# Thread: confused on second part q

1. ## confused on second part q

really confused on this exercise question, i kind of get the gist of question 1 but im not sure how to show the solution and question 2 is really confusing, any help please?

2. ## Re: confused on second part q

Originally Posted by SashaP
Suppose that $\displaystyle\sum\limits_{k = 1}^3 {{\alpha _k}f\left( {{v_k}} \right)}$ is the zero vector and at least one alpha is not zero.

Then $\displaystyle{\sum\limits_{k = 1}^3 {{\alpha _k}f\left( {{v_k}} \right)} = \sum\limits_{k = 1}^3 {f\left( {{\alpha _k}{v_k}} \right)} = f\left( {\sum\limits_{k = 1}^3 {{\alpha _k}{v_k}} } \right)}$ is also the zero vector.

YOU explain:
1. each step in the sequence of sums
2. what does it mean for the kernel to be only the zero vector?
3. what is the contradiction there?

3. ## Re: confused on second part q

1a) Asks you to find "the image by f of the vector $\begin{pmatrix}r \\ s\\ t \\ u\end{pmatrix}$". You do that by the matrix multiplication
$\begin{pmatrix}-1 & -2 & 0 & 1 \\ 0 & 3 & 1 & 1 \\ 2 & 0 & 2 & -4 \\ 0 & -1 & 0 & 0 \end{pmatrix}$ $\begin{pmatrix}r \\ s\\ t\\ u\end{pmatrix}$

1b) Asks you to find the determinant of the matrix: use row operations to find $\left|\begin{array}{cccc}-1 & -2 & 0 & 1 \\ 0 & 3 & 1 & 1 \\ 2 & 0 & 2 & -4 \\ 0 & -1 & 0 & 0 \end{array}\right|$.

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 $af(v_1)+ bf(v_2)+ cf(v_3)= 0$. That is, such that $f(av_1+ bv_2+ cv_3)= 0$. That is, that $av_1+ bv_2+ cv_3$ is in rhe kernel of f. Now if the kernel consists only of the 0 vector, we must have $av_1+ bv_2+ cv_3= 0$. Since you are given that $v_1$, $v_2$, and $v_3$ 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.

4. ## Re: confused on second part q

Originally Posted by HallsofIvy
1a) Asks you to find "the image by f of the vector $\begin{pmatrix}r \\ s\\ t \\ u\end{pmatrix}$". You do that by the matrix multiplication
$\begin{pmatrix}-1 & -2 & 0 & 1 \\ 0 & 3 & 1 & 1 \\ 2 & 0 & 2 & -4 \\ 0 & -1 & 0 & 0 \end{pmatrix}$ $\begin{pmatrix}r \\ s\\ t\\ u\end{pmatrix}$

1b) Asks you to find the determinant of the matrix: use row operations to find $\left|\begin{array}{cccc}-1 & -2 & 0 & 1 \\ 0 & 3 & 1 & 1 \\ 2 & 0 & 2 & -4 \\ 0 & -1 & 0 & 0 \end{array}\right|$.

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 $af(v_1)+ bf(v_2)+ cf(v_3)= 0$. That is, such that $f(av_1+ bv_2+ cv_3)= 0$. That is, that $av_1+ bv_2+ cv_3$ is in rhe kernel of f. Now if the kernel consists only of the 0 vector, we must have $av_1+ bv_2+ cv_3= 0$. Since you are given that $v_1$, $v_2$, and $v_3$ 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.

Thank you.
Is question 1a (-r-2s+u)
(3s+t+u)
(2r+2t-4u)
(-s. )

And is determinant of m -4. Therefore it's invertible. Also is there any example question same as 2 you know?