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Thread: confused on second part q

  1. #1
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    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?
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  2. #2
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    Re: confused on second part q

    Quote Originally Posted by SashaP View Post
    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?
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  3. #3
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    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.
    Last edited by HallsofIvy; Dec 8th 2016 at 09:38 AM.
    Thanks from topsquark and SashaP
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    Re: confused on second part q

    Quote Originally Posted by HallsofIvy View Post
    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?
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