<br>由AX=b有唯一解知r(A)=r(A,b)=n,因此AX=0只有零解.<br>若r(A<sup>T</sup>A)<n,则方程组A<sup>T</sup>AX=0有非零解,即存在X<sub>0</sub>≠0使A<sup>T</sup>AX<sub>0</sub>=0.所以有<img src="/s/tiw/p3/UpLoadImage/2013-05-07/018484be-658a-49d8-aead-8150c760d4f3.png" width="192" height="25">,即AX<sub>0</sub>=0.于是方程组AX=0有非零解,这与AX=0只有零解矛盾,故r(A<sup>T</sup>A)=n,即A<sup>T</sup>A可逆.<br>由AX=b得,A<sup>T</sup>AX=A<sup>T</sup>b,有X=(A<sup>T</sup>A)<sup>-1</sup>A<sup>T</sup>b.<br>如果η<sub>1</sub>,η<sub>2</sub>,…,η<sub>t</sub>是线性方程组AX=b的解,则u<sub>1</sub>η<sub>1</sub>+u<sub>2</sub>η<sub>2</sub>+…+u<sub>t</sub>η<sub>t</sub>也是AX=b的一个解.其中u<sub>1</sub>+u<sub>2</sub>+…+u<sub>t</sub>=1.<br>因为η<sub>1</sub>,η<sub>2</sub>,…,η<sub>t</sub>是AX=b的解,所以η<sub>2</sub>-η<sub>1</sub>,η<sub>3</sub>-η<sub>1</sub>,…,η<sub>t</sub>-η<sub>1</sub>是AX=0的解.<br>由u<sub>1</sub>+u<sub>2</sub>+…+u<sub>t</sub>=1,得u<sub>1</sub>=1-u<sub>2</sub>-u<sub>3</sub>…-u<sub>t</sub>,所以有<br>u<sub>1</sub>η<sub>1</sub>+u<sub>2</sub>η<sub>2</sub>+…+u<sub>t</sub>η<sub>t</sub>=(1-u<sub>2</sub>-u<sub>3</sub>-…-u<sub>t</sub>)η<sub>1</sub>+u<sub>2</sub>η<sub>2</sub>+…+u<sub>t</sub>η<sub>t</sub>=η<sub>1</sub>+u<sub>2</sub>(η<sub>2</sub>-η<sub>1</sub>)+ u<sub>3</sub>(η<sub>3</sub>-η<sub>1</sub>)+…+u<sub>t</sub>(η<sub>t</sub>-η<sub>1</sub>)<br>即μ<sub>1</sub>η<sub>1</sub>+μ<sub>2</sub>η<sub>2</sub>+…+μ<sub>t</sub>η<sub>t</sub>也是AX=b的解.
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