# Thread: Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

1. ## Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

Hello, I would like you to read the attached image where I have presented the entire problem and my two questions

*

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

2. ## Re: Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

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:
$\left \{ \begin{array}{ll} x_1+2x_3=0 \\ x_2-3x_3=0 \end{array} \right.$
Choose $x_3=t$ so we obtain:
$\left \{ \begin{array}{ll} x_1=-2t \\ x_2=3t \\x_3=t \end{array} \right.$
The solution set is:
$V=\left \{ \left ( \begin{array}{ccc} -2t \\ 3t \\ t \end{array} \right) | t \in \mathbb{R} \right \}$

3. ## Re: Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

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.

4. ## Re: Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

We are given the system:
$\left \{\begin{array}{lll} 2x_1+4x_3=0 \\ x_1+x_2-x_3=0 \\ -x_1+3x_2-11x_3=0\end{array} \right.$

We can solve the system if we apply row operations on the following matrix:
$\left ( \begin{array}{ccc} 2&0&4 \\ 1&1&-1 \\ -1&3&-11 \end{array} \right)$
$=\left ( \begin{array}{ccc} 1&0&2 \\ 1&1&-1 \\ -1&3&-11 \end{array} \right)=\left ( \begin{array}{ccc} 1&0&2 \\ 0&1&-3 \\ 0&3&-9 \end{array} \right)=\left ( \begin{array}{ccc} 1&0&2 \\ 0&1&-3 \\ 0&1&-3 \end{array} \right)=\left ( \begin{array}{ccc} 1&0&2 \\ 0&1&-3 \\ 0&0&0 \end{array} \right)$

Therefore we have reduced the given system to:
$\left \{ \begin{array}{ll} x_1+2x_3=0 \\ x_2-3x_3=0 \end{array} \right.$

We can express $x_1$ and $x_2$ in terms of $x_3$ so choose $x_3=t$.

5. ## Re: Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

I get what you mean by a one zero row now, thanks

I must ask though if you know the reason for choosing dim (Vf) = 2.

Are you guys familiar with this terminology?

6. ## Re: Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

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 $x_1+ 2x_3= 0$ and $x_2- 3x_3= 0$ which are, of course, equivalent to $x_1= -2x_3$ and $x_2= 3x_3$. 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 $R^3$ into a 2 dimensional subspace of $R^3$.

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 $2x_1+ 4x_3= a$, $x_1+ x_2- x_3= b$, and $-x_1+ 3x_2- 11x_3= c$ for some $x_1, x_2, x_3$. We can put that into an augmented matrix and row reduce:
$\begin{bmatrix}2 & 0 & 4 & a \\ 1 & 1 & -1 & b \\ -1 & 3 & -11 & c\end{bmatrix}$
$\begin{bmatrix}1 & 1 & -1 & b \\ 0 & 1 & -3 & b- \frac{a}{2}\\ 0 & 0 & 0 & c- 2b+ 2a\end{bmatrix}$
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.

7. ## Re: Finding the N(F) and V(f) the kernel and a values of a matrix, Questions

Thank you, I am studying linear alg in another language than English, thats why am confused about the terminology