Question

3 1. Let A = 0 (a) Compute the eigenvalues of A and specify their algebraic multiplicities. (b) For every eigenvalue 1, determine the eigenspace Ex and specify its dimension. (c) Is A a defective matrix? Why or why not? (d) Is A a singular matrix? Why or why not? (e) Determine the eigenvalues of (74) + 5.

Answer 1

Following is the attached paper work for the first four parts:

A = 3 1 -7 -1 - 1 1 4 -4 characteristic Eigen values are the roots of the polynomial of the matrix. f(t) = det A . t I -7 بح t o i det -1 -1 0 f ܓܙ o 4 - 4 O 0 f 1-€ 7 بیا è det 1 -1-t 1 4 -4 t = (1-4) (+1-4). (-4) + 4) +7 ( + - 4) +3 64 + +(106) 5. #41-4') + 4 (1-4) + 74-28 + 24 +12 = ++++4'-*+74-28 +2 4+12+ at 8 는 3 .t + 16 + Factorize the polynomial : Secame wife out of 116 - )

0,-4 and 4. -4,0, 4 with Thus, the roots are So, the eigen values are algebraic multiplicities 1. eigen value -4. The eigen space Corresponding Noll A - (-+)) A + 4I) Take the b ) to it is Null 7 6.00 بیا 4 04 0 Null 4 4 . - 4 40 Null 5 7. 3 - سد 4 - 4 الیا00 Perform the elementary now operations : 3 R, » -1R, 5 - 7 R2 R w 3 5 IT -| w 4 4 4 4 -4 - 3 1 w R, →-58,R/1 R3 > -42, + Ri 8 8 o - 7 3 5 4 - 4 CSScanned with CamScanner 19

1 - 3 R&R R₃ | RB - - 3 -1 R3 → +RE+R3 o 1 1 Ô ܥܙ 1 o 1 1 The equations obtained are 8, -322-83 =o H2 + H₂=0 Hence, the eigen space is :3 € Its dimension is -2 -1 KER ex } H Using the same procedure, it follows that eigen spaces the remaining two are E. VER Its dimension is 2 Eq 6): <,483.945 83. Its dimensert 1 . CSScanned with CamScanner

three distinet eigen values hence, it is c) No, A is not defectiver because A has diagonalizable. d) Yes, A is singular since o is an eigen Value which implies that det (A) is zero. CS Scanned with CamScanner