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Let T: V !V be a linear operator. If S: V !V is another operator and Tcommutes with S, i.e., if [S;T] := ST TS= 0; (1) then both the range, R[S], and the eigenspaces, E [S] := fv2V jSv= vg (2) are T-invariant for any 2F. Indeed, if v2R[S], then v= Sufor some u2V and Tv= T(Su) = (TS)u= (ST)u= S(Tu) 2R[S]: Similarly, if v2E [S], then
Let \(V\) and \(W\) be vector spaces such that both have dimension \(n\) and let \(T: V \mapsto W\) be a linear transformation. Suppose \(B_1\) is an ordered basis of \(V\) and \(B_2\) is an ordered basis of \(W\).
Overview. Today we will discuss the algebra-geometry dictionary. 1. Vanishing ideals of subsets of n. A. 2. The Zariski topology. 3. Radical ideals and the strong Nullstellensatz. 4. Trick of Rabinowitch. 5. Prime ideal, maximal ideals and varieties. 6. Coordinate ring. 7. Morphisms between algebraic sets and varieties.
As a modest generalization of the invariance of T(V), we observe that if W ˆV is an invariant subspace, then T(W) ˆW so T(T(W)) ˆT(W) and thus T(W) is also an invariant subspace.
Solution: Let $\nu_1$, $\cdots$, $\nu_m$ be a basis of $V$, denote the first row of $\ca M(T)$ with respect to the bases $\nu_1$, $\cdots$, $\nu_m$ and $w_1$, $\cdots$, $w_n$ by $(a_1,\cdots,a_m)$. If $(a_1,\cdots,a_m)=0$, then we can choose $v_i=\nu_i$, $i=1,\cdots,m$.
F n, where V is a vector space of dimension n over F , given by T (v) = [v]B for any ordered basis B of B, is an isomorphism. Example. The coordinate vector of vj with respect to an ordered basis B = fv1; : : : ; vng of V is the standard basis vector [vj]B = ej of Rn.
Let V be a vector space over a ̄eld F and T 2 L(V; V ) be linear operator. A scalar ̧ 2 F is said to be a characterisitic value of T; if. (e) = ̧e for some e 2 V with e 6= 0: characterisitic value is also known an eigen value.
