Subspaces and dimensions
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Exercise:
Let V be a finite dimensional vector space over K. Then every subspace mathcalU subseteq V is also finite dimensional. Moreover dim mathcalU leq dimV with equality iff mathcalUV.
Solution:
Proof. Denote n:dimV. Let mathcalU subseteq V be a subspace. We'll first show that mathcalU has a finite basis. If mathcalU then varnothing is the basis of mathcalU and we are done. So ase now mathcalUneq. Take v_ in mathcalU backslash. If mathcalU Spv_ we get a basis for mathcalU. If mathcalUneq Spv_ then take v_in mathcalUbackslash Spv_. By Lemma E v_v_ are linearly indepent. We continue this process and after j steps we obtain a list v_...v_j in mathcalU that is linearly indepent. If Spv_...v_jmathcalU we are done. Otherwise we choose v_j+in mathcalUbackslash Spv_...v_j. Again by Lemma E v_...v_jv_j+ are linearly indepent. However in V a list of linearly indepent vectors can have length n at most Longrightarrow The procedure described above must after not more than n steps say kleq n steps. And we obtain a list v_...v_k in mathcalU kleq n of linearly indepent vectors s.t. Spv_...v_kmathcalU. By definition v_...v_k is a basis for mathcalU Longrightarrow dim mathcalUkleq ndimV. Now if mathcalUV then clearly dim mathcalUdimV. Conversely suppose dim mathcalUdimV Longrightarrow kn. The list of vectors v_...v_nin mathcalU is a basis for mathcalU hence are linearly indepent also when considered in V. By the theorem shown before v_...v_n form a basis for V Longrightarrow mathcalUSpv_...v_nV.
Let V be a finite dimensional vector space over K. Then every subspace mathcalU subseteq V is also finite dimensional. Moreover dim mathcalU leq dimV with equality iff mathcalUV.
Solution:
Proof. Denote n:dimV. Let mathcalU subseteq V be a subspace. We'll first show that mathcalU has a finite basis. If mathcalU then varnothing is the basis of mathcalU and we are done. So ase now mathcalUneq. Take v_ in mathcalU backslash. If mathcalU Spv_ we get a basis for mathcalU. If mathcalUneq Spv_ then take v_in mathcalUbackslash Spv_. By Lemma E v_v_ are linearly indepent. We continue this process and after j steps we obtain a list v_...v_j in mathcalU that is linearly indepent. If Spv_...v_jmathcalU we are done. Otherwise we choose v_j+in mathcalUbackslash Spv_...v_j. Again by Lemma E v_...v_jv_j+ are linearly indepent. However in V a list of linearly indepent vectors can have length n at most Longrightarrow The procedure described above must after not more than n steps say kleq n steps. And we obtain a list v_...v_k in mathcalU kleq n of linearly indepent vectors s.t. Spv_...v_kmathcalU. By definition v_...v_k is a basis for mathcalU Longrightarrow dim mathcalUkleq ndimV. Now if mathcalUV then clearly dim mathcalUdimV. Conversely suppose dim mathcalUdimV Longrightarrow kn. The list of vectors v_...v_nin mathcalU is a basis for mathcalU hence are linearly indepent also when considered in V. By the theorem shown before v_...v_n form a basis for V Longrightarrow mathcalUSpv_...v_nV.