Question

7. Let A be a 5 x 5 matrix such that 1 2 .40 3 3 6 0 9 3 • det(A+15) = 0 • Nul(A) is 3 dimensional. (a) (5 points) What is rank(A)? Explain the reason why. (b) (5 points) What are the cigenvalues of A? (c) (5 points) Write down the characteristic polynomial of A. (d) (5 points) Is A diagonalizable? Why or Why not?

Answer 1

--- Given Å iu 5x5 matrix. such that 1) Nulla) = 3 2.) det (A-(-1) Is) = 0 - 1 3.) A 3 w 1 2 3 ㅋㅋ a) Using rank-nullity theorem MarkCA) + Nul(A) = no. of colums of A Mank (A) + 3 = 5 rank (A) = 2 Sum of rank and mellity is equal to dimension of the matrix 6) As det (A-G-IIS) = 0 le. det (A-dis) = 0 x=-1 is eigenvelue of A Also get any ergen value & and eigen -vector X AX=ax from 3) 11

. 2 2 3 roma i t= 3 is another eigen-verkose volue. Also rank (A) = 2 < dimension of A : der 14 ) = 0 -1 del CA-OIS) = 0 Hence, doo is also an eigen-value. Therefore, eigen-values of A are = 0,-1,3 () As Nul (A) =3 and d=0 is egenvalue and other eigenvalues are d=-1,3 3. .. Characteristic polynomial of A to B (H1) (1-3) = 0 X 244-33- . 2A4-342= =) A5

i d) Adgebuc nultiplicite of t=0 is 3 Geometric multiplicity = Nul(A)= 3 :: Algebric = Geometric multiplicity for da Also, for t= -1 and 3 Algebric multiplicity - Geometric multiplicity - 1 As, Algebric multiplicity = Geometric multiplicity for all eigen-values A is diagnalizable Therefore,
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