An n×n matrix A has a fixed point if there is a vector v in Rn with A*v = v.

Question

1)Show that every such A has at least one fixed point.

2) The identity matrix has infinitely many fixed points. For all arbitrary 2 × 2 matrices A, show that if A has two distinct fixed points v and w in R2 , then A has infinitely many fixed points.

3) In the same situation as 2), show that if A is also invertible and neither vector v nor vector w is a multiple of the other, then A must be the identity matrix.

Answer 1

1. Consider v=0 (the 0 vector). Then for any nxn matrix A, Av = A0 = 0 = v.
2. Av=v and Aw=w => A(v+w)=Av+Aw=v+w=(v+w) thus the sum of any two fixed points is a fixed point. Furthermore, since v and w are distinct, then at most one of them is the 0 vector. Suppose v is the 0 vector, then w is not the 0 vector, so w+w=2w is also a fixed point and 2w≠w or v. So A has infinitely many fixed points.
3. If neither v nor w is a multiple of the other then it must be true that span{v,w} is all of R². Thus any vector x in R² can be written as linear combination of v and w: x = c₁v+c₂w where c₁ and c₂ are fixed real numbers. Therefore, for every x in R², Ax = A(c₁v+c₂w) = A(c₁v)+A(c₂w) = c₁(Av)+c₂(Aw) = c₁v+c₂w = x. Furthermore, since A is invertible, the following is also true: Ax = x => A^-1(Ax)=A^-1x => A^-1x= x. Thus, A has the exact properties of the Identity matrix on R², and since the Identity matrix is unique, A must equal the identity matrix.

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