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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

It follows from Exercise 14 that, for a diagonalizable matrix, the number of nonzero eigenvalues (counted according to multiplicity) equals the rank of the matrix. Give an example of a defective matrix whose rank is not equal to the number of nonzero eigenvalues.

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

Show that any \(3 \times 3\) matrix of the form \\[ \left(\begin{array}{lll} a & 1 & 0 \\ 0 & a & 1 \\ 0 & 0 & b \end{array}\right) \\] is defective.

We can show that, for an \(n \times n\) stochastic matrix, \(\lambda_{1}=1\) is an eigenvalue and the remaining eigenvalues must satisfy \\[ \left|\lambda_{j}\right| \leq 1 \quad j=2, \ldots, n \\] (See Exercise \(24 \text { of Chapter } 7, \text { Section } 4 .)\) Show that if \(A\) is an \(n \times n\) stochastic matrix with the property that \(A^{k}\) is a positive matrix for some positive integer \(k,\) then \\[ \left|\lambda_{j}\right|<1 \quad j=2, \ldots, n \\]

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 \(|\lambda|=1\)
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