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Question

Let $A$ be a $4\times 4$ matrix with real entries such that $-1,1,2,-2$ are eigen values.If $B=A^4-5A^2+5I$ then trace of $A+B$ is...........

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Put every eigenvalue of A in the equation of B to get the respective eigenvalue of B [ The Cayley-Hamilton Theorem  ]

let b1, b2, b3, and b4 are the eigenvalues of B and a1, a2, a3 and a4 are the eigenvalues of A

a1 = -1, a2 = 1, a3 = 2, a4 = -2

b1 = (a1)^4 - 5(a1)^2 + 5 

b2 = (a2)^4 - 5(a2)^2 + 5

b3 = (a3)^4 - 5(a3)^2 + 5

b4 = (a4)^4 - 5(a4)^2 + 5

on putting values of a1,a2,a3 and a4, we get

b1 = 1, b2 = 1, b3 = 1 and b4 = 1

trace of a matrix = sum of eigenvalues 

trace(A + B ) = trace(A) + trace(B) 

trace(A) = -1 + 1 + 2 -2 = 0

trace(B) = 1 + 1 + 1 + 1 = 4

trace(A + B ) = 0 + 4 = 4

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is this correct or not?

Answer 1

Put every eigenvalue of A in the equation of B to get the respective eigenvalue of B [ The Cayley-Hamilton Theorem  ]

let b1, b2, b3, and b4 are the eigenvalues of B and a1, a2, a3 and a4 are the eigenvalues of A

a1 = -1, a2 = 1, a3 = 2, a4 = -2

$b_{1} = a_{1}^{4} - 5a_{1}^{2} + 5$

$b_{2} = a_{2}^{4} - 5a_{2}^{2} + 5$

$b_{3} = a_{3}^{4} - 5a_{3}^{2} + 5$

$b_{4} = a_{4}^{4} - 5a_{4}^{2} + 5$

on putting values of a1,a2,a3 and a4, we get

b1 = 1, b2 = 1, b3 = 1 and b4 = 1

trace of a matrix = sum of eigenvalues 

trace(A + B ) = trace(A) + trace(B) 

trace(A) = -1 + 1 + 2 -2 = 0

trace(B) = 1 + 1 + 1 + 1 = 4

trace(A + B ) = 0 + 4 = 4