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Math Help - Linear transformation proof

  1. #1
    M.R
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    Linear transformation proof

    Hi,

    I am trying to prove the following are not a linear transformation:

    1. T: R^3 -> R defined by T(x) = x_{1}x_{2}x_{3}

    2. P -> P, T(p)=p+2p'+3p'', where P is the vector space of polynomials (of any degree)

    3. det: M_{22} -> R, det \left(\begin{array}{cc}a&b\\c&d\end{array}\right) = ad-bc


    My attempt:

    I have tried the 0 vector & -1 scaler multiplication and it staisfies all the 3 eqations. Furthermore they satisfy addition as well. So why are they not linear transformation?
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  2. #2
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    Re: Linear transformation proof

    Quote Originally Posted by M.R View Post
    Hi,

    I am trying to prove the following are not a linear transformation:

    1. T: R^3 -> R defined by T(x) = x_{1}x_{2}x_{3}

    2. P -> P, T(p)=p+2p'+3p'', where P is the vector space of polynomials (of any degree)

    3. det: M_{22} -> R, det \left(\begin{array}{cc}a&b\\c&d\end{array}\right) = ad-bc


    My attempt:

    I have tried the 0 vector & -1 scaler multiplication and it staisfies all the 3 eqations. Furthermore they satisfy addition as well. So why are they not linear transformation?
    The first one is not linear because it does not satisfy

    T(\mathbf{v}_1+\mathbf{v_2})=T(\mathbf{v}_2)+T (\mathbf{v}_2)

    Let v_1=e_1+e_2+e_3 and v_2=-e_1

    This gives

    T(v_1)=1 \quad T(v_2)=0 but

    T(v_1+v_2)=T(e_2+e_3)=0 \ne T(v_1)+T(v_2)=1+0=1

    For number 2. It is most definitely a linear operator. Differentiation is one of the first linear operators people learn.

    For number 3. Try addition again

    Note that

    \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}+\begin{bmatrix}0 & 0 \\ 0 & 1 \end{bmatrix}

    use the same idea as in number 1
    Thanks from M.R
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