1. If U and W are both 6-dimensional subspaces of the rel vector space \mathbb{R}^{10}， find the minimal and maximal value of \dim(U\cap W).

  • Consider that \dim(U)+\dim(W)=\dim(U+W)+\dim(U\cap W) and U\subseteq U+W\subseteq \mathbb{R}^{10}, therefore 2\leq\dim(U\cap W)=6+6-\dim(U+W)\leq6. So the minimal value is 2, where U+W=\mathbb{R}^{10}, and maximal value is 6, where U\subseteq W or W\subseteq U.

2. Suppose that V_1,V_2,V_3\subseteq V.

(i). Given counter-examples to the following statements:

(a). V_1+(V_2\cap V_3)=(V_1+V_2)\cap(V_1+V_3)

  • V_1=<i>,V_2=<j>,V_3=<i+j>, so V_1+(V_2\cap V_3)=<i>,(V_1+V_2)\cap(V_1+V_3)=<i,j>.

(b). (V_1+V_2)\cap V_3=(V_1\cap V_3)+(V_2\cap V_3)

  • V_1=<i>,V_2=<j>,V_3=<i+j>, so (V_1+V_2)\cap V_3=<i+j>,(V_1\cap V_3)+(V_2\cap V_3)=0.

(ii). Show that each equlaity in (i) can be replaced by a valid inclusion of one side in the other.

  • (a). V_1+(V_2\cap V_3)\subseteq(V_1+V_2)\cap(V_1+V_3)
    \forall x=\alpha+\beta\in V_1+(V_2\cap V_3),\alpha\in V_1,\beta\in V_2\cap V_3, notice that \beta\in V_2 and \beta\in V_3, we get x\in V_1+V_2,x\in V_1+V_3\Rightarrow x\in(V_1+V_2)\cap(V_1+V_3).
    Q.E.D
    (b). (V_1+V_2)\cap V_3\supseteq(V_1\cap V_3)+(V_2\cap V_3)
    \forall x=\alpha+\beta\in (V_1\cap V_3)+(V_2\cap V_3),\alpha\in V_1\cap V_3,\beta\in V_2\cap V_3, notice that \alpha,\beta\in V_3\Rightarrow\alpha+\beta\in V_3 and \alpha\in V_1,\beta\in V_2\Rightarrow\alpha+\beta\in V_1+V_2, we get x\in(V_1+V_2)\cap V_3.
    Q.E.D

3. Let F be a field with \mathrm{char}(F)\neq2. Let \{v_1,v_2,v_3,v_4,v_5\} be a basis for a vector space V over F. Prove that \beta=\{v_1+v_2,v_2+v_3,v_3+v_4,v_4+v_5,v_5+v_1\} is also a basis for V. Is the statement still true if \mathrm{char}(F)=2?

  • Notice that:
    (v_1,v_2,v_3,v_4,v_5)=\beta\frac{1}{2} \begin{bmatrix} 1&1 & -1 &1 &-1 \\ -1& 1 &1 &-1 &1 \\ 1&-1 &1 &1 &-1 \\ -1& 1 & -1 & 1 & 1\\ 1&-1 &1 &-1 &1\end{bmatrix}
    Q.E.D.
  • No, because we can't get 2 in F.

4. Suppose m is a positive integer. For 0\leq k\leq m, let p_k(x)=x^k(1-x)^{m-k}. Show that p_0,\cdots, p_m is a basis of F_m[x].

  • Obviously, when m=1, we have p_0=1-x,p_1=x,F_1[x]=<1,x>=<p_0,p_1>.
    Suppose that when m=k the statement is true, when m=k+1, F_{k+1}[x]\ni p(x)=\sum_{i=0}^{k+1}a_ix^i=a_{k+1}p_{k+1}(x)+q(x),q(x)\in F_k[x]. And with our supposition, q(x)=\sum_{i=0}^kb_ip_i'(x). Notice that p_i(x)+p_{i+1}(x)=p_i'(x), so p(x)=a_{k+1}p_{k+1}(x)+\sum_{i=0}^kb_i(p_i(x)+p_{i+1}(x))=\sum_{i=0}^{k+1}c_ip_i(x),c_i\in F.
    Addtionally, \dim F_{k+1}[x]=k+1, so when m=k+1, p_0,\cdots,p_{k+1} is still a basis of F_m[x].
    In this way, according to the Mathematical Induction, for all m\in \mathbb{Z}^+, p_0,\cdots,p_m is a basis of F_m[x].

5. If V_1,V_2, and V_3 are subspaces of a finite dimensional vector space, then either prove the formula \dim(V_1+V_2+V_3)=\dim V_1+\dim V_2+\dim V_3-\dim(V_1\cap V_2)-\dim(V_1\cap V_3)-\dim(V_2\cap V_3)+\dim(V_1\cap V_2\cap V_3) or give a counter-example.

  • Consider V_1=<i>,V_2=<j>,V_3=<i+j>, so \dim(V_1+V_2+V_3)=2,\dim V_1+\dim V_2+\dim V_3-\dim(V_1\cap V_2)-\dim(V_1\cap V_3)-\dim(V_2\cap V_3)+\dim(V_1\cap V_2\cap V_3)=1+1+1-0-0-0+0=3\neq2. So the formula is false.

6. Prove that if V_1,V_2, and V_3 are subspaces of a finite dimensional vector space, then \dim(V_1+V_2+V_3)=\dim V_1+\dim V_2+\dim V_3-\frac{\dim(V_1\cap V_2)+\dim(V_1\cap V_3)+\dim(V_2\cap V_3)}{3}-\frac{\dim((V_1+V_2)\cap V_3)+\dim((V_1+V_3)\cap V_2)+\dim((V_2+V_3)\cap V_1)}{3}.

  • Notice that:
    \dim(V_1+V_2+V_3)=\dim(V_1+V_2)+\dim(V_3)-\dim((V_1+V_2)\cap V_3)=\dim V_1+\dim V_2+\dim V_3-\dim(V_1\cap V_2)-\dim((V_1+V_2)\cap V_3)
    \dim(V_1+V_2+V_3)=\dim(V_1+V_3)+\dim(V_2)-\dim((V_1+V_3)\cap V_2)=\dim V_1+\dim V_2+\dim V_3-\dim(V_1\cap V_3)-\dim((V_1+V_3)\cap V_2)
    \dim(V_1+V_2+V_3)=\dim(V_2+V_3)+\dim(V_1)-\dim((V_2+V_3)\cap V_1)=\dim V_1+\dim V_2+\dim V_3-\dim(V_2\cap V_3)-\dim((V_2+V_3)\cap V_1)
    Add them together, we get 3\dim(V_1+V_2+V_3)=3\dim V_1+3\dim V_2+3\dim V_3-\dim(V_1\cap V_2)-\dim(V_1\cap V_3)-\dim(V_2\cap V_3)-\dim((V_1+V_2)\cap V_3)-\dim((V_1+V_3)\cap V_2)-\dim((V_2+V_3)\cap V_1), which is what we want to prove.

7. Let V be a vector space over \mathbb{Q}. Prove that the union of three subspaces of V is a subspace of V if and only if one of the subspaces contains the other two.

  • Let V_1,V_2,V_3 are subspaces of V. Without loss of generality, let V_1 contains V_2,V_3. Therefore V_1\cup V_2\cup V_3=V_1 is the subspace of V apparently.
    Now suppose that U=V_1\cup V_2\cup V_3 is a subspace of V. As we all know, the union of finate subspaces can't cover the whole space unless one of them is the whole space (we suppose all spaces are over \mathbb{Q}), so \exist V_i=U, which means V_i contains the other two subspaces.
    Combine the two conclusion, we have proved the statement.