assign5-sol

# assign5-sol - Math 235 Assignment 5 Due 9:15am Wednesday 1...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 235 Assignment 5 Due 9:15am, Wednesday Feb 28, 2007. 1. From the Text Â§ 5.3 #31. Construct a nonzero 2 Ã— 2 matrix that is invertible but not diag- onalizable. Solution. Consider a matrix 1 1 1 . The determinant of this matrix is one, implying this matrix is invertible. It is easy to see that the matrix has only one eigenvalue Î» = 1 with algebraic multiplicity two. The eigenspace corresponding to the eigenvalue is spanned by 1 , thus geometric multiplicity of Î» is one which is less than algebraic multiplicity of Î» . Therefore the matrix is not diagonalizable. #32. Construct a nondiagonal 2 Ã— 2 matrix that is diagonalizable but not invertible. Solution. Consider 1 1 1 1 . The determinant of the matrix is zero, implying this matrix is not invertible. The characteristic equation of the matrix is given by Î» ( Î»- 2) = 0 . Thus there are two distinct eigenvalues, Î» 1 = 0 and Î» 2 = 2 . Since the dimension of the matrix is also two, then the matrix is diagonalizable. Â§ 5.4 #10. Define T : P 3-â†’ R 4 by T ( p ) = p (- 3) p (- 1) p (1) p (3) . a. Show that T is a linear transformation. b. Find the matrix for T relative to the basis { 1 , t, t 2 , t 3 } for P 3 and the standard basis for R 4 . Solution. a. We need to show that T ( c p + d q ) = cT ( p ) + dT ( q ) , where c and d are constants. 1 T ( c p + d q ) = ( c p + d q )(- 3) ( c p + d q )(- 1) ( c p + d q )(1) ( c p + d q )(3) = c ( p (- 3)) + d ( q (- 3)) c ( p (- 1)) + d ( q (- 1)) c ( p (1)) + d ( q (1)) c ( p (3)) + d ( q (3)) = c p (- 3) p (- 1) p (1) p (3) + d q (- 3) q (- 1) q (1) q (3) = cT ( p ) + dT ( q ) b. Let B = { 1 , t, t 2 , t 3 . } Then p ( t ) = a + bt + ct 2 + dt 3 = a b c d B . We want to find a matrix A such that A a b c d B = p (- 3) p (- 1) p (1) p (3) = a + b (- 3) + c (- 3) 2 + d (- 3) 3 a + b (- 1) + c (- 1) 2 + d (- 1) 3 a + b (1) + c (1) 2 + d (1) 3 a + b (3) + c (3) 2 + d (3) 3 . It is easy to see that A = 1- 3 9- 27 1- 1 1- 1 1 1 1 1 1 3 9 27 . Â§ 5.5 #4. Let the following matrix act on C 2 . Find the eigenvalues and a basis for each eigenspace in C 2 . 5- 2 1 3 . Solution. 5- Î»- 2 1 3- Î» = (5- Î» )(3- Î» ) + 2 = Î» 2- 8 Î» + 17 = 0 2 Î» 1 , 2 = 8 Â± âˆš 64- 4 * 17 2 = 8 Â± 2 i 2 = 4 Â± i. For Î» 1 = 4 + i we have 5- (4 + i )- 2 1 3- (4 + i ) a + bi c + di = ( a + b- 2 c ) + (- a + b- 2 d ) i ( a- c + d ) + ( b- c- d ) i = . This gives us four equations: a + b- 2 c = 0- a + b- 2 d = 0 a- c + d = 0 b- c- d = 0 We can now set up the matrix corresponding to the above system of linear equations 1 1- 2- 1 1- 2 1- 1 1 1- 1- 1 -â†’ 1 1- 2 2- 2- 2- 1 1 1 1- 1- 1...
View Full Document

### Page1 / 11

assign5-sol - Math 235 Assignment 5 Due 9:15am Wednesday 1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online