# Invariant subspace of linear mapping

Let $$T$$ be a linear transformation on a complex vector space $$V$$ of dimension $$4$$ and let $$\lambda_1,\lambda_2,\lambda_3$$ be distinct eigenvalues of $$T$$.

Eigenvectors are: $$\lambda_1\rightarrow \{v_1,v_2\}$$, $$\lambda_2\rightarrow \{v_3\}$$, and $$\lambda_3\rightarrow \{v_4\}$$.

I want to find how many invariant subspaces I have.

I know that $$V(\lambda_2)$$ , $$V(\lambda_3)$$ , $$V(\lambda_2)+V(\lambda_3)$$ are distinct invariant subspaces.

Whats about the $$V(\lambda_1)$$ how many invariant subspaces I have?

If $$\alpha\in\mathbb{C}$$, then $$\langle v_1 + \alpha v_2\rangle$$ is invariant subspace?

• $V_1=\{v\in\operatorname{dom}T:Tv=x_1v\}$ is invariant Aug 13, 2021 at 12:34
• Yes , but whats about $\alpha\in\mathbb(C)$ $V_1 + \alpha*V_2$ ? Aug 13, 2021 at 12:42
• I am unfamiliar with this "$C$" notation Aug 13, 2021 at 12:42
• I meant complex numbers Aug 13, 2021 at 12:47
• First list all invariant subspaces of dimension 1, then dimension 2, and so on. Aug 13, 2021 at 12:52

Hints: The one-dimensional invariant subspaces: $$L(v_3)$$, $$L(v_4)$$, $$L(v_2)$$, $$L(v_1+\alpha v_2)$$, $$\alpha\in\mathbb{C}$$.
The two-dimensional invariant subspaces: $$V(\lambda_1)$$, $$L(w,v_3)$$, $$L(w,v_4)$$, $$L(v_3,v_4)$$, $$w\in V(\lambda_1)$$, $$w\neq0$$.
The three-dimensional invariant subspaces: $$L(v_1,v_2,v_3)$$, $$L(v_1,v_2,v_4)$$, $$L(w,v_3,v_4)$$, $$w\in V(\lambda_1)$$, $$w\neq0$$.
$$L(w_1,\ldots,w_k)$$ is a linear envelope of vectors $$w_1,\ldots,w_k$$, i.e. the subspace spanned by $$w_1,\ldots,w_k$$.