Let A = I 2 - 2 M M T , where M is a real matrix of order 2 × 1 such that the relation M T M = I 1 holds. If is a real number such that the relation A X = X holds for some non-zero real matrix X of order 2 × 1 , then the sum of squares of a

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Solution

Q.
Let $A = I_{2} - 2 M M^{T},$ where $M$ is a real matrix of order $2 \times 1$ such that the relation $M^{T} M = I_{1}$ holds. If $λ$ is a real number such that the relation $A X = λ X$ holds for some non-zero real matrix $X$ of order $2 \times 1,$ then the sum of squares of all possible values of $λ$ is equal to _______ . [2024]

Ans.
(2)

Let $M = [\begin{matrix} a \\ b \end{matrix}]$

Then $M^{T} M = I_{1} \Rightarrow [\begin{matrix} a & b \end{matrix}] [\begin{matrix} a \\ b \end{matrix}] = [1]$

$\Rightarrow a^{2} + b^{2} = 1$ ...(i)

Now, $A = - 2 M M^{T} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] - 2 [\begin{matrix} a^{2} & a b \\ a b & b^{2} \end{matrix}]$

$\Rightarrow A = [\begin{matrix} 1 - 2 a^{2} & - 2 a b \\ - 2 a b & 1 - 2 b^{2} \end{matrix}]$

Also, Let $X = [\begin{matrix} x \\ y \end{matrix}]$

Then, $A X = λ X$

$\Rightarrow [\begin{matrix} 1 - 2 a^{2} & - 2 a b \\ - 2 a b & 1 - 2 b^{2} \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} λ x \\ λ y \end{matrix}]$

$\Rightarrow [\begin{matrix} (1 - 2 a^{2}) x - 2 a b y \\ - 2 a b x + (1 - 2 b^{2}) y \end{matrix}] = [\begin{matrix} λ x \\ λ y \end{matrix}]$

On comparing, we get

$(1 - 2 a^{2}) x - 2 a b y = λ x$ ...(ii)

$- 2 a b x + (1 - 2 b^{2}) y = λ y$ ...(iii)

From (ii) & (iii), we get

$(1 - 2 a^{2} - λ) (1 - 2 b^{2} - λ) = 4 a^{2} b^{2}$

$\Rightarrow 1 - 2 a^{2} - λ - 2 b^{2} + 4 a^{2} b^{2} + 2 b^{2} λ - λ + 2 a^{2} λ + λ^{2} = 4 a^{2} b^{2}$

$\Rightarrow λ^{2} - 2 λ (1 - a^{2} - b^{2}) + (1 - 2 a^{2} - 2 b^{2}) = 0$

$\Rightarrow λ = \frac{2 (1 - a^{2} - b^{2}) \pm \sqrt{4 {(1 - a^{2} - b^{2})}^{2} - 4 (1 - 2 a^{2} - 2 b^{2})}}{2}$

$\Rightarrow λ = \frac{2 (1 - a^{2} - b^{2}) \pm 2 (a^{2} + b^{2})}{2}$

$\Rightarrow λ = 1 - a^{2} - b^{2} + a^{2} + b^{2} or λ = 1 - a^{2} - b^{2} - a^{2} - b^{2}$

$\Rightarrow λ = 1 or λ = 1 - 2 (a^{2} + b^{2}) = 1 - 2 = - 1$ ( $∵$ from (i))

So, sum of squares of all possible values of $λ = {(1)}^{2} + {(- 1)}^{2} = 2$

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