Have you tried to solve the system? If you do you'll obtain one zero row that means you have to choose one parameter. Solving the system:
Choose so we obtain:
The solution set is:
Hello, I would like you to read the attached image where I have presented the entire problem and my two questions
*
Please see the attached file
To clarify question 2 a bit more: How can i judge by the one parametric value of t:s which dim to take?
How do i know its a line,vectors parallel to a plane, or vectors in a room?
Is there any analytical way to check this?
Thanks
Have you tried to solve the system? If you do you'll obtain one zero row that means you have to choose one parameter. Solving the system:
Choose so we obtain:
The solution set is:
Actually I am studying linear alg on my own, i have never been to a lecture on linear algebra, thats why i have limited knowledge, and I have an exam that I am trying to pass very soon, i dont have a particular book either, im trying to read through some free books online sometimes, and i borrowed one book from the library where I saw the dim values., however they did not explain how to set the dim values.
* Thanks for your answer Siron, but could you specify what a one zero row looks like?
* I dont understand why you pick x3 = t, could you pick anyone to be t?
Could someone answer that second question? How to know whether something is a dim vl =1, dim vpi = 2 , dim V = 3
In The exam problem that i presented to you above which was studying they got dim v(F) = 2. I still dont get it though.
We are given the system:
We can solve the system if we apply row operations on the following matrix:
Therefore we have reduced the given system to:
We can express and in terms of so choose .
Terminology? Even if you are studying Linear Algebra on your own, you should have seen the definition of "dimension of a vector space". Here, as has been pointed out, row reduction shows that the original system of equations, giving the kernel of F, is equivalent to and which are, of course, equivalent to and . That means that any vector in the solution space (the kernel of the linear transformation) is of the form [tex]<x_1, x_2, x_3>= <-2x_3, 3x_3, x_3>= x_3<-2, 3, 1>. That is, any vector in the kernel is simply <-2, 3, 1> times a constant. <-2, 3, 1> spans the entire space and, of course, a set containing a single vector is "independent" so it is a basis for the kernel- the kernel has dimension 1.
There is a theorem, that perhaps you have not met yet, that says that if F is a linear transformation from vector space U to vector space V, the sum of the dimensions of the kernel and image is equal to the dimension of U. Here U has dimension 3 and the kernel has dimension 1 so the image has dimension 2. That is, F maps all of into a 2 dimensional subspace of .
If you have not had that theorem, you can prove that the image has dimension 2 directly: let <a, b, c> be some vector in the image. Then , , and for some . We can put that into an augmented matrix and row reduce:
which is solvable if and only if c- 2b+ 2a= 0 or c= 2b- 2a. That is, any such vector can be written <a, b, c>= <a, b, 2b- 2a>= <a, 0, -2a>+ <0, 1, 2b>= a<1, 0, -2>+ b<0, 1, 2>. Those two vectors span the image of F and are independent so are a basis- the image of F has dimension 2.