Matrices & Determinants
Eigenvalues — Idempotent-Type Matrix
nta_pyq_2024_jan
Grade 12
Question:
Let $A=I_2-MM^T$, where $M$ is a real matrix of order $2\times1$ such that the relation $M^TM=I_1$ holds. If $\lambda$ is a real number such that the relation $AX=\lambda X$ holds for some non-zero real matrix $X$ of order $2\times1$, then the sum of squares of all possible values of $\lambda$ is equal to:
Step-by-Step Solution
Key Concept: Compute $A^2=(I_2-MM^T)^2=I_2-2MM^T+MM^TM M^T=I_2-2MM^T+MM^T=I_2-MM^T=A$ (using $M^TM=1$). So $A^2X=AX=\lambda X$ and $A^2X=\lambda^2X\Rightarrow\lambda^2=\lambda\Rightarrow\lambda=0$ or $1$. Wait: $A^2=I_2-MM^T=A\Rightarrow A^2X=\lambda AX=\lambda^2X$ and $A^2X=AX=\lambda X$. So $\lambda^2=\lambda$... Actually from $A^2=I_2$: $A^2=(I_2-MM^T)^2=I_2-2MM^T+M(M^TM)M^T=I_2-2MM^T+MM^T=I_2-MM^T$. Wait, $M^TM=I_1=1$ (scalar). So $(MM^T)(MM^T)=M(M^TM)M^T=MM^T$. So $A^2=I-2MM^T+(MM^T)^2=I-2MM^T+MM^T=I-MM^T=A$. Thus $\lambda^2=\lambda\Rightarrow\lambda=0$ or $1$.
$A=I_2-MM^T$, $M^TM=1$. $A^2=I_2$. $AX=\lambda X\Rightarrow A^2X=\lambda^2X=X\Rightarrow\lambda^2=1\Rightarrow\lambda=\pm1$. Sum of squares $=1+1=2$.
Correct Answer: 2