# A linear map $T: \mathbb{R^3 \to \mathbb{R^3}}$ has a two dimensional invariant subspace.

Let $$T: \mathbb{R^3 \to \mathbb{R^3}}$$ be an $$\mathbb{R}$$-linear map. Then I want to show that $$T$$ has a $$2$$ dimensional invariant subspace of $$\mathbb{R^3}.$$

I considered all possible minimal polynomial of $$T$$ and applying canonical forms I found some obvious $$2$$ dimensional invariant subspaces.

I stuck when the minimal polynomial is of the form $$(X-a)^3$$ for some real number $$a.$$

In this situation since the minimal polynomial and the characteristic polynomial coincides $$T$$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.

• Juanra's proof is basically Gelfand's proof in his book Lectures on Linear Algebra, although Gelfand worked over $\mathbb C$. Menezio's proof is along the same vein but it replaces the inner product by a bilinear pairing. Thus it is cleaner and more general, as it works over other vector spaces without inner products. However, both proofs fail to mention that they work because any linear operator on an odd-dimensional real vector space has an eigenvalue. (In contrast, consider the $2\times2$ rotation matrix for the angle $\pi/2$; it has not any one-dimensional invariant subspace.) Nov 13, 2021 at 11:39
• math.stackexchange.com/questions/4310618/…
– Paul
Nov 19, 2021 at 18:14

There is a simpler (and cleaner) solution: the adjoint has a real eigenvalue with associated eigenvector $$v$$. Now, the orthogonal complement of $$span(v)$$ is a $$2$$ dimensional invariant subspace of $$T$$.

• Indeed it is simpler but what do you mean by cleaner (than the other) ? Dec 28, 2020 at 10:47
• Don't have to consider several cases. That is what I mean by "cleaner". Dec 28, 2020 at 20:02

If the minimal polynomial is of the form $$(X-a)^3$$, then there is some basis $$B=\{e_1,e_2,e_3\}$$ of $$\mathbb{R}^3$$ such that the matrix of $$T$$ with respect to $$B$$ is$$\begin{bmatrix}a&1&0\\0&a&1\\0&0&a\end{bmatrix}.$$So, consider the space spanned by $$e_1$$ and $$e_2$$.

• How can it be a diagonal matrix ? also second matrix how do you get ? Jan 10, 2019 at 11:41
• I wrote “minimal polynomial” but I was thinking about characteristic polynomials. I hope that everything is correct now. Jan 10, 2019 at 11:43

Consider the dual linear map $$f^t : \text{Hom}(\mathbb{R}^3,\mathbb R) \longrightarrow \text{Hom}(\mathbb R^3, \mathbb R), \qquad g\mapsto f^t(g) = g\circ f$$ This map has an eingenvector $$g\in \text{Hom}(\mathbb R^3, \mathbb R)$$ different from $$0$$. The kernel of $$g$$ is a two dimensional subspace of $$\mathbb R^3$$ and it is easy to show that $$f\left(\ker(g)\right) \subseteq \ker(g)$$.

The characteristic polynomial of the transpose map $$f^T$$ is cubic with real coefficients. Hence it has a real root $$\alpha$$, and there is a real eigenvector $$v=(a,b,c)$$ such that $$f(v)=\alpha v$$. Then the 2-dim space given by the equation

$$ax+by+cz=0$$ is invariant under $$f$$. That is because

$$\langle v, f(w)\rangle=\langle f^T(v),w\rangle=\alpha\langle v, w\rangle=0$$

for any vector $$w$$ in the subspace.