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linear-algebra-kalika102pages-24.pdf

S3 · 中三 questions. Answer in your own words, then submit to get a score.

  1. 1. Given that X1={u1,u2,…,un}X_1 = \{u_1, u_2, \ldots, u_n\} and X2={v1,v2,…,vn}X_2 = \{v_1, v_2, \ldots, v_n\} are subspaces of a vector space VV, explain why αiui∈X1\alpha_i u_i \in X_1 for any scalar αi\alpha_i and index ii.

  2. 2. Using the expression for αu+βv\alpha u + \beta v shown in the derivation, identify the coefficient of uku_k (for a fixed kk between 1 and nn) in the final linear combination, and write it in terms of α\alpha, β\beta, and the original coefficients.

  3. 3. Why does the fact that αiui∈X1\alpha_i u_i \in X_1 and βivi∈X2\beta_i v_i \in X_2 for all ii imply that αiui+βivi∈X1+X2\alpha_i u_i + \beta_i v_i \in X_1 + X_2?

  4. 4. Show how the vector αu+βv\alpha u + \beta v is expressed as a linear combination of the vectors u1,…,un,v1,…,vnu_1, \dots, u_n, v_1, \dots, v_n, using the intermediate steps provided.

  5. 5. What property of subspaces ensures that [X1∪X2][X_1 \cup X_2] is itself a subspace of VV, according to the given reasoning?

Transcript
## Page 1 Now, αu+βv=α(a1u1+a2u2+⋯+anun+b1v1+b2v2+⋯+bnvn)+β(c1u1+c2u2+⋯+cnun+d1v1+d2v2+⋯+dnvn)\alpha u + \beta v = \alpha (a_1 u_1 + a_2 u_2 + \cdots + a_n u_n + b_1 v_1 + b_2 v_2 + \cdots + b_n v_n) \\ + \beta (c_1 u_1 + c_2 u_2 + \cdots + c_n u_n + d_1 v_1 + d_2 v_2 + \cdots + d_n v_n) =αa1u1+αa2u2+⋯+αanun+αb1v1+αb2v2+⋯+αbnvn+βc1u1+βc2u2+⋯+βcnun+βd1v1+βd2v2+⋯+βdnvn= \alpha a_1 u_1 + \alpha a_2 u_2 + \cdots + \alpha a_n u_n + \alpha b_1 v_1 + \alpha b_2 v_2 + \cdots + \alpha b_n v_n \\ + \beta c_1 u_1 + \beta c_2 u_2 + \cdots + \beta c_n u_n + \beta d_1 v_1 + \beta d_2 v_2 + \cdots + \beta d_n v_n =p1u1+p2u2+⋯+pnun+q1v1+q2v2+⋯+qnvn= p_1 u_1 + p_2 u_2 + \cdots + p_n u_n + q_1 v_1 + q_2 v_2 + \cdots + q_n v_n =(αa1+βc1)u1+(αa2+βc2)u2+⋯+(αan+βcn)un+(αb1+βd1)v1+(αb2+βd2)v2+⋯+(αbn+βdn)vn= (\alpha a_1 + \beta c_1) u_1 + (\alpha a_2 + \beta c_2) u_2 + \cdots + (\alpha a_n + \beta c_n) u_n \\ + (\alpha b_1 + \beta d_1) v_1 + (\alpha b_2 + \beta d_2) v_2 + \cdots + (\alpha b_n + \beta d_n) v_n =p1u1+p2u2+⋯+pnun+q1v1+q2v2+⋯+qnvn= p_1 u_1 + p_2 u_2 + \cdots + p_n u_n + q_1 v_1 + q_2 v_2 + \cdots + q_n v_n ∈[X1∪X2]\in [X_1 \cup X_2] ∴[X1∪X2]\therefore [X_1 \cup X_2] is a subspace of VV. 2) Let X1={u1,u2,…,un}X_1 = \{u_1, u_2, \ldots, u_n\} X2={v1,v2,…,vn}X_2 = \{v_1, v_2, \ldots, v_n\} Given that X1X_1 and X2X_2 are subspaces of VV, ∵X1\because X_1 is a subspace of VV ∴ui+uj∈X1\therefore u_i + u_j \in X_1, ∀ 1≤i,j≤n\forall\ 1 \le i, j \le n and αiui∈X1\alpha_i u_i \in X_1, ∀ 1≤i≤n\forall\ 1 \le i \le n ∵X2\because X_2 is a subspace of VV ∴vi+vj∈X2\therefore v_i + v_j \in X_2, ∀ 1≤i,j≤n\forall\ 1 \le i, j \le n and βivi∈X2\beta_i v_i \in X_2, ∀ 1≤i≤n\forall\ 1 \le i \le n αiui∈X1\alpha_i u_i \in X_1, βivi∈X2\beta_i v_i \in X_2, ∀ 1≤i≤n\forall\ 1 \le i \le n ⇒αiui+βivi∈X1+X2\Rightarrow \alpha_i u_i + \beta_i v_i \in X_1 + X_2, ∀ 1≤i≤n\forall\ 1 \le i \le n