/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Show that if an orthogonal (unit... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 35

Show that if an orthogonal (unitary) matrix is triangular, then it is diagonal.

Short Answer

Expert verified
If \(A\) is an orthogonal (unitary) triangular matrix, we have \(A^T A = I\). We consider the upper and lower triangular cases separately. In both cases, we find that the diagonal elements satisfy the sum of their squared equals to 1. Thus, if a matrix is both orthogonal (unitary) and triangular, it must be diagonal.

Step by step solution

01

Write down the matrix A

Let's denote the matrix in question by \(A\), so \(A\) is both orthogonal (unitary) and triangular. Let \(a_{ij}\) indicate the elements of \(A\). From the definitions, we have: - Orthogonal (unitary): \(A^T A = I\) (or \(A^* A = I\) for complex-valued matrices) - Triangular: either \(a_{ij} = 0~~\forall i > j\) (upper triangular) or \(a_{ij} = 0~~\forall i < j\) (lower triangular). We will make separate cases for upper and lower triangularity and show that in both cases, the matrix must be diagonal, which will complete the proof. #Step 2: Upper triangular case#
02

Multiply the transpose with the matrix

Let's first consider the upper triangular case. In this case, all elements below the diagonal are zero. Write \(A^T A\), and use the property \(A^T A=I\) to deduce the relationship between the diagonal elements: For the upper triangular case, a typical element of \(A^T A\) is: \((A^T A)_{ij} = \sum_k a_{ki}^T a_{kj}\). Now, for the elements on the diagonal (\(i=j\)), we have: \((A^T A)_{ii} = \sum_k a_{ki}^T a_{ki} = 1\). Using the fact that \(a_{ki} = 0\) for \(k>i\), we can rewrite the sum as: \((A^T A)_{ii} = \sum_{k=1}^i a_{ki}^T a_{ki} = 1\). This indicates that the diagonal elements satisfy the sum of their squares equal to 1. #Step 3: Lower triangular case#
03

Multiply the transpose with the matrix

Now, let's consider the lower triangular case. In this case, all elements above the diagonal are zero. Write \(A^T A\), and use the property \(A^T A=I\) to deduce the relationship between the diagonal elements: For the lower triangular case, a typical element of \(A^T A\) is: \((A^T A)_{ij} = \sum_k a_{ki}^T a_{kj}\). Now, for the elements on the diagonal (\(i=j\)), we have: \((A^T A)_{ii} = \sum_k a_{ki}^T a_{ki} = 1\). Using the fact that \(a_{ki} = 0\) for \(k<i\), we can rewrite the sum as: \((A^T A)_{ii} = \sum_{k=i}^n a_{ki}^T a_{ki} = 1\). This indicates that the diagonal elements satisfy the sum of their squared equals to 1, similar to the upper triangular case. #Step 4: Conclusion#
04

Combining the upper and lower triangular cases

Combining the results from the upper and lower triangular cases, we observe that the diagonal elements must satisfy the sum of their squared equals to 1, which corresponds to the property of being an orthogonal (unitary) matrix. Since this holds for both upper and lower triangular cases, we have shown that if a matrix is both orthogonal (unitary) and triangular, its diagonal elements must satisfy this property. Hence, the matrix must be diagonal. This completes the proof.

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