Matrices Question 2
Question 2 - 2024 (01 Feb Shift 2)
Let $A=I_{2}-M M^{T}$, where $M$ 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 $\mathrm{X}$ of order $2 \times 1$, then the sum of squares of all possible values of $\lambda$ is equal to :
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Answer (2)
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 $=2$
