Linear Algebra II

September 18, 2014

Lecture 9

Filed under: 2014 Fall — Y.K. Lau @ 9:31 PM

We give a brief summary of the vector space theory covered today.

 

Next, we recap the key ideas in the proof of the dimension formula:

{\dim (U+W)=\dim U+\dim W-\dim (U\cap W)}.

If {\{x_1,\cdots, x_d\}} is a basis for {U\cap W}, {\{x_1,\cdots, x_d, u_1,\cdots, u_m\}} and {\{x_1,\cdots, x_d, w_1,\cdots, w_p\}} are bases for {U} and {W} resp., then {\{x_1,\cdots, x_d, u_1,\cdots, u_m, w_1,\cdots, w_p\}} is a basis for {U+W}.

 

To show the linear independence, we consider

\displaystyle  r_1x_1+\cdots+r_d x_d +s_1 u_1+\cdots+ s_mu_m {+} \displaystyle t_1 w_1+\cdots+t_p w_p {=\underline{0}.}

From this we deduce

\displaystyle  r_1x_1+\cdots+r_d x_d + s_1 u_1+\cdots+ s_mu_m \displaystyle = \displaystyle (-t_1)w_1+\cdots +(-t_p)w_p           {(*)}

Keypoints:

  1. The LHS is in {U} while the RHS is {W}.
    Thus {(-t_1)w_1+\cdots +(-t_p)w_p\in W} {\cap} {U}.
  2. As {\{x_1,\cdots, x_d\}} is a basis for {U\cap W},
            {(-t_1)w_1+\cdots +(-t_p)w_p = a_1x_1+\cdots a_dx_d}.
    Rearranging, we get

    \displaystyle a_1x_1+\cdots a_dx_d+t_1w_1+\cdots +t_pw_p =\underline{0}.

  3. As {\{x_1,\cdots, x_d, w_1,\cdots, w_p\}} is a basis for {W}, we infer {a_1=\cdots =a_d=t_1=\cdots =t_p=0}.
  4. Put {t_1=\cdots =t_p=0} into {(*)}, {r_1x_1+\cdots+r_d x_d+s_1 u_1+\cdots+ s_mu_m=\underline{0}}.
  5. As {\{x_1,\cdots, x_d, u_1,\cdots, u_m\}} is a basis for {U}, we infer {a_1=\cdots =a_d=s_1=\cdots =s_m=0}.

 

 

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