No One-to-One Linear Transformation: V to W

hannahlu92
Messages
4
Reaction score
0

Homework Statement



Prove: If V and W are finite-dimensional vector spaces such that dim(W)<dim(V), then there is no one-to-one linear transformation T:V-->W




The Attempt at a Solution


I don't know how to do a well thought out proof.
 
Physics news on Phys.org


Hi hannahlu92! :smile:

The first thing you should do with such a statement is trying to find concrete examples. Can you find examples of V and W such that dim(V)<dim(W). Is it true that there doesn't exist such a one-to-one map for these examples? (I.e. is it inuitively true).

Then, to actually start proving it, you'll need to unwind the concept. What does dimension mean? What does one-to-one mean? Can we find some connection between the definition of dimension and the concept of one-to-one maps?
 


thank you for taking the time to try and help me. My final is tomorrow and I still can't understand Linear Algebra
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top