Question

1 Compute and completely factor the characteristic polynomial of the following matrix: 0 A= -4 5 0 1 1 For credit, you have to factor the polynomial and show work for each step. B In the following, use complex numbers if necessary. For each of the following matrices: • compute the characteristic polynomial; • list all the eigenvalues (possibly complex) with their algebraic multiplicity; • for each eigenvalue, find a basis (possibly complex) of the corresponding eigenspace, and write the geometric multiplicity of the eigenvalue; • state whether the matrix is diagonalizable. If yes, write down an eigenbasis and the corresponding diagonal matrix. If not, motivate why. A = (131) B= 3 -1

Answer 1

A. We have the matrix A A= 5 1-4 ol Let it be the eigen value of A. Then the characteristic polynomial of PI A is given by 1-) Io 5 1-4 -4. I-d D 1 3 = (1-x) [(-1) 44] -1 (5-5) = (1->) (142-22+4) -56- = (-1) 012-8845-5 =(1-2) (22-27) = 1 (l-) (1-2). x +37=20 Hence, the chana eteristic polynomial. is given by -x+31-21 and its complete factors 1 (1-1) (A-2) and hence, characteristic equation of . - ² +3 1²-2x=0 of is given by A i

B For Matrix .A-(, - Then Let & be an eigen value of the matrix A. det (A - 1Ix) = 0 where Iz is 2x 2 identity matarix. -> 3-1 -4 3-) =) (3-2)²+4=0 T x= 3 + 2 of A is "There fore, the characteristic polynomial. is (3-1)44 = 42-62412 and eigen values ane (3+2i), (3-21) Algebric multiplicity of (2+2i) is 1. Algebric multiplicity of 8–2i) is 1. də (3 tai) the corresponding eigen vector. Then Fors Let X= xj) be can be AX=dx.

خی 시 w 22 -4. (3+21)x1 (3 +26) *2 we get . 3x1+2ix, From this, 34-4862 Xyz 222, t cy Also, x, = 2i xe [... from x+3x2= (3+21)42 E=F x=ding. Taking naac (50) we get x=2ic zie Hence, X= we get 382 () -- (85) nz Therefore, Basis of eigen space of eigen value (3+21) is {(3:)] whose dimension is 1. so, geometrie multiplicity = dimension. of the basis of eigen.space=1 .. Algebnic multiplicity of (3+2i)=1 geometric multiplicity of (3+21) = 1 .

Agam, for x= (3-2i) , Let Y= (22) be the eigen vector. AY= 19.5 = (3-21 Then 3-4/ 3 Y Y (34,-442 (3-ai)y, y, +342 (3-21) % ) from this, we get 34,-4Y2 = (3-2i)y, -492=-2iyi -> y = -2i 42. 4, +342= (3-21942. get: y = y=d, then , =) Also, from we -QiY2 if l-aid y= se 2 42 d d/-22 • So, the eigen space. Coppesponding to the eigen (3-2i) is -22 {{ :) value whose dimension. is 1. So, Algebraic multiplicity of (3-ai) is 1 end geometric mulltiplicity of (9-ai) is 1.

• Hence, We conclude that, for the matrix A Algebric multiplicity= Geometrie Multiplicity for all eigen values of A. Hence, the matrix is diagonlisable. the diagonal mateix is (* 3+ 2i o 3-2i Since, diagonal matrice coppesponding to. a matrix consists of the eigen values. as its diagonal elements and othera. elements are are zeno 0 Fors Matraix B= Tw 3 ) : Let d be the eigen value, then the Chanacteristie polynomial is . det (B-1 Iz) and equation is de+ (B-1I2) = 8 3-) -7 -1 1- = => 13-1 (1-1)+1 =) 3-x-3x+2²+1=0 =) x²-42+4=0 =) (1-2) (+-2)=0

• Therefore the characteristic polynomial. of B is 2-42+420 . All eigen values of B are. 2.9.2 Algebooie multiplicity of the eigen value a=2 is 2 T it's pepea tation is 2 Now, if xa al be the eigen vectons. Connesponding the eigen value d= 2. then BX = 2x - کر = ) 06) 지 2 ㅓ 22 3x17 22 () 224 222 -X, +42 .. we get 321+22=axi = x=-22 from -x+22=222 we get = x=-22 Taking C, we get X= AI SU (2)

c.e, X= .@ • (-1) [+ {(:) • the basis of the eigen space corresponding to 1= 2 is on its dimension is = 1 So, we can conclude that. mit Algebooie multplieity of 2 = 2 geometrie multiplieity of 2 = 1 Hence, they are not equal.. Therefore, the matrix B= is. not diagonalisatie 3 1 ) -1 Х

To show diagonalisability of a matrix, we have used the fact that

" A matrix is diagonalisable if and only if its algebraic multiplicity = geometric multiplicity corresponding to each eigen value of the matrix"

Using this fact we conclude that matrix A is diagonalisable but B is not.

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