1. Oct 25, 2012

### Artusartos

Proposition: Let V be an A-invariant subspace of $F^n$ $A \in M_(F)$. Let $B= v_1, ... , v_k$ be a basis of V.

Extend it to a basis $\bar{B} = v_1, ... , v_n$ of F^n.

Then $[T_A]_{\bar{B}} = \begin{bmatrix} [T_A|_v:V \rightarrow V]_B & X \\zeros & C \end{bmatrix}$

Where C is the matrix of $\mu_x : F_A^n/V \rightarrow F_A^n/V$ with respect to $\bar{v_{k+1}} , ... , \bar{v_n}$, which is a basis of $F_A^n /V$

EXAMPLE:

For $\begin{bmatrix} 2 & 1 \\0 & 3 \end{bmatrix}$, V=span(e_1) (already invariant)

$\mu _x : F^2/V \rightarrow F^2/V$has matrix [3] with respect ot hte basis e_2.

Why does it have matrix [3]? Where did that come from?

A-3I = $\begin{bmatrix} -1 & 1 \\0 & 0 \end{bmatrix}$

N(A-3I) = span(e_1 + e_2)

$A(e_1 + e_2) = \begin{bmatrix} 3 \\3 \end{bmatrix} = 3(e_1 + e_2)$

Why did they find the null space? And why is A being multiplied by the vector that spans the null space?

The extended basis is $\bar{B} = e_1, e_1 + e_2$

$[T_A]_{\bar{B}} = \begin{bmatrix} 2 & 0 \\0 & 3 \end{bmatrix}$