Basis, row-rank, Gauss
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Exercise:
Let B in M_mtimes n be a matrix which is in reduced row echelon form. Then the rows of B that are not totally form a basis for Row SB. In particular row-rankB equals the number of non-zero rows in B number of pivots in B.Also the pivot columns of B form a basis for Col SB. In particular for matrices B that are in reduced row echelon form we have row-rankBcol-rankB.
Solution:
Proof. Denote by j_ ... j_r the column numbers of the pivots. Denote by u_...u_r the non-zero rows of B. We'll first show that u_...u_r are lineary indepent. Indeed ase that _i^r lambda_i u_i this is a vector in K^n for some lambda_...lambda_r in K. The entry number j_k in this vector above is precisely lambda_k because in col#j_k we have everything except a in entry k that comes from the vector u_k. Since _i^r lambda_i u_i we obtain that lambda_k. This holds forall leq k leq r Longrightarrow u_...u_rr are lineary indepent. But Row SB Spu_...u_r Longrightarrow u_...u_r are a basis for Row SB. We now turn to Col SB. Clearly textColSBsubseteq leftleftarrayc x_ . . . x_m arrayright in K^m | x_r+...x_mright K^rtimes ^m-r. At the same time the pivot columns of B are of the form: e_leftarrayc . . arrayright e_leftarrayc . . arrayright e_rleftarrayc . . arrayright So e_...e_r make a basis for K^rtimes ^m-r. Longrightarrow K^rtimes ^m-r Spe_...e_rsubseteq Col SB subseteq K^rtimes ^m-r. We have equality everywhere hence e_...e_r form a basis for Col SB.
Let B in M_mtimes n be a matrix which is in reduced row echelon form. Then the rows of B that are not totally form a basis for Row SB. In particular row-rankB equals the number of non-zero rows in B number of pivots in B.Also the pivot columns of B form a basis for Col SB. In particular for matrices B that are in reduced row echelon form we have row-rankBcol-rankB.
Solution:
Proof. Denote by j_ ... j_r the column numbers of the pivots. Denote by u_...u_r the non-zero rows of B. We'll first show that u_...u_r are lineary indepent. Indeed ase that _i^r lambda_i u_i this is a vector in K^n for some lambda_...lambda_r in K. The entry number j_k in this vector above is precisely lambda_k because in col#j_k we have everything except a in entry k that comes from the vector u_k. Since _i^r lambda_i u_i we obtain that lambda_k. This holds forall leq k leq r Longrightarrow u_...u_rr are lineary indepent. But Row SB Spu_...u_r Longrightarrow u_...u_r are a basis for Row SB. We now turn to Col SB. Clearly textColSBsubseteq leftleftarrayc x_ . . . x_m arrayright in K^m | x_r+...x_mright K^rtimes ^m-r. At the same time the pivot columns of B are of the form: e_leftarrayc . . arrayright e_leftarrayc . . arrayright e_rleftarrayc . . arrayright So e_...e_r make a basis for K^rtimes ^m-r. Longrightarrow K^rtimes ^m-r Spe_...e_rsubseteq Col SB subseteq K^rtimes ^m-r. We have equality everywhere hence e_...e_r form a basis for Col SB.