QR 量化化

When we discussed diagonalization we essentially taught ourselves how to factor matrices. In this lesson we are going to apply our knowledge of the Gram-Schmidt process and orthonormalization in order to factor matrices by another method, known as QR factorization.
::当我们讨论对等化时,我们基本上教自己如何计算矩阵。在这个教训中,我们将运用我们关于Gram-Schmidt过程和正正正化的知识,以便用另一种方法,即QR系数化来计算矩阵。

Theorem: Given an  $\begin{array}{r}m\times n\end{array}$  matrix  $\begin{array}{r}A\end{array}$  with linearly independent columns one can factor that matrix into the form  $\begin{array}{r}A=QR\end{array}$  where  $\begin{array}{r}Q\end{array}$  is also an  $\begin{array}{r}m\times n\end{array}$  matrix with columns which form an orthonormal basis for the column space of  $\begin{array}{r}A\end{array}$  and  $\begin{array}{r}R\end{array}$  is an  $\begin{array}{r}n\times n\end{array}$  upper triangular matrix which is invertible and has positive entries on the diagonal.
::论理:如果有一个有线性独立的列的 mxn 矩阵A,那么可以将该矩阵纳入表AR,其中Q也是mxn 矩阵,其列构成A和R列空间的正正态基数,是 nxn 的上三角矩阵,不可逆,在对角有正条目。

Try to prove this on your own using properties about orthogonal matrices and the Gram-Schmidt process. I will just run through  an example of this technique being used.
::尝试使用关于矩形矩阵的属性和 Gram-Schmidt 进程来证明这一点。 我将尝试一个使用这种技术的例子 。

Example: Let  $\begin{array}{r}A=\left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\end{array}$
::示例:让我们A=[100110111111]

Applying the Gram-Schmidt process to get an orthonormal basis for the column space of A, we get that
::应用 Gram- Schmidt 进程来获得 A 列空间的正正异常值, 我们知道

$\begin{array}{r}Q=\left[\begin{array}{ccc}\frac{1}{2}& -\frac{3}{\sqrt{12}}& 0\\ \frac{1}{2}& \frac{1}{\sqrt{12}}& -\frac{2}{\sqrt{6}}\\ \frac{1}{2}& \frac{1}{\sqrt{12}}& \frac{1}{\sqrt{6}}\\ \frac{1}{2}& \frac{1}{\sqrt{12}}& \frac{1}{\sqrt{6}}\end{array}\right]\end{array}$
::[12-312012112-2612112161212161216]

Now, we can solve for  $\begin{array}{r}R\end{array}$  taking  $\begin{array}{r}{Q}^{T}A=Q^T(QR)=IR=R\end{array}$  because Q is an orthogonal matrix.
::现在,我们可以解决 R 使用 QTAT( QR) =IR=R 的问题, 因为 Q 是正对矩阵 。

Then, we get that 
::然后,我们得到

$\begin{array}{r}R=\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& \frac{1}{2}\\ -\frac{3}{\sqrt{12}}& \frac{1}{\sqrt{12}}& \frac{1}{\sqrt{12}}& \frac{1}{\sqrt{12}}\\ 0& -\frac{2}{\sqrt{6}}& \frac{1}{\sqrt{6}}& \frac{1}{\sqrt{6}}\end{array}\right]\cdot \left[\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\right]\\ R=\left[\begin{array}{ccc}2& \frac{3}{2}& 1\\ 0& \frac{3}{\sqrt{12}}& \frac{2}{\sqrt{12}}\\ 0& 0& \frac{2}{\sqrt{6}}\end{array}\right]\end{array}$  
::R=[121212121212-3121121212120-261616]][100111111]R=[232103122120026]

Hence, we have found our Q and R.
::因此,我们找到了我们的Q和R。

For more information look at these sources which will give you valuable insight on QR.
::欲了解更多信息,请查看这些来源,这些来源将为您提供关于QR的宝贵见解。