Given that $\chi_A(x)=x^3-ax^2+bx-c$
Find $\chi_B(x)$ For:
a)B=A-2I
b)$B=A^2$
For a) would you put x+2 in for x in the $\chi_A(x)$. As Det(XI-B)=Det(XI+2I-A)= det((X+2)I-A)
And b: Im not sure how to do but you have to square the eigenvalues?
Given that $\chi_A(x)=x^3-ax^2+bx-c$
Find $\chi_B(x)$ For:
a)B=A-2I
b)$B=A^2$
For a) would you put x+2 in for x in the $\chi_A(x)$. As Det(XI-B)=Det(XI+2I-A)= det((X+2)I-A)
And b: Im not sure how to do but you have to square the eigenvalues?
From $A^3-aA^2+bA-cI=0$ you get $$ A(A^2+bI)=aA^2+cI $$ Squaring both sides, $$ A^2(A^4+2bA^2+b^2I)=a^2A^4+2acA^2+c^2I \\ B(B^2+2bB+b^2I)=a^2B^2+2acB+c^2I \\ B^3+2bB^2+b^2B-a^2B^2-2acB-c^2I=0 \\ B^3+(2b-a^2)B^2+(b^2-2ac)B-c^2I=0. $$