查看完整版本 : vector span

雷帝杜倫 2015-10-31 06:03 PM

vector span

Consider the following vectors in R3:x1 =(1,−1,2)^Tx2 =(−2,3,1)^Tx3 =(-1,3,8)^TLet S be the subspace of R^3 spanned by x1, x2, x3. Actually, [color=#ff0000]S can be represented in[/color][color=#ff0000]terms of the two vectors x1 and x2, since the vector x3 is already in the span of x1 and[/color]x2; that is, [color=#0000ff]x3 = 3x1 + 2x2 (1)[/color]Any linear combination of x1, x2, and x3 can be reduced to a linear combination of x1and x2:α1x1 + α2x2 + α3x3 = α1x1 + α2x2 + α3(3x1 + 2x2)= (α1 + 3α3)x1 + (α2 + 2α3)x2Thus,S = Span(x1, x2, x3) = Span(x1, x2)

我想問紅色個句,點x3 already in the span?同埋藍色條式係點整出黎?

top11 2015-10-31 06:28 PM

[quote]原帖由 [i]雷帝杜倫[/i] 於 2015-10-31 06:03 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=428982738&ptid=25241405][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
Consider the following vectors in R3:x1 =(1,−1,2)^Tx2 =(−2,3,1)^Tx3 =(-1,3,8)^TLet S be the subspace of R^3 spanned by x1, x2, x3. Actually, S can be represented interms of the two vectors ... [/quote]



要試計

consider
x3 = a (x1) + b(x2) 看看能否解出

(-1, 3, 8) = a(1, -1, 2) + b(-2, 3, 1)
(-1, 3, 8) = (a, -a, 2a) + (-2b, 3b, b)
(-1, 3, 8) = (a - 2b, -a + 3b, 2a + b)

{a - 2b = -1
{-a + 3b = 3
{2a + b = 8

使用首兩條式得 b = 2, a = 3,看看也符合 第三條式。
故 x3 只是 x1 和 x2 的 linear combination。

即 由 x1 和 x2 也可得出 x3。
即是 span(x1, x2, x3) = span(x1, x2)
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查看完整版本: vector span