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Math Help - Vectors and invertible matrix

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    Vectors and invertible matrix

    Consider two vectors v1 and v2 in R^n. They form the matrix G =

    [(v1 dot v1) (v1 dot v2)]
    [(v2 dot v1) (v2 dot v2)]. For which choices of v1 and v2 is the matrix G invertible?
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  2. #2
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    Quote Originally Posted by noles2188 View Post
    Consider two vectors v1 and v2 in R^n. They form the matrix G =

    [(v1 dot v1) (v1 dot v2)]
    [(v2 dot v1) (v2 dot v2)]. For which choices of v1 and v2 is the matrix G invertible?
    As long as ||v_1 \times v_2|| \ne 0.

    The determinant of your matrix is

    (v_1 \cdot v_1)(v_2 \cdot v_2) - (v_1 \cdot v_2)^2 \ne 0 for invertibility.

    Thus,

     <br />
||v_1|||^2 ||v_2||^2 - ||v_1||^2 ||v_2||^2 \cos^2 \theta \ne 0<br />

     <br />
= ||v_1||^2 ||v_2||^2 \sin^2 \theta \ne 0<br />

     <br />
= ||v_1 \times v_2||^2 \ne 0<br />
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    Quote Originally Posted by noles2188 View Post
    Consider two vectors v1 and v2 in R^n. They form the matrix G =

    [(v1 dot v1) (v1 dot v2)]
    [(v2 dot v1) (v2 dot v2)]. For which choices of v1 and v2 is the matrix G invertible?

    The matrix G is NOT invertible iff its rows are linearly independent iff the second row is a multiple scalar of the first one iff v_2\cdot v_1=k(v_1\cdot v_1)=kv_1\cdot v_1\,,\,\,v_2\cdot v_2=k(v_1\cdot v_2)=kv_1\cdot v_2  \Longrightarrow (v_2-kv_1)\cdot v_1=0\,,\,\,(v_2-kv_1)\cdot v_2=0

    So if both vectors are non-zero then the above means that it must be...

    Tonio
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