Let \(M \in M_5(\RR)\) be a \(5\times 5\) square matrix with real coefficients defining a linear map \(L: \Bbb{R}^5 \to \Bbb R^5\). Assume that when considered as an element of \(M_5(\Bbb C)\), then the scalars \(0, 1+i, 1+2i\) are eigenvalues of \(M\).
Show that the associated linear map \(L\) is neither injective nor surjective.
Compute the characteristic polynomial and minimal polynomial of \(M\).
How many fixed points can \(L\) have?
(That is, how many solutions are there to the equation \(L(v) = v\) with \(v\in \Bbb R^5\)?)
Solution
Because \(M\) has real coefficients, every nonreal eigenvalue occurs together with its complex conjugate. Thus, besides the given eigenvalues \[ 0,\qquad1+i,\qquad1+2i, \] the numbers \[ 1-i,\qquad1-2i \] are also eigenvalues. These five eigenvalues are distinct.
1. The map \(L\) is neither injective nor surjective.
Since \(0\) is an eigenvalue, there exists \(0\ne v\in\RR^5\) with \[ L(v)=0. \] Hence \(\ker L\ne0\), so \(L\) is not injective. A linear map from a finite-dimensional vector space to itself is injective if and only if it is surjective, so \(L\) is not surjective either.
2. The characteristic polynomial is\[ \chi_M(x) =x(x-(1+i))(x-(1-i))(x-(1+2i))(x-(1-2i)). \]
A \(5\times5\) matrix has a degree-\(5\) characteristic polynomial counting eigenvalues with algebraic multiplicity. We already have five distinct eigenvalues, so these are all roots and each has multiplicity \(1\).
3. Over \(\RR[x]\), this becomes\[ \chi_M(x)=x(x^2-2x+2)(x^2-2x+5). \]
Pair the conjugate roots: \[ (x-(1+i))(x-(1-i))=(x-1)^2+1=x^2-2x+2, \] \[ (x-(1+2i))(x-(1-2i))=(x-1)^2+4=x^2-2x+5. \]
4. The minimal polynomial equals the characteristic polynomial.
2. Hence it has degree at least \(5\). But the minimal polynomial divides the degree-\(5\) characteristic polynomial, so equality holds. Therefore, over \(\RR\),\[ \mu_M(x)=x(x^2-2x+2)(x^2-2x+5). \]
5. The only fixed point of \(L\) is \(0\).
A fixed point satisfies \[ L(v)=v, \] i.e. \[ (M-I)v=0. \] A nonzero fixed point would make \(1\) an eigenvalue of \(M\). But the complete eigenvalue list above does not contain \(1\); equivalently, \[ \chi_M(1)=1\cdot1\cdot4=4\ne0. \] Thus \(\ker(M-I)=0\), so the only fixed point is the zero vector. Hence there is exactly one fixed point.