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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?

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

1 Answer 1

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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 $w_1,\ldots,w_k$.

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