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Now explain why we may choose eigenvectors V₁ for ₁ and v₂ for ₂ that each have unit length and are orthogonal. Finally, define P and D using the data you've conjured so far such that A = PDP ¹ where P is an orthogonal matrix and D is diagonal. Delete this red text.

Now explain why we may choose eigenvectors V₁ for ₁ and v₂ for ₂ that each have unit length and are orthogonal. Finally, define P and D using the data you've conjured so far such that A = PDP ¹ where P is an orthogonal matrix and D is diagonal. Delete this red text.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Now explain why we may choose eigenvectors V₁ for X₁ and V₂ for X₂ that each have unit length and are orthogonal. Finally, define P and D using the data you've conjured so far such that A = PDP ¹ where P is an orthogonal matrix and D is diagonal. Delete this red text.
Transcribed Image Text:Now explain why we may choose eigenvectors V₁ for X₁ and V₂ for X₂ that each have unit length and are orthogonal. Finally, define P and D using the data you've conjured so far such that A = PDP ¹ where P is an orthogonal matrix and D is diagonal. Delete this red text.
Solution. Let A be a symmetric 2 x 2 matrix. Then, A has the form a b [81] b d A = If b = 0, then A is orthogonally diagonalizable because... Replace ... above with the reason why A is orthogonally diagonalizable when b = 0. (What can you take P and D to be in this case?) Delete this red text. Now suppose b +0. The characteristic polynomial of A is (a-X) (d-X) - b² = λ² − (a + d)λ + ad-b². From this it follows that the discriminant is (a+d)² - 4(ad-b²) = (a - d)² + 4b². Since b = 0, this quantity is always positive and hence the characteristic polynomial has two distinct real roots. Thus A has two distinct real eigenvalues, X₁ and X₂.
Transcribed Image Text:Solution. Let A be a symmetric 2 x 2 matrix. Then, A has the form a b [81] b d A = If b = 0, then A is orthogonally diagonalizable because... Replace ... above with the reason why A is orthogonally diagonalizable when b = 0. (What can you take P and D to be in this case?) Delete this red text. Now suppose b +0. The characteristic polynomial of A is (a-X) (d-X) - b² = λ² − (a + d)λ + ad-b². From this it follows that the discriminant is (a+d)² - 4(ad-b²) = (a - d)² + 4b². Since b = 0, this quantity is always positive and hence the characteristic polynomial has two distinct real roots. Thus A has two distinct real eigenvalues, X₁ and X₂.
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