My question is a finding basis.I used the definition but I am not sure if it did work or not.

Let $ { v_1 ,..., v_n} $ denote a set of linearly independent vectors in V. Let

U={u $ \in V $ : u= $ \sum_{i=1}^n\left (a_iv_i\right )$ where $a_i \in F $ for all i and $ \sum_{i=1}^n\left (a_i\right )$=0

I did the following:

if i=1, u=$a_1v_1$ , $a_1$ =0

if i=2 , u=$a_1v_1+a_2v_2$ , $a_1+a_2$=0



if i=n , u=$a_1v_1+a_2v_2+...+a_nv_n$ , $a_1+a_2+...+a_n=0 $

I use the definition of a basis: $ (v_1, v_2,...,v_n) $ is a basis of U if every $u \in U $ can be written uniquely in the form u= $a_1v_1+a_2v_2+...+a_nv_n$

To show that every $u \in U $ can be written uniquely in the form u= $a_1v_1+a_2v_2+...+a_nv_n$ I used the matrix A has the following form;

\begin{matrix} v_1 & 0 & \ldots & 0 \\ v_1 & v_2 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ v_1 & v_2 & \ldots & v_n \end{matrix}

since the matrix has a pivot position in every row then its columns are linearly independent. Can we say that every $u \in U $ can be written uniquely in the form u= $a_1v_1+a_2v_2+...+a_nv_n$

any help will be appreciated thanks

  • $\begingroup$ $U$ has dimension $n-1$, so if you have found a basis for $U$ with $n$ elements (which is what it looks like, although I don't understand everything you have written), you have not understood the concepts. $\endgroup$ – Gerry Myerson Aug 31 '12 at 3:34

There are many choices of basis for $U$. One I might suggest is $(v_2 - v_1, v_3 - v_1, \ldots, v_n - v_1)$. Note that $U$ has dimension $n - 1$, not $n$. You can show that this is actually a basis of $U$ by showing that

  1. These vectors are elements of $U$.
  2. They are linearly independent.
  3. $\dim(U) < n$.

The first one is obvious.

The second one: Suppose $a_2, \ldots a_n$ satisfy $\sum_{i=2}^n a_i(v_i - v_1) = \sum_{i=2}^n a_i v_i - \sum_{i=2}^n a_i v_1 = 0$. By linear independence of $v_i$, we must have $a_i = 0$.

The third one: Let $v = \sum_{i=1}^n v_i$. Show that this is not in $U$.

  • $\begingroup$ thank you so much Tunococ. It is really helpful. Can n-2 or a number less than n be a dimension of U? $\endgroup$ – yagmur Aug 31 '12 at 3:47
  • $\begingroup$ No provided that you can find $n-1$ linearly independent vectors. In particular if you prove (1) then you know that $dim(U) \geq n-1$. $\endgroup$ – shoda Aug 31 '12 at 4:31
  • $\begingroup$ I think shoda meant (2). In (1), you define $n - 1$ vectors in $U$, namely $v_2 - v_1$, $v_3 - v_1, \ldots, v_n - v_1$. If you know they are linearly independent, then $\dim(U) \ge n - 1$. You need to show that they actually span $U$, which can be done by showing $\dim(U) = n - 1$ as in (3). Therefore, those vectors form a basis of $U$. $\endgroup$ – Tunococ Aug 31 '12 at 6:23

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