4
$\begingroup$

In Friedberg's Linear Algebra, the following theorem is presented:

"Let V and W be vector spaces over F, and suppose that V is finite-dimensional with a basis $\left\{v_1,...,v_n\right\}$. For any vectors $w_1,...,w_n$ in W there exists exactly one linear transformation $T:V\rightarrow W$ such that $T(v_i)=(w_i)$ for $i=1,...,n$."

I think I understand what this is saying, but then in the exercises there is a true/false question that is supposedly false that is as follows:

"Given $x_1,x_2\in V$ and $y_1,y_2\in W$, there exists a linear transformation $T:V\rightarrow W$ such that $T(x_1)=y_1$ and $T(x_2)=y_2$."

My question is why is this false? Doesn't it follow from the theorem?

$\endgroup$
1
  • 4
    $\begingroup$ $x_1$ and $x_2$ may be linearly dependent. $\endgroup$ Commented Jun 5, 2017 at 18:02

3 Answers 3

3
$\begingroup$

The easiest counterexample for the statement you quote is to take $x_1 = x_2 = 0$ and (assuming $\dim W \geq 1$) and $y_1 = 0$ while $y_2 \neq 0$. Then we must have both $T(x_1) = T(0) = y_1 = 0$ and $T(x_2) = T(0) = y_2 \neq 0$ which clearly cannot hold.

$\endgroup$
2
$\begingroup$

Let $V=W=\mathbb{R}^2$. Let $x_1 = (1,0), x_2 = (2,0), y_1=(0,1), y_2=(1,3)$. Suppose there exits a linear transformation $T$ as required. Then $$T(2,0)=(1,3).$$ Also $$T(2,0)=2T(1,0)=2(0,1)=(0,2).$$

This gives a contradiction. The reason this happens, as mentioned in the comments, is that $x_1$ and $x_2$ are linearly dependent.

$\endgroup$
2
$\begingroup$

The theorem applies to a set of basis vectors in $V$. Hence, they are linearly independent. The $x_1$ and $x_2$ in the problem may not be linearly independent.

$\endgroup$
1
  • $\begingroup$ how do i prove the theorem? $\endgroup$
    – reyna
    Commented Dec 25, 2018 at 11:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .