Originally Posted by
HallsofIvy 1a) Asks you to find "the image by f of the vector $\displaystyle \begin{pmatrix}r \\ s\\ t \\ u\end{pmatrix}$". You do that by the matrix multiplication
$\displaystyle \begin{pmatrix}-1 & -2 & 0 & 1 \\ 0 & 3 & 1 & 1 \\ 2 & 0 & 2 & -4 \\ 0 & -1 & 0 & 0 \end{pmatrix}$$\displaystyle \begin{pmatrix}r \\ s\\ t\\ u\end{pmatrix}$
1b) Asks you to find the determinant of the matrix: use row operations to find $\displaystyle \left|\begin{array}{cccc}-1 & -2 & 0 & 1 \\ 0 & 3 & 1 & 1 \\ 2 & 0 & 2 & -4 \\ 0 & -1 & 0 & 0 \end{array}\right|$.
1c) A matrix is "invertible" if and only if its determinant is not 0.
For 2a, I would use "indirect proof"- assume that the result is not true. That is, assume there exist a, b, and c, not all 0, such that $\displaystyle af(v_1)+ bf(v_2)+ cf(v_3)= 0$. That is, such that $\displaystyle f(av_1+ bv_2+ cv_3)= 0$. That is, that $\displaystyle av_1+ bv_2+ cv_3$ is in rhe kernel of f. Now if the kernel consists only of the 0 vector, we must have $\displaystyle av_1+ bv_2+ cv_3= 0$. Since you are given that $\displaystyle v_1$, $\displaystyle v_2$, and $\displaystyle v_3$ are independent, what does that tell you about a, b, and c?
A basis, S, for an n dimensional vector space, V, has 3 properties:
1) the vectors in S are linearly independent.
2) the vectors in S span V.
3) there are n vectors in S.
And any two of those imply the third! So if you know (from 2a) that these vectors are independent and you know that there are 3 vector while the dimension of the space is 3, then it follows that they also span V and form a basis.