線性代數考題

線性代數考古題,請高手幫忙解答。多謝了!

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1 個解答

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  • 7 年前
    最佳解答

    Assume, for some real numbers A,B,C,

    Av1 + Bv2 + Cv3 = 0 --- (1)

    then T(Av1 + Bv2 + Cv3) = T(0) = 0

    Note that T(Av1 + Bv2 + Cv3) = T(Av1) + T(Bv2) + T(Cv3)

    T(Av1) + T(Bv2) + T(Cv3) = 0

    A*T(v1) + B*T(v2) + C*T(v3) = 0

    Av1 + 2Bv2 - Cv3 = 0 --- (2)

    Repeating yields

    Av1 + 4Bv2 + Cv3 = 0 --- (3)

    (3)-(1),

    3Bv2 = 0

    since v2 is nonzero, so B=0

    Then, (1)+(2) gives 2Av1 = 0 since B=0,

    therefore A=0 as, again, v1 is nonzero

    Finally, putting A=B=0 into (1) gives C=0 as v3 is also nonzero.

    Hence A=B=C=0, ie, {v1, v2, v3} is linearly independent.

    2014-01-22 00:19:08 補充:

    linear independence 和 linear transformation 的部分我不熟, 我是照著參考書做的。

    如有錯誤, 請多多包涵 @@

    2014-01-22 02:58:00 補充:

    Finally, putting A=B=0 into (1) gives C=0 as v3 is also nonzero.

    Hence A=B=C=0. Note that the trivial solution is the only solution, which implies {v1, v2, v3} is linearly independent.

    2014-01-22 02:58:52 補充:

    我想 trivial solution is the only solution 是重點

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