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Math Help - Linear Independence and bases help

  1. #1
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    Linear Independence and bases help

    Hi:
    I was reading over my linear algebra book, and trying to study for a midterm, but they didn't provide these conceptual answers in the back of the book.
    So I was wondering if anyone could show me how the following three questions are done.

    Thanks.

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  2. #2
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    bump
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  3. #3
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    A. suppose\ \vec{x}\in V, \mathfrak{B}=(\vec{v}_1,\vec{v}_2);then\ \vec{x}=span(\vec{v}_1,\vec{v}_2)
    If (\vec{v}_1,c\vec{v}_1+\vec{v}_2) linear dependence, then k\vec{v}_1=c\vec{v}_1+\vec{v}_2\rightarrow (k-c)\vec{v}_1=\vec{v}_2, it conflict with (\vec{v}_1,\vec{v}_2) linear independence, \therefore (\vec{v}_1,c\vec{v}_1+\vec{v}_2) linear independence.
    \vec{x}=k_1\vec{v}_1+k_2\vec{v}_2=(k_1-k_2 c)\vec{v}_1+k_2(c\vec{v}_1+\vec{v}_2)=span(\vec{v}  _1,c\vec{v}_1+\vec{v}_2), so (\vec{v}_1,c\vec{v}_1+\vec{v}_2) is basis of V.
    It's the same to rest proof.

    B. L:R^4\rightarrow R^5, A\in R^{5\times 4},L(\vec{x})=A\vec{x}.
    c_1L(\vec{x}_1)+c_2L(\vec{x}_2)+c_3L(\vec{x}_3)=0\  rightarrow c_1A\vec{x}_1+c_2A\vec{x}_2+c_3A\vec{x}_3=0 \rightarrow A(c_1\vec{x}_1+c_2\vec{x}_2+c_3\vec{x}_3)=0
    \because ker(L)={\vec{0}},\therefore c_1\vec{x}_1+c_2\vec{x}_2+c_3\vec{x}_3=0
    \because (\vec{x}_1,\vec{x}_2,\vec{x}_3)\ is\ linearly\ independent,\therefore c_1=c_2=c_3=0
    \therefore \{L(\vec{x}_1),L(\vec{x}_2),L(\vec{x}_3)\}\ is\ linearly\ independent

    C.
    \mathfrak{B}_a=(\begin{bmatrix}  1&0&0\\0&0&0\\0&0&0  \end{bmatrix},\begin{bmatrix}  0&0&0\\0&1&0\\0&0&0  \end{bmatrix},\begin{bmatrix}  0&0&0\\0&0&0\\0&0&1  \end{bmatrix})

    \mathfrak{B}_b=(\begin{bmatrix}  1&0&0\\0&0&0\\0&0&0  \end{bmatrix},\begin{bmatrix}  0&0&0\\0&1&0\\0&0&0  \end{bmatrix},\begin{bmatrix}  0&0&0\\0&0&0\\0&0&1  \end{bmatrix},\begin{bmatrix}  0&1&0\\0&0&0\\0&0&0  \end{bmatrix}\begin{bmatrix}  0&0&1\\0&0&0\\0&0&0  \end{bmatrix}\begin{bmatrix}  0&0&0\\0&0&1\\0&0&0  \end{bmatrix})
    Last edited by math2009; February 18th 2009 at 10:04 PM.
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  4. #4
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    Cheers Math2009!
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