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Math Help - range of linear transformation

  1. #1
    Member Jskid's Avatar
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    range of linear transformation

    Let L : R^4 \rightarrow R^3 be defined by L(x,y,z,w)=(x+y,z+w,x+z) Find a basis for range L.

    What I did:
    {<br />
\left[\begin{array}{c}1\\0\\1\end{array}\right]<br />
\left[\begin{array}{c}1\\0\\0\end{array}\right]<br />
\left[\begin{array}{c}0\\1\\1\end{array}\right]<br />
\left[\begin{array}{c}0\\1\\0\end{array}\right]<br />
} spans range L

    but rref of the coefficent matrix has a row of zeros so it's not a spanning set.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Jskid View Post
    Let L : R^4 \rightarrow R^3 be defined by L(x,y,z,w)=(x+y,z+w,x+z) Find a basis for range L.

    What I did:
    {<br />
\left[\begin{array}{c}1\\0\\1\end{array}\right]<br />
\left[\begin{array}{c}1\\0\\0\end{array}\right]<br />
\left[\begin{array}{c}0\\1\\1\end{array}\right]<br />
\left[\begin{array}{c}0\\1\\0\end{array}\right]<br />
} spans range L

    but rref of the coefficent matrix has a row of zeros so it's not a spanning set.
    I may be crazy but isn't the image \mathbb{R}^3? Namely, isn't \ker L=\left\{(x,-x,-x,x):x\in\mathbb{R}\right\}=\text{span}_\mathbb{R}  \{(1,-1,-1,1)\} so that \dim \text{im}(L)=4-1=3 etc.
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