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Exercise:
Let T:Vlongrightarrow W be a linear map where V is finite dimensional. Then textdim KerT+textdim ImTtextdimV.

Solution:
Proof. Write n:textdimV. Let u_...u_k be a basis for textKerT. Since textKerT is a subspace of V it is also finite dimensional. Ext u_...u_k to a basis u_...u_kv_...v_n-k of V. bf By a previous Lemma: textImTtextSpTu_...Tu_kv_...v_n-ktextSpTv_...Tv_n-k We claim that Tv_...Tv_n-k form a basis for textImT. Indeed we have just proven that Tv_...Tv_n-k span textImT so it remains to show that Tv_...Tv_n-k are linearly indepent. To see this let a_...a_n-kin K and ase _i^n-ka_iTv_i Since T is linear Tleft_i^n-ka_iv_iright_i^n-ka_iTv_i &Longrightarrow _i^n-ka_iv_i in textKerT textNow u_...u_kquad textis a basis of KerT. &Longrightarrow exists b_...b_k in K : a_v_+...+a_n-kv_n-kb_u_+...+b_ku_k &Longrightarrow a_v_+...+a_n-kv_n-k+-b_u_+...+-b_ku_k. But u_...u_kv_...v_n-k are linearly indepent they form a basis of V. Longrightarrow a_...a_n-k b_...b_k. This proves that Tv_...Tv_n-k form a basis for textImT. bf Conclusion: textdim ImTn-ktextdimV-textdim KerT &Longrightarrow textdim KerT+textdim ImTtextdimV
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Exercise:
Let T:Vlongrightarrow W be a linear map where V is finite dimensional. Then textdim KerT+textdim ImTtextdimV.

Solution:
Proof. Write n:textdimV. Let u_...u_k be a basis for textKerT. Since textKerT is a subspace of V it is also finite dimensional. Ext u_...u_k to a basis u_...u_kv_...v_n-k of V. bf By a previous Lemma: textImTtextSpTu_...Tu_kv_...v_n-ktextSpTv_...Tv_n-k We claim that Tv_...Tv_n-k form a basis for textImT. Indeed we have just proven that Tv_...Tv_n-k span textImT so it remains to show that Tv_...Tv_n-k are linearly indepent. To see this let a_...a_n-kin K and ase _i^n-ka_iTv_i Since T is linear Tleft_i^n-ka_iv_iright_i^n-ka_iTv_i &Longrightarrow _i^n-ka_iv_i in textKerT textNow u_...u_kquad textis a basis of KerT. &Longrightarrow exists b_...b_k in K : a_v_+...+a_n-kv_n-kb_u_+...+b_ku_k &Longrightarrow a_v_+...+a_n-kv_n-k+-b_u_+...+-b_ku_k. But u_...u_kv_...v_n-k are linearly indepent they form a basis of V. Longrightarrow a_...a_n-k b_...b_k. This proves that Tv_...Tv_n-k form a basis for textImT. bf Conclusion: textdim ImTn-ktextdimV-textdim KerT &Longrightarrow textdim KerT+textdim ImTtextdimV
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eth, hs22, lineare algebra, proof, rank, vector space
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(3, default)
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ENG (English)
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