By Emmanuel Kowalski
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Extra resources for Linear Algebra
Example text
The solution sets of the equations Ax “ b and A1 x “ b1 are the same. There exists a matrix B P Mm,m pKq such that b1 “ Bb. Proof. It suffices to check this for a single elementary operation. For a row exchange, this is easy because we are only permuting the equations. Now consider pA1 , b1 q “ rowi,j,t ppA, bqq. Only the j-th equation is changed. The “old” pair of i-th and j-th equations is ai,1 x1 ` ¨ ¨ ¨ ` ai,n xn “ bi aj,1 x1 ` ¨ ¨ ¨ ` aj,n xn “ bj . The “new” pair is ai,1 x1 ` ¨ ¨ ¨ ` ai,n xn “ bi paj,1 ´ tai,1 qx1 ` ¨ ¨ ¨ ` paj,n ´ tai,n qxn “ bj ´ tbi .
Vi´1 , vi ` w, vi`1 , . . , vn q “ f pv1 , . . , vi´1 , vi , vi`1 , . . , vn q ` f pv1 , . . , vi´1 , w, vi`1 , . . , vn q. The element w satisfies 1¨w´ ÿ tj vj “ 0, j“i so by (2) the second term is equal to 0W . 5. For K “ Q, or R or C, or most other fields, one can in fact that the property (1) as definition of alternating multilinear maps. Indeed, if (1) holds, then when vi “ vj with i “ j, we get by exchanging the i-th and j-th arguments the relation f pv1 , . . , vn q “ ´f pv1 , .
6. Let V1 and V2 be two finite-dimensional vector spaces with dimpV1 q “ n and dimpV2 q “ m. Let Bi be an ordered basis of Vi . The map " HomK pV1 , V2 q ÝÑ Mm,n pKq TB1 ,B2 f ÞÑ Matpf ; B1 , B2 q is an isomorphism of vector spaces. In particular: (1) We have dim HomK pV1 , V2 q “ mn “ dimpV1 q dimpV2 q. (2) If two linear maps f1 and f2 coincide on the basis B1 , then they are equal. Proof. We write B1 “ pe1 , . . , en q, B2 “ pf1 , . . , fm q. The linearity of the map TB1 ,B2 is left as exercise.