Consider the natural action of on , the set of  real matrices, by left multiplication.

(i) Under this action, if , show that the stabiliser of   is {I}, where I is the 2 × 2 identity matrix.

(ii) Suppose that det(x) = 0 in the remaining parts of this exercise. We will show that the stabiliser of x is infinite. If x = 0, the stabiliser of x is . So suppose x ≠ 0. Let us write . Then,   for non-zero . Why ?

(iii) Let  be a vector that is not a scalar multiple of  . Show that there is a matrix  such that  and  .(Hint: Set up two sets of simultaneous equations in two unknowns and argue why they have a solution.)

(iv) Check that I−b is in the stabiliser of x. Also, show that there are infinitely many choices of   for which I − b is invertible.

Let's address each part of the exercise:

(i) To show that the stabilizer of \( x \) is \({I}\) when \( \text{det}(x) \neq 0 \), let \( g \in GL_2(\mathbb{R}) \) be such that \( gx = x \). Since \( gx = x \), then \( gx - x = 0 \), which implies \( (g - I)x = 0 \). Since \( \text{det}(x) \neq 0 \), \( x \) is invertible. Thus, \( (g - I)x = 0 \) implies \( g - I = 0 \), i.e., \( g = I \). Therefore, the ___ ___ _______ ___ ___ ____ ______ ___.
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