from this chapter

Which of the matrices that follow are positive definite? Negative definite? Indefinite? (a) \(\left(\begin{array}{ll}3 & 2 \\ 2 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}3 & 4 \\ 4 & 1\end{array}\right)\) (c) \(\left(\begin{array}{rr}3 & \sqrt{2} \\ \sqrt{2} & 4\end{array}\right)\) (d) \(\left(\begin{array}{rrr}-2 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & -2\end{array}\right)\) (e) \(\left(\begin{array}{lll}1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2\end{array}\right)\) (f) \(\left(\begin{array}{lll}2 & 0 & 0 \\ 0 & 5 & 3 \\ 0 & 3 & 5\end{array}\right)\)

Let \(A\) be an \(n \times n\) matrix with positive real eigenvalues \(\lambda_{1}>\lambda_{2}>\cdots>\lambda_{n} .\) Let \(\mathbf{x}_{i}\) be an eigenvector belonging to \(\lambda_{i}\) for each \(i,\) and let \(\mathbf{x}=\) \(\alpha_{1} \mathbf{x}_{1}+\cdots+\alpha_{n} \mathbf{x}_{n}\) (a) Show that \(A^{m} \mathbf{x}=\sum_{i=1}^{n} \alpha_{i} \lambda_{i}^{m} \mathbf{x}_{i}\) (b) Show that if \(\lambda_{1}=1,\) then \(\lim _{m \rightarrow \infty} A^{m} \mathbf{x}=\alpha_{1} \mathbf{x}_{1}\)

Let \(A\) be a nonnegative irreducible \(3 \times 3\) matrix whose eigenvalues satisfy \(\lambda_{1}=2=\left|\lambda_{2}\right|=\left|\lambda_{3}\right|\) Determine \(\lambda_{2}\) and \(\lambda_{3}\)

Show that if \(\sigma\) is a singular value of \(A,\) then there exists a nonzero vector x such that \\[ \sigma=\frac{\|A \mathbf{x}\|_{2}}{\|\mathbf{x}\|_{2}} \\]

Let \(A\) be an \(n \times n\) matrix with singular value decomposition \(U \Sigma V^{T}\) and let \\[ B=\left(\begin{array}{cc} O & A^{T} \\ A & O \end{array}\right) \\] Show that if \\[ \mathbf{x}_{i}=\left(\begin{array}{c} \mathbf{v}_{i} \\ \mathbf{u}_{i} \end{array}\right], \quad \mathbf{y}_{i}=\left[\begin{array}{r} -\mathbf{v}_{i} \\ \mathbf{u}_{i} \end{array}\right], \quad i=1, \ldots, n \\] then the \(\mathbf{x}_{i}\) 's and \(\mathbf{y}_{i}\) 's are eigenvectors of \(B\). How do the eigenvalues of \(B\) relate to the singular values of \(A ?\)
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