Cayley hamilton theorem for the following [7 4
-5 -2]
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Hope it will help you....
Step-by-step explanation:
Claim: Let A be an
n\times n matrix with characteristic equation
f(x)=0. Then
f(A)=0.
Proof: We prove it in the special case where
A is diagonalisable.
A is diagonalisable, so there exists a basis of eigenvectors
e_1,e_2,...,e_n. It will suffice to show that the linear map corresponding to
L= f(A) sends everything to zero.
Any vector can be written
v=v_1e_1+v_2e_2+....+v_ne_n, so by the linearity of
L it suffices to show that each component is mapped to the zero vector. But clearly
L(v_je_j) = v_jL(e_j) = v_j f(\lambda_j)e_j = 0e_j = 0, where
\lambda_j is the eigenvalue of
e_j.
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