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Let $A=I_2-2 \mathrm{MM}^{\mathrm{T}}$, where $\mathrm{M}$ is real matrix of order $2 \times 1$ such that the relation $M^T M=I_1$ holds. If $\lambda$ is a real number such that the relation $\mathrm{AX}=\lambda \mathrm{X}$ holds for some non-zero real matrix $X$ of order $2 \times 1$, then the sum of squares of all possible values of $\lambda$ is equal to:
$1$
$2$
$3$
$4$
Solution
$ \mathrm{A}=\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}} $
$ \mathrm{A}^2=\left(\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}}\right)\left(\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}}\right) $
$ =\mathrm{I}_2-2 \mathrm{MM}^{\mathrm{T}}-2 \mathrm{MM}^{\mathrm{T}}+4 \mathrm{MM}^{\mathrm{T}} \mathrm{MM}^{\mathrm{T}} $
$ =\mathrm{I}_2-4 \mathrm{MM}^{\mathrm{T}}+4 \mathrm{MM}^{\mathrm{T}} $
$ =\mathrm{I}_2 $
$ \mathrm{AX}=\lambda \mathrm{X} $
$ \mathrm{A}^2 \mathrm{X}=\lambda \mathrm{AX}$
$ \mathrm{X}=\lambda(\lambda \mathrm{X}) $
$ \mathrm{X}=\lambda^2 \mathrm{X} $
$\mathrm{X}\left(\lambda^2-1\right)=0 $
$ \lambda^2=1 $
$ \lambda= \pm 1$
Sum of square of all possible values $=2$
