<br>设k<sub>1</sub>α<sub>1</sub>+k<sub>2</sub>(α<sub>1</sub>+α<sub>2</sub>)+…+k<sub>n-1</sub>(α<sub>1</sub>+α<sub>2</sub>+…+α<sub>n-1</sub>)+k<sub>n</sub>(α<sub>1</sub>+α<sub>2</sub>+…+α<sub>n</sub>)=0,整理得k<sub>1</sub>α<sub>1</sub>+k<sub>2</sub>k<sub>n</sub>α<sub>n</sub>+(k<sub>n-1</sub>+k<sub>n</sub>)α<sub>n-1</sub>+…+(k<sub>2</sub>+k<sub>3</sub>+…+k<sub>n-1</sub>+k<sub>n</sub>)α<sub>2</sub>+(k<sub>1</sub>+k<sub>2</sub>+…+k<sub>n</sub>)α<sub>1</sub>=0.<br>因为α<sub>1</sub>,α<sub>2</sub>,…,α<sub>n</sub>线性无关,所以有<br><img src="/s/tiw/p3/UpLoadImage/2013-05-07/5424c85c-8440-4595-9b5c-c4adf6a1042a.jpg" width="410" height="125"><br>所以向量组α<sub>1</sub>,α<sub>1</sub>+α<sub>2</sub>,α<sub>1</sub>+α<sub>2</sub>+α<sub>3</sub>,…,α<sub>1</sub>+α<sub>2</sub>+α<sub>3</sub>+…+α<sub>n</sub>线性无关.
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