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Chapter 6: Problem 20

Let \(A\) be a \(n \times n\) matrix with Schur decomposition \(U T U^{H} .\) Show that if the diagonal entries of \(T\) are all distinct, then there is an upper triangular matrix \(R\) such that \(X=U R\) diagonalizes \(A\)

Short Answer

Expert verified
In conclusion, we have found that if A is a given matrix with a known Schur decomposition that has distinct diagonal entries, then A can be diagonalized using an upper-triangular matrix R (equal to T) and a diagonal matrix D. The diagonal matrix D can be found using the equation \( D = U^H A U \) and the required upper-triangular matrix R is equal to T. Therefore, the matrix X = UR diagonalizes A with A = XDX^(-1).

Step by step solution

01

Rewrite the Given Schur Decomposition

Recall the Schur decomposition of A: \[ A = UTU^H \]
02

Express A as a Diagonal Matrix

We want to find a diagonal matrix D and an upper-triangular matrix R such that A = XDX^(-1), where X = UR. We know that A = UTU^H, so let's find a matrix \(U^H R^{鈭�1} U\) that makes UTU^H diagonal. Rearrange the Schur decomposition equation to resemble the diagonalization form: \[ D = U^H A U \] Since T is an upper-triangular matrix with distinct diagonal entries and U is a unitary matrix, we can use this equation to find D.
03

Find Matrix R

Now, we have X = UR and A = XDX^(-1). We can rewrite this as A = (UR)D(UR)^(-1). Let's find R by comparing this equation to the Schur decomposition (A = UTU^H). From A = (UR)D(UR)^(-1), note that U is the same in both equations if R is chosen to be equal to T (since U has to be unitary): \[ R = T \] Now we have X = UT.
04

Express Matrix A in Terms of Matrix X

Since X = UT and A = UTU^H, we can plug in X to express A in terms of X: \[ A = XU^H \]
05

Confirm A is Diagonalized

Given A = XDX^(-1), we have done the following: 1. Found the diagonal matrix D = U^H A U. 2. Found the upper-triangular matrix R = T. 3. Shown that A = XU^H, where X = UR. We've shown that A can be diagonalized with a diagonal matrix D and an upper-triangular matrix R. Therefore, there is an upper-triangular matrix R (equal to T) such that X = UR diagonalizes A.

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Most popular questions from this chapter

Show that \(A\) and \(A^{T}\) have the same eigenvalues. Do they necessarily have the same eigenvectors? Explain.

Let \(A\) be a singular \(n \times n\) matrix. Show that \(A^{T} A\) is positive semidefinite, but not positive definite.

Let \(U\) be a unitary matrix. Prove that (a) \(U\) is normal. (b) \(\|U \mathbf{x}\|=\|\mathbf{x}\|\) for all \(\mathbf{x} \in \mathbb{C}^{n}\) (c) if \(\lambda\) is an eigenvalue of \(U,\) then \(|\lambda|=1\)

Let \(Q\) be a \(3 \times 3\) orthogonal matrix whose determinant is equal to 1 (a) If the eigenvalues of \(Q\) are all real and if they are ordered so that \(\lambda_{1} \geq \lambda_{2} \geq \lambda_{3},\) determine the values of all possible triples of eigenvalues \(\left(\lambda_{1}, \lambda_{2}, \lambda_{3}\right)\) (b) In the case that the eigenvalues \(\lambda_{2}\) and \(\lambda_{3}\) are complex, what are the possible values for \(\lambda_{1} ?\) Explain. (c) Explain why \(\lambda=1\) must be an eigenvalue of \(Q\)

Let \(p(\lambda)=(-1)^{n}\left(\lambda^{n}-a_{n-1} \lambda^{n-1}-\cdots-a_{1} \lambda-a_{0}\right)\) be a polynomial of degree \(n \geq 1\), and let \\[ C=\left(\begin{array}{ccccc} a_{n-1} & a_{n-2} & \cdots & a_{1} & a_{0} \\ 1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & 0 \\ \vdots & & & \\ 0 & 0 & \cdots & 1 & 0 \end{array}\right) \\] (a) Show that if \(\lambda_{i}\) is a root of \(p(\lambda)=0,\) then \(\lambda_{i}\) is an eigenvalue of \(C\) with eigenvector \(\mathbf{x}=\) \(\left(\lambda_{i}^{n-1}, \lambda_{i}^{n-2}, \ldots, \lambda_{i}, 1\right)^{T}\) (b) Use part (a) to show that if \(p(\lambda)\) has \(n\) distinct roots, then \(p(\lambda)\) is the characteristic polynomial of \(C\). The matrix \(C\) is called the companion matrix of \(p(\lambda)\)
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