# Basis for a linear map

Let $V$ be a vector space over $\mathbb C$ with $\dim V=n$ and $F\colon V\to V$ be a linear map.

(a) Show that there always exists a basis $\{v_1,\ldots,v_n\}$ such that $F(v_j)$ is in space $W_j$ which is generated by $\{v_1,...,v_j\}$

(b) Is part (a) true if we consider a linear map $F\colon V\to V$ where $V$ is a vector space over $\mathbb R$? Justify your answer by giving a counterexample or a proof.

I want to solve this question T T

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Are you familiar with A) eigenvectors, B) Jordan canonical form, C) rotations? – Jyrki Lahtonen Apr 28 '12 at 19:31
Use induction. See for instance Schur decomposition. – lhf Apr 28 '12 at 19:52

In general: if there exists such a basis , then the first element of it is an eigenvector of the map, since it must be $\,Fv_1\in \langle\,v_1\,\rangle :=Span\{v_1\}\Longleftrightarrow \exists\,\lambda\in\mathbb F\,\,s.t.\,\,Fv_1=\lambda v_1\Longleftrightarrow v_1\,$ is an eigenvector of $\,F\,$ , so we begin with the easy one:
$(b)\,\,$ Take the map $$F:\mathbb R^2\longrightarrow \mathbb R^2\,\,,\,\,F\binom{x}{y}:=\binom{\;y}{\!\!-x}$$whose characteristic polynomial is $\,x^2+1\,$. As this pol. is irreducible over $\,\mathbb R\,$ the map $\,F\,$ has not eigenvalues and by the first part we cannot have a basis as required.
$(a)\,\,$ Induction on $\,n:=\dim_\mathbb{C}V\,\,$ : for $\,n=1\,$ any basis of $\,V:=\langle\,v\,\rangle\,$ (to be sure, $\,0\neq v\in V\,$) will do.
Let us now assume the claim is true for any vector space of dimension $\,k\leq n-1\,$ and let us take a complex vector space $\,V\,$ of dimension $\,n\,$ . Since $\,F\in\operatorname{End}_\mathbb{C}(V)\,$ and $\,\mathbb C\,$ is algebraically closed, the characteristic polynomial of $\,F\,$ has a root, say $\,\lambda\,$, corresponding to a (non-zero, of course) eigenvector $\,v_1\,\,:\,\,Fv_1=\lambda v_1\,$ .
Complete now the linearly independent set $\,\{v_1\}\,$ to a basis $\,\mathcal B:=\{v_1,w_2,\ldots,w_n\}\,$ of $\,V\,$ , and take $\,U:=\langle\,w_2,\ldots,w_n\,\rangle\,$ ($\,U\,$ is, in fact, the kernel of the element of the dual basis of $\mathcal B\,$ not vanishing on $\,v_1\,$) . By the inductive hypothesis, there exists a basis $\,\{v_2,\ldots\,v_n\}\,$ of $\,U\,$ s.t. $$Fv_j\in\langle v_2,\ldots,v_j\,\rangle\,,\,\forall\,j=2,3,...,n$$...and so we're done, since $\,\langle v_2,\ldots,v_j\,\rangle\,\leq\,\langle\,v_1,v_2,\ldots,v_j\,\rangle$