{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Math 304Solutions 7

# Math 304Solutions 7 - Math 304 Solutions 7 1 Section 6.1 4...

This preview shows pages 1–3. Sign up to view the full content.

Math 304 Solutions 7 1 Section 6.1 4. Let A be a nonsingular matrix and let λ be an eigenvalue of A . Show that 1 is an eigenvalue of A - 1 . Answer: Let v be the eigenvector for A corresponding to the eigen- value λ . Geometrically, the result of multiplying a multiple of v by A is to multiply its length by λ . Thus, multiplying by A - 1 will multiply the length of the vector by 1 . Here is a more algebraic proof: Consider A - 1 ( A v ): A - 1 ( A v ) = v Since A v = λ v , A - 1 ( A v ) = A - 1 ( λ v ) Thus, we have that A - 1 ( λ v ) = v Thus, λ v is an eigenvector for A with eigenvalue 1 . 10. Show that the matrix A = cos θ - sin θ sin θ cos θ will have complex eigenvalues if θ is not a multiple of π . Give a geo- metric interpretation of this result. Answer: det( λI - A ) = ( λ - cos θ ) 2 + sin 2 θ 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
If det( λI - A ) = 0, then ( λ - cos θ ) 2 = - sin 2 θ If λ is real, the left side of this equation is always greater than or equal to 0, and the right side of this equation is always less than or equal to zero. The only way for this to be valid is if both sides equal 0. Thus,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Math 304Solutions 7 - Math 304 Solutions 7 1 Section 6.1 4...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online