# Invariant Subsapce of Real vector space

Let $$V$$ be a real vector space of dimension at least $$3$$ and let $$T\in \operatorname{End}_{\mathbb{R}}(V)$$. Prove that there is a non-zero subspace $$W$$, $$W\neq V$$, such that $$T(W)\subseteq W$$.

I can prove this using module theory by make $$V$$ into an $$\mathbb{R}[x]$$-module by defining $$x\cdot v=T(v)$$ for all $$v\in V$$. Then show that it can't be a simple $$\mathbb{R}[x]$$-module.

However, I would like to see more basic prove without using module stuffs.

• What if we take advantage of the fact that a real polynomial of odd degree has at least one real root? Dec 7 '21 at 10:40
• I think @kabenyuk that we also need to take into account the Cayley--Hamilton Theorem ;-) Dec 7 '21 at 10:41
• @kabenyuk I use that fact to argue that $V$ cant be a simple $\mathbb{R}[x]$-module. But I want to avoid using module stuffs.
– Edix
Dec 7 '21 at 10:45

The easiest way to prove the statement, to me, is to note that there must exist some basis for $$V$$ such that, in that basis, the matrix representing the mapping $$T$$ has a Jordan normal form, which for real matrices looks like this.

Note that for a single Jordan block $$J$$ of the form $$J=\begin{bmatrix}C & I &\\ &C&\ddots\\ &&\ddots&I\\ &&&C\end{bmatrix}$$

where the matrix $$C$$ is a $$2\times 2$$ matrix, you can see immediatelly that both $$Je_1$$ and $$Je_2$$ are linear combinations of $$e_1, e_2$$. In other words, if $$W$$ is the subspace, generated by the first two basis vectors, then $$T(W)\subseteq W$$.

• Why it is the case? I mean, what guarantees the existence?
– Edix
Dec 7 '21 at 10:55
• @Edix The existence of a Jordan normal form is a theorem of linear algebra.
– 5xum
Dec 7 '21 at 10:56
• Thank you, I got it!
– Edix
Dec 7 '21 at 13:07

In my view if you don't want the machinery of modules, then you don't want the machinery of the Jordan Normal Form, as that is best developed under module theory (which e.g. is how Artin's Algebra does it).

The simplest solution is to select a basis and then work with $$A\in \mathbb R^{n\times n}$$.

Now do a field extension and work over $$\mathbb C$$. By Fundamental Theorem of Algebra, $$A$$ has an eigenvalue $$\lambda$$. If $$\lambda \in \mathbb R$$ then this means there is some non-zero $$\mathbf v$$ such that $$A\mathbf v = \lambda \mathbf v$$ and you have uncovered a 1-dim $$A-$$invariant subspace and you are done. This of course includes the special case where $$\lambda =0$$-- so if $$\det\big(A\big)=0$$ then we are done.

Accordingly the rest of the proof assumes $$\det\big(A\big)\neq 0$$ and $$\lambda$$ is non-real. Then you have

$$A\mathbf v_1 = \lambda \cdot \mathbf v_1$$
where $$\mathbf v_1 \in \mathbb C^{n}$$ with at least one non-real component.
(If they were all real then $$\lambda \in \mathbb R$$. Similarly not all components are purely imaginary.) Now define
$$\mathbf v_2: = \overline{\mathbf v_1}$$
This is the component-wise complex conjugate of $$\mathbf v_1$$ and
$$A\mathbf v_1 = \lambda \cdot \mathbf v_1 \implies \mathbf v_2^TA^* = \mathbf v_1^*A^* = \bar{\lambda_1}\cdot \mathbf v_1^*= \bar{\lambda_1}\cdot \mathbf v_2^T\implies A \mathbf v_2 =\bar{\lambda_1}\cdot \mathbf v_2$$
since $$A$$ is real

$$\mathbf B:= \bigg[\begin{array}{c|c|c|c|c} \mathbf v_1 & \mathbf v_2\end{array}\bigg]\displaystyle \left[\begin{matrix}1 & i \\ 1 & -i\end{matrix}\right]$$

which has two linearly independent real vectors.
(Viewed as a basis for a subspace:) This is obviously $$A$$-invariant since
$$A\mathbf B=A\left(\bigg[\begin{array}{c|c|c|c|c} \mathbf v_1 & \mathbf v_2\end{array}\bigg]\displaystyle \left[\begin{matrix}1 & i \\ 1 & -i\end{matrix}\right]\right)=\left(A\bigg[\begin{array}{c|c|c|c|c} \mathbf v_1 & \mathbf v_2\end{array}\bigg]\right)\displaystyle \left[\begin{matrix}1 & i \\ 1 & -i\end{matrix}\right]$$
$$=\bigg[\begin{array}{c|c|c|c|c} \mathbf v_1 & \mathbf v_2\end{array}\bigg]\displaystyle \left(D\left[\begin{matrix}1 & i \\ 1 & -i\end{matrix}\right]\right)= \mathbf B C$$
for some $$C\in GL_2(\mathbb R)$$

To confirm this, check $$D\left[\begin{matrix}1 & i \\ 1 & -i\end{matrix}\right]= \left[\begin{matrix}1 & i \\ 1 & -i\end{matrix}\right]C$$ for some $$2\times 2$$ matrix $$C$$, and $$\dim \ker \big(A\mathbf B\big) =0$$ hence $$C$$ must be injective and thus invertible. Further $$\mathbf B$$ is real and injective so let $$\mathbf B_L^{-1}$$ be a left inverse, then $$C= \mathbf B_L^{-1}A\mathbf B$$ i.e. a product of 3 real matrices hence $$C\in GL_2(\mathbb R)$$

So by considering any $$\mathbf x \in \mathbb R^n$$, we see $$\mathbf B\mathbf x$$ generates a 2 dimensional $$A-$$invariant real sub-space.