I know what is expected, yet I have troubles to calculate or proof it when a rather big matrix is given with numbers and letters.
This is the function I have:
How can I check if it's one to one? Do I need to check for everything that is in my function T, so let's say 3x-z, and take a random number for x and z, and check whether the value that rolls out, is unique? I did this, but I don't think this is correct to be honest.
What is the way to approach this? I'd really appreciate some help and tips...
$\endgroup$ 53 Answers
$\begingroup$T is one-to-one function if the kernel is trivial, so you you neeed to check implication $T(x,y,z)=0\implies x=y=z=0$.
$\endgroup$ 4 $\begingroup$Find the REF of the standard matrix (it's not necessary to get to RREF). Then, look at the pivots (the leading 1's of the rows).
If we have a pivot in every column, then the nullspace of the matrix (and hence the kernel of $T$) is zero-dimensional. So, $T$ is one-to-one if and only if the REF has pivot in every column.
If we have a pivot in every row, then $T(x,y,z) = \mathbf b$ has a solution for any $\mathbf b$. That is, the dimension of the image is equal to the dimension of the space. So, $T$ is onto if and only if the REF has a pivot in every row.
$\endgroup$ 4 $\begingroup$alans has given you the method for b)
One of the many ways to solve for c) :
Find the range of the linear transformation. i.e. $R(T) = span\{T(1,0,0),T(0,1,0),T(0,0,1)\} = span\{(3,1,1,0),(0,2,-1,1),(-1,0,1,2)\}$
Note: You can use any other basis you like. Also, i would strongly recommend you to prove that the above statement is true in general.
Suppose on the contrary that $R(T)$ is equal to $\mathbb{R}^{4}$.
Then there exist $\lambda_{1},\lambda_{2},\lambda_{3}\in \mathbb{R}$ such that $(0,0,0,1) = \lambda_{1}\cdot (3,1,1,0) + \lambda_{2}\cdot (0,2,-1,1) + \lambda_{3}\cdot (-1,0,1,2)$
Now, this system of linear equation gives you no solution, which is a contradiction.
Note: Suppose you are given another example which $T$ is indeed onto, then you will also find $R(T)$ and show that it is equal to the vector space that $T$ is mapped to.
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