Answer to Question #198128 in Linear Algebra for sabelo Zwelakhe Xu

Question

Suppose V1,V2,VM is linearly independent in V and W€V.Prove that dim span(V1+W1,V2+W,.....VM+W)> or equal to m-1.

2.Suppose U1,U2,.....Um are finite dimensional subspaces of V.Prove that U1+U2+.....Um is finite dimensional and dim(U1+U2+.....Um)<or equal to dim U1+dim U2+.......dim Um.

Answer 1

$PROOF1\\Prove\thickspace that\\dim\thickspace Span(v1+w,...,vm+w)≤m−1.\\It\thickspace should\thickspace be\thickspace dim\thickspace Span(v1+w,...,vm+w)≥m−1.\\By\thickspace pluggingin\thickspace w=0,\\you\thickspace could\thickspace have\thickspace seenth\thickspace at\thickspace the\thickspace statement\thickspace of\thickspace the\thickspace problem\thickspace is\thickspace impossible.\\Suppose\thickspace dimSpan(v1+w,...,vm+w)≤m−2.\\\thickspace Then\thickspace the re\thickspace are\thickspace at\thickspace least\thickspace two\thickspace vectors\thickspace that\thickspace can\thickspace be\thickspace written\thickspace as\thickspace a\thickspace linear\thickspace combination\thickspace of\thickspace the\thickspace others.\\Without\thickspace loss\thickspace of\thickspace generality,\thickspace let\thickspace them\thickspace be\thickspace v1+w\thickspace and\thickspace v2+w. \\Thus\thickspace we\thickspace can\thickspace write:\\ v1+w=a3(v3+w)+···+am(vm+w)\\ v2+w=b3(v3+w)+···+bm(vm+w)\\ Isolate\thickspace w\thickspace to\thickspace get:\\ w(1−a3−...−am)=−v1+a3v3+···+amvm\\ w(1−b3−...−bm)=−v2+b3v3+···+bmvm\\ Note\thickspace that\thickspace 1−a3−...−am=0\thickspace and\thickspace 1−b3−...−bm=0\thickspace since\thickspace otherwise\thickspace v1\thickspace and\thickspace v3 would\thickspace both\thickspace be\thickspace written\thickspace as\thickspace linear\thickspace combinations\thickspace of\thickspace v3,...,vm.\\ Thus:\\ w=\dfrac{1}{1−a3−...−am}(−v1+a3v3+···+amvm)\\ w=\dfrac{1}{1−b3−...−bm}(−v2+b3v3+···+bmvm)\\ Subtracting\thickspace one\thickspace from\thickspace the\thickspace other:\thickspace \\ \dfrac{1}{1−a3−...−am}(−v1+a3v3+···+amvm)−\dfrac{1}{1−b3−...−bm}(−v2+b3v3+···+bmvm)=0\\ with\thickspace not\thickspace all\thickspace coefficients\thickspace zero\thickspace (in\thickspace particular,\thickspace the\thickspace coefficients\thickspace of\thickspace v1\thickspace and\thickspace v2\thickspace are\thickspace non\thickspace zero.\thickspace \\This\thickspace contradicts\thickspace the\thickspace linear\thickspace independence\thickspace of\thickspace v1,...,vm.\\ Thus\thickspace dim\thickspace Span(v1+w,...,vm+w)≥m−1. \\ PROOF2\\ Proof\thickspace by\thickspace induction\thickspace on\thickspace m.\\ Base\thickspace case.\\\thickspace In\thickspace the\thickspace m = 1\thickspace case,\thickspace this\thickspace formula\thickspace reduces\thickspace to\thickspace dim(U1) ≤ dim(U1), which\thickspace is\thickspace trivial.\\ Inductive\thickspace step.\\ We\thickspace assume\thickspace that\thickspace dim(U1 +··· +Um) ≤ dim(U1)+···+dim(Um) and\thickspace we\thickspace will\thickspace prove\thickspace that\thickspace dim(U1 +··· +Um +Um+1) ≤ dim(U1)+···+dim(Um)+dim(Um+1).\\\thickspace Let\thickspace W =U1+···+Um.\\\thickspace By\thickspace a\thickspace theorem\thickspace in\thickspace Axler\thickspace and\thickspace our\thickspace inductive\thickspace hypothesis,\thickspace dim(W +Um+1) = dim(W)+dim(Um+1)−dim(W ∩Um+1) ≤dim(W)+dim(Um+1) ≤(dim(U1) +··· +dim(Um))+dim(Um+1), as\thickspace desired.\\\thickspace Therefore,\thickspace by\thickspace the\thickspace PMI,\thickspace the\thickspace inequality\thickspace holds\thickspace for\thickspace every\thickspace m ≥ 1.\\$