線性代數考題

Question

線性代數考古題，請高手幫忙解答。多謝了！

圖片參考：http://imgcld.yimg.com/8/n/AD01848228/o/2014012122...

Answer 1

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