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Consider an orthonormal basisu1→,u2→,…,un→ in Rm. Find the least square solutions of the system Ax→=u→nwhere A=[...u→1...u→n-1...]..

Short Answer

Expert verified

The solution isATAx→k=ATu→n

Step by step solution

01

Step:1 Definition of least square

Consider a linear system as Ax→=u→n.

Here A is an matrix, a vectorx→ inRn called a least square solution of this system ifu→n-Ax→k≤u→n-Ax→ for allx→ in Rm.

02

Step:2 Explanation of the solution

Consider a linear system as Ax→=u→nwhere [...u→1...u→n-1...]and A is A is an n×mmatrix, a vector x→in Rncalled a least square solution of this system if u→n-Ax→k≤u→n-Ax→for all x→in Rm.

The term least square solution reflects the fact that minimizing the sum of the squares of the component of the vector u→n-Ax→.

To show x→kof a linear system Ax→=u→n.

Consider the following string of equivalent statements.

The vector x→kis a least square solution of the system Ax→=u→n.

Then by the definition as follows.

⇔u→n-Ax→k≤u→n-Ax→for allx→inRm

Also by the theorem 5.4.3 as follows.

⇔Ax→=proj(b→)where V=imA.

Similarly by the theorem 5.1.4 and 5.4.1 as follows.

⇔u→n-Ax→is in V⊥=(imA)⊥=keR(AT)

⇕AT(u→n-Ax→)=0→⇕ATAx→k=ATu→n

Therefore, the least square of the systemAx→=u→n are the exact solution of the system ATAx→k=ATu→n.

Thus, the system is called the normal equation of Ax→=u→n.

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