Proposition: Let V be an A-invariant subspace of [itex]F^n[/itex] [itex]A \in M_(F)[/itex]. Let [itex]B= v_1, ... , v_k[/itex] be a basis of V.(adsbygoogle = window.adsbygoogle || []).push({});

Extend it to a basis [itex] \bar{B} = v_1, ... , v_n [/itex] of F^n.

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

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

EXAMPLE:

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

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

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

A-3I = [itex]\begin{bmatrix} -1 & 1 \\0 & 0 \end{bmatrix}[/itex]

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

[itex]A(e_1 + e_2) = \begin{bmatrix} 3 \\3 \end{bmatrix} = 3(e_1 + e_2)[/itex]

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 [itex] \bar{B} = e_1, e_1 + e_2 [/itex]

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

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# A question about modules

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