Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do we prove that if $\{v_1,...,v_m\}$ and $\{w_1,...,w_m\}$ are bases for a real vector space, then there are at most $m$ real numbers $\lambda$ such that $v_1+\lambda w_1,...,v_m+\lambda w_m$ are linearly dependent, using the following fact?

If $A=(a_{ij})$ is an $m\times n$ matrix with no all-zero column and $n>m$, and $k$ is minimal such that the first $k$ columns of A are linearly dependent, then there exist scalars $b_1,...,b_k$ such that $\Sigma_{j=1}^k a_{ij}b_j =0 \:\forall i$ but for any distinct scalars $\lambda_1,...,\lambda_k$, there exists $i$ such that $\Sigma_{j=1}^k a_{ij}b_j\lambda_j \not= 0$.

I think we should assume $n>m$ and let $A=(a_{ij})$ be such that $\Sigma_{i=1}^m a_{ij}(v_i+\lambda w_i) =0 \:\forall j$. Then we've got a matrix $A$ as in the second paragraph, but I can't see how to use the scalars $b_j$. Can they be interpreted as components of $v_i$ or $w_i$? And how do we get around the fact that we're summing over $i$ here and over $j$ in the second paragraph?

(The problem can be solved easily using determinants, but I need to do it using the second paragraph!)

Many thanks for any help with this!

share|improve this question
    
But this isn't a matrix, because the $v_i$'s and $w_i$'s are vectors, not scalars. –  Harry Macpherson May 30 '12 at 19:26
    
Answering my own question here... Use $\Sigma_i a_{ij}(v_i + \lambda_j w_i) = 0\:\forall j$ to get $\Sigma_{i,j} a_{ij}b_j(v_i + \lambda_j w_i) = 0$, which implies $\Sigma_{i,j} a_{ij}b_j \lambda_j w_i = 0$. Since the $w_i$ form a basis, this means $\Sigma_{i,j} a_{ij}b_j \lambda_j = 0$, which is a contradiction by the second paragraph. –  Harry Macpherson May 30 '12 at 21:18
    
And how do you show such a matrix $A$ exists? –  Keivan May 30 '12 at 21:27

1 Answer 1

up vote 0 down vote accepted

Let's suppose there exist more than $m$ $\lambda_j$ such that the set $\{v_i - \lambda_jw_i \}_{1\le i\le m}$ is linearly dependent.

By definition, for every such $\lambda_j$ there exist scalars $\alpha_{ij}$ such that $\sum_{i=1}^m \alpha_{ij}(v_i - \lambda_jw_i) = 0$.

Now we have our matrix. The elements $b_j$ of your preposition are not important and we can, without loss of generality, suppose that $\sum_{j=1}^k \alpha_{ij} = 0$ (the equation $\sum_{i=1}^m \alpha_{ij}(v_i - \lambda_jw_i) = 0$ is still valid if we multiply the left side by $b_j$ ).

So, let's use the preposition:

\begin{align} 0 &= \Biggl(\sum_{j=1}^k \alpha_{ij}\Biggr)(v_i - \lambda_jw_i) = \sum_{j=1}^k \alpha_{ij}(v_i - \lambda_jw_i) =\\&= \sum_{j=1}^k \alpha_{ij}v_i - \sum_{j=1}^k \alpha_{ij}\lambda_jw_i = -\Biggl(\sum_{j=1}^k \alpha_{ij}\lambda_j\Biggr)w_i \end{align}

so $w_i = 0$ which contradict the fact that $\{w_i\}_{1\le i\le m}$ is a base.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.