Exercise 1.3.11

Answers

(i): $S \cap T$ : the possible dimension is 7 if $S \subset T$ , or 0 if no such vector exists.
(ii): $S + T$ : 7 if $S \subset T$ , 8,9 if there’s vector in $S$ but not in $T$ . If we think about matrices $A$ and $B$ , both with 10 rows. If $rank (A) = 2$ and $rank (B) = 7$ , then $A$ ’s column space is a dimension 2 subspace in $R^{10}$ , and $B$ ’s column space is dimension 7 subspace in $R^{10}$ . So for $Ax$ and $By$ are vectors in subspaces $S$ and $T$ separately. Then $Ax + By = [\begin{matrix} A & B \end{matrix}] [\begin{matrix} x \\ y \end{matrix}]$ is a vector of $s + t$ . So we need find the column space of $[\begin{matrix} A & B \end{matrix}]$ . The maximum number of independent columns is thus 9, and minimum number of independent columns is 7.
(iii): All vectors in $R^{10}$ that are perpendicular to every vector in $S$ : If we think about $S$ as column space in $R^{10}$ , the space that perpendicular to a column space is the left nullspace, whose dimension is $m - r = 10 - 2 = 8$ .

niuers

2020-03-20 00:00