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Now suppose b +0. The characteristic polynomial of A is (a-X)(d-A)-b² = x² − (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. Now explain why we may choose eigenvectors V₁ for ₁ 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.

Now suppose b +0. The characteristic polynomial of A is (a-X)(d-A)-b² = x² − (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. Now explain why we may choose eigenvectors V₁ for ₁ 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.

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 suppose b +0. The characteristic polynomial of A is (a-X)(d-A)-b² = x² − (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. Now explain why we may choose eigenvectors V₁ for ₁ 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 suppose b +0. The characteristic polynomial of A is (a-X)(d-A)-b² = x² − (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. Now explain why we may choose eigenvectors V₁ for ₁ 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.
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