# Do all linear operators with an invariant space, other than the kernel and range, have an upper triangular matrix from?

I was reading Axler's book, and was wondering why he starts a chapter discussing Eigenvalues and Upper Triangular matrices with a discussion about invariant subspaces. I was trying to understand the deep connection here.

Say $T$ is some linear operator $T: \mathscr{L}(V)$ on some vector space $V$. Let $U \subset V$. Then $T$ is invariant on $U$ if $Tu \in U$.

So all linear operators such as $T: \mathscr{L}(V)$ have atleast two invariant subspaces: the kernel and the range of the operator. But that is not what I am interested in.

So obviously the existence of some invariant subspace other than the null and kernel is causing some operators to have eigenvalues. I was just wondering if all operators with some invariant subspace other than the null and kernel also can take an upper triangular form?

Again, I am just trying to understand the relationship between invariant subspaces and upper triangular matrices.

In Linear Algebra Done Right, I started the chapter on eigenvectors and eigenvalues with a brief discussion of invariant subspaces to provide motivation. To understand an operator from a finite-dimensional vector space to itself, one can try to decompose the vector space into a direct sum of subspaces of smaller dimension and then study the operator restricted to each of those subspaces. Those restrictions are operators (meaning that they map each subspace into itself) if and only if the subspaces are invariant under the operator.

The simplest possible nonzero invariant subspaces are one-dimensional subspaces. Thinking about what it means for a one-dimensional subspace to be invariant immediately leads to the notion of eigenvector and eigenvalue.

For operators on complex finite-dimensional vector spaces, this approach culminates in a beautiful decomposition of the vector space as a direct sum of generalized eigenspaces; see Theorem 8.21 (in the third edition).

• Cool, Prof. Axler. Thanks for responding. Yes, I can see the connection now. While not all linear operators generate invariant subspaces, the chapter shows how the the structure of the invariant subspaces affects the structure of the operator--or the matrix generated by the operator(or vice versa). So the theorem every operator in a complex finite-dimensional vector space has an upper triangular matrix depends on the existence of at least one invariant subspace in the operator--as given by the fundamental theorem of algebra. – krishnab Feb 23 '17 at 9:40

The relation between invariant subspaces and having an upper triangular form is best expressed using the notion of a flag. Given a finite dimensional vector space $V$ of dimension $n$, a (full) flag on $V$ is just a sequence $F = (V_k)_{k=0}^n$ of subspaces

$$\{ 0 \} = V_0 \subsetneq V_1 \subsetneq \dots \subsetneq V_n = V$$

such that $\dim V_i = i$. That is, we choose a nested sequence of subspaces of $V$ that starts with $\{ 0 \}$, ends with $V$ and such that the dimension of the subspaces jumps by one each step. A flag $F$ is called $T$-invariant if each $V_i$ is $T$-invariant.

Now assume you know that $T$ takes an upper triangular form with respect to the basis $(v_1,\dots,v_n)$ of $V$. Define a flag $F$ on $V$ by $V_i := \operatorname{span}(v_1,\dots,v_i)$. Then the fact that $T$ is upper triangular immediately implies that each $V_i$ is $T$-invariant so $F$ is $T$-invariant. Conversely, if there exists a $T$-invariant flag $F = (V_i)_{i=0}^n$ then by choosing arbitrary $v_i \in V_i \setminus V_{i-1}$ we get a basis $(v_1,\dots,v_n)$ of $V$ with respect to which $T$ is upper triangular.

In general, if you know that $T$ has a non-trivial invariant subspace, this is not enough to deduce that $T$ can be represented by an upper triangular matrix. It's not even enough to know that $T$ has $T$-invariant subspaces of all possible dimensions. For counterexamples, think about orthogonal $3 \times 3$ real matrices.

• This is really helpful. I see how the nesting of subspaces allows us to develop some very nice proofs. Thanks for the insight. I had not heard of flags before. – krishnab Feb 23 '17 at 9:41