# find the subspaces of $\mathbb{C}^2$ invariant under $T$

Let $$T$$ be the linear operator on $$\mathbb{C}^2$$ ,the matrix of which in the standard order basis is

$$A= \begin{bmatrix} 1 & -1 \\ 2 & 2 \\ \end{bmatrix}$$

find the subspaces of $$\mathbb{C}^2$$ invariant under $$T$$

My attempt : The characteristic polynomial of $$A= |A-Ix|=(x-1)(x-2) + 2=x^2-3x+4$$

This is a parabola opening upwards with vertex $$(\frac{3}{2},\frac{7}{4})$$, so it has no real root

Here root of $$x^2-3x +4$$ is $$\frac{3 \pm \sqrt 7i}{2}$$

Also, we know that if $$v$$ is an eigenvector of $$T$$ with eigenvalue $$\lambda$$ ,then its span will be an invariant subspace of $$T$$

$$\implies$$ If $$W$$ is an invariant subspace of $$\mathbb{C}^2$$ then dim $$W=1= \dim(\mathbb{C})$$

Therefore the only subspaces of $$\mathbb{C}^2$$ invariant under $$T$$ are $$\mathbb{C}$$ and the zero susbspace

Is it true ?

One dimensional subspaces of $$\mathbb C^{2}$$ are isomorphic to $$\mathbb C$$ but they are not equal to $$\mathbb C$$. In fact, $$\mathbb C$$ is not a subset of $$\mathbb C^{2}$$. You have to find the eigen vectors corresponding to the eigen values you have found and the spans of these vectors give you the one dimensional subspaces which are invariant. You should also observe that $$\mathbb C^{2}$$ is itself an invariant subspace.
• okay kavi sir .That means all of the invariant subspaces are $\{(0,0)\},\mathbb{C}$ and $\mathbb{C^2}$. Am i right ? Commented Sep 25, 2021 at 4:48