# Determining all invariant subspaces of an operator.

I saw the following question from M. Artin's book, Algebra.

I need to find all invariant subspaces of the real linear operator T whose matrix has column vectors $(1,0)$ and $(1,1)$ as its first and second columns.

I think that I need to find the eigenvector corresponding to eigenvalue $1$. Then the matrix of $T$ will have the block form described above. Then the eigenvector will be a basis for the invariant subspace.

I have no idea how to find all such invariant subspaces.

Any help is appreciated. Also, I don't know how to draw a matrix. If anyone can explain how to do that. $:)$

Given a linear operator $T$ acting on a vector space $V$ an invariant subspace of $T$ is a subspace $W$ of $V$ such that $T(W)\subseteq W$. So, We always have two invariant subspaces: $Ker (T)$ and $Range(T)$.

In this case we can easily see that $Ker (T)=\{0\}$ and $Range(T)=\mathbb{R}^2$ that are two trivial invariant subspaces of dimension $0$ and $2$.

Invariant subspaces of dimension $1$ are spanned by a vector $x$ such that $T(x)=\lambda x$, i.e. eigenvectors of the matrix that represents $T$:

$$\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} = \lambda \begin{bmatrix} x_1\\ x_2 \end{bmatrix}$$

The matrix is exactly in Jordan canonical form, so we see immediately that it has an eigenvalue $\lambda=1$ with algebraic multiplicity $2$ and only one proper eigenvector $[1,0]^T$ that span the eigenspace $W=\{[k,0]^T\}$, so this is an invariant subspace of dimension $1$. And there are no other invariant subspaces.

About the last question: edit this answer and look how I've write the matrices.

Without using Jordan form.

An eigenvector of the eigenvalue $\lambda=1$ satisfies the equation: $$\begin{bmatrix} 1&1\\ 0&1 \end{bmatrix} \begin{bmatrix} x_1\\ x_2 \end{bmatrix} = 1 \begin{bmatrix} x_1\\ x_2 \end{bmatrix}$$ that is : $$\begin{cases} x_1+x_2=x_1\\ x_2=x_2 \end{cases}$$ that has solution $x_2=0 \;\land\;x_1=k \, \forall k \in \mathbb{R}$. This means that all vectors of the form $[k,0]^T$ are solutions, i.e. are elements of the eigenspace, and as an eigenvector we can chose $[1,0]^T$.

• I understood that the Kernel and Range are invariant subspaces of dimension 0 and 2 respectively. I also get that eigenvectors span invariant subspaces of dimension 1. But since I haven't studied Jordan form, I was thinking of an explanation which doesn't use it. This question is from the section of eigenvectors. Is there a simpler method to find invariant subspaces of dimension 1? – user264750 Oct 19 '15 at 12:55
• I've added something to my answer. I hope it's usefull :) – Emilio Novati Oct 19 '15 at 13:12
• How did you so easily see that the kernel is trivial? I had to use the matrix to write down the transformation explicitly as T(x,y)=(x+y,y) and hence find the kernel. Is there a way we can tell the kernel is trivial from the matrix directly? – user264750 Nov 14 '15 at 6:43
• I see that the matrix has non null determinant so $T(x)=0$ has only the null solution. – Emilio Novati Nov 14 '15 at 8:30

A common way to find eigenvektors $v$ for an eigenvalue $\lambda$ is:

$$T v = \lambda I v \Longleftrightarrow (T - \lambda I) v = 0$$

Calculate $T - \lambda I$ and solve the equation. Can you take it from here?

• I may be wrong but going by the question the OP knows how to find the null space eigenvector. – Chinny84 Oct 19 '15 at 8:22
• That null space (and all its subspaces) should be exactly what he's looking for. Something missing? – Niklas Oct 19 '15 at 8:34
• I wasn't saying what you did was wrong - but I think he wants to find all invariant subspaces. Can you show that? (ps. This is not my area of expertise) – Chinny84 Oct 19 '15 at 9:01