Let A=I_2-2 MM^{T},where M is a real matrix o | vedclass

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(B) Given $A=I_2-2 MM^{T}$,where $M^T M=I_1=1$.First,we calculate $A^2$:$A^2 = (I_2-2 MM^{T})(I_2-2 MM^{T})$$= I_2 - 2 MM^{T} - 2 MM^{T} + 4 MM^{T} MM

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$2$

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(B) Given $A=I_2-2 MM^{T}$,where $M^T M=I_1=1$.First,we calculate $A^2$:$A^2 = (I_2-2 MM^{T})(I_2-2 MM^{T})$$= I_2 - 2 MM^{T} - 2 MM^{T} + 4 MM^{T} MM^{T}$Since $M^T M = 1$,we have $M^T M M^T = (M^T M) M^T = 1 \cdot M^T = M^T$.So,$A^2 = I_2 - 4 MM^{T} + 4 M(M^T M) M^T = I_2 - 4 MM^{T} + 4 MM^{T} = I_2$.Given $AX = \lambda X$ for a non-zero matrix $X$,we have:$A^2 X = A(\lambda X) = \lambda(AX) = \lambda^2 X$.Since $A^2 = I_2$,we have $I_2 X = \lambda^2 X$,which implies $X = \lambda^2 X$.Since $X 
eq 0$,we must have $\lambda^2 = 1$,so $\lambda = 1$ or $\lambda = -1$.The possible values for $\lambda$ are $1$ and $-1$.The sum of the squares of all possible values of $\lambda$ is $(1)^2 + (-1)^2 = 1 + 1 = 2$.

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