[问题2014S13] 解答
(1) 先证必要性：若 A=LUA=LU
<script id="MathJax-Element-1" type="math/tex">A=LU</script> 是 非异阵 A<script id="MathJax-Element-2" type="math/tex">A</script> 的 LU<script id="MathJax-Element-3" type="math/tex">LU</script> 分解，则 L<script id="MathJax-Element-4" type="math/tex">L</script> 是主对角元全部等于 1 的下三角阵，U<script id="MathJax-Element-5" type="math/tex">U</script> 是主对角元全部非零的上三角阵. 由 Cauchy-Binet 公式知 |A_k|=|L_k|\cdot|U_k|=|U_k|\neq 0,\,\,k=1,2,\cdots,n,
<script id="MathJax-Element-6" type="math/tex; mode=display">|A_k|=|L_k|\cdot|U_k|=|U_k|\neq 0,\,\,k=1,2,\cdots,n,</script> 其中 |A_k|,|L_k|,|U_k|<script id="MathJax-Element-7" type="math/tex">|A_k|,|L_k|,|U_k|</script> 分别表示 A,L,U<script id="MathJax-Element-8" type="math/tex">A,L,U</script> 的第 k<script id="MathJax-Element-9" type="math/tex">k</script> 个顺序主子式.再证充分性以及分解的唯一性：我们对 A<script id="MathJax-Element-10" type="math/tex">A</script> 的阶数 n<script id="MathJax-Element-11" type="math/tex">n</script> 进行归纳. n=1<script id="MathJax-Element-12" type="math/tex">n=1</script> 时, 结论显然成立. 设阶数 <n<script id="MathJax-Element-13" type="math/tex">A
<script id="MathJax-Element-14" type="math/tex">A</script> 的第
n-1<script id="MathJax-Element-15" type="math/tex">n-1</script> 个顺序主子阵
A_{n-1}<script id="MathJax-Element-16" type="math/tex">A_{n-1}</script> 满足条件: 它的
n-1<script id="MathJax-Element-17" type="math/tex">n-1</script> 个顺序主子式全部非零，故由归纳假设，
A_{n-1}<script id="MathJax-Element-18" type="math/tex">A_{n-1}</script> 存在唯一的
LU<script id="MathJax-Element-19" type="math/tex">LU</script> 分解：
A_{n-1}=L_{n-1}U_{n-1},
<script id="MathJax-Element-20" type="math/tex; mode=display">A_{n-1}=L_{n-1}U_{n-1},</script> 其中
L_{n-1}<script id="MathJax-Element-21" type="math/tex">L_{n-1}</script> 是主对角元全部等于 1 的
n-1<script id="MathJax-Element-22" type="math/tex">n-1</script> 阶下三角阵，
U_{n-1}<script id="MathJax-Element-23" type="math/tex">U_{n-1}</script> 是主对角元全部非零的
n-1<script id="MathJax-Element-24" type="math/tex">n-1</script> 阶上三角阵. 设
A=\begin{bmatrix} A_{n-1} & \alpha \\ \beta‘ & a_{nn} \end{bmatrix}=\begin{bmatrix} L_{n-1} & 0 \\ x‘ & 1 \end{bmatrix}\begin{bmatrix} U_{n-1} & y \\ 0 & z \end{bmatrix}=\begin{bmatrix} L_{n-1}U_{n-1} & L_{n-1}y \\ x‘U_{n-1} & x‘y+z \end{bmatrix},
<script id="MathJax-Element-25" type="math/tex; mode=display">A=\begin{bmatrix} A_{n-1} & \alpha \\ \beta‘ & a_{nn} \end{bmatrix}=\begin{bmatrix} L_{n-1} & 0 \\ x‘ & 1 \end{bmatrix}\begin{bmatrix} U_{n-1} & y \\ 0 & z \end{bmatrix}=\begin{bmatrix} L_{n-1}U_{n-1} & L_{n-1}y \\ x‘U_{n-1} & x‘y+z \end{bmatrix},</script> 其中
\alpha,\beta,x,y<script id="MathJax-Element-26" type="math/tex">\alpha,\beta,x,y</script> 为
n-1<script id="MathJax-Element-27" type="math/tex">n-1</script> 维列向量,
z<script id="MathJax-Element-28" type="math/tex">z</script> 为数. 由此可得：
\alpha=L_{n-1}y,\,\, \beta‘=x‘U_{n-1},\,\,a_{nn}=x‘y+z.
<script id="MathJax-Element-29" type="math/tex; mode=display"> \alpha=L_{n-1}y,\,\, \beta‘=x‘U_{n-1},\,\,a_{nn}=x‘y+z.</script> 因为
L_{n-1},U_{n-1}<script id="MathJax-Element-30" type="math/tex">L_{n-1},U_{n-1}</script> 为非异阵, 由上式可唯一解得：
y=L_{n-1}^{-1}\alpha,\,\,x‘=\beta‘U_{n-1}^{-1},\,\,z=a_{nn}-\beta‘U_{n-1}^{-1}L_{n-1}^{-1}\alpha=a_{nn}-\beta‘A_{n-1}^{-1}\alpha.
<script id="MathJax-Element-31" type="math/tex; mode=display">y=L_{n-1}^{-1}\alpha,\,\,x‘=\beta‘U_{n-1}^{-1},\,\,z=a_{nn}-\beta‘U_{n-1}^{-1}L_{n-1}^{-1}\alpha=a_{nn}-\beta‘A_{n-1}^{-1}\alpha.</script> 令
L=\begin{bmatrix} L_{n-1} & 0 \\ \beta‘U_{n-1}^{-1} & 1 \end{bmatrix},\,\,U=\begin{bmatrix} U_{n-1} & L_{n-1}^{-1}\alpha \\ 0 & a_{nn}-\beta‘A_{n-1}^{-1}\alpha \end{bmatrix},
<script id="MathJax-Element-32" type="math/tex; mode=display">L=\begin{bmatrix} L_{n-1} & 0 \\ \beta‘U_{n-1}^{-1} & 1 \end{bmatrix},\,\,U=\begin{bmatrix} U_{n-1} & L_{n-1}^{-1}\alpha \\ 0 & a_{nn}-\beta‘A_{n-1}^{-1}\alpha \end{bmatrix},</script> 则
A=LU<script id="MathJax-Element-33" type="math/tex">A=LU</script> 即为
A<script id="MathJax-Element-34" type="math/tex">A</script> 的唯一的
LU<script id="MathJax-Element-35" type="math/tex">LU</script> 分解.
(2) 我们对 A<script id="MathJax-Element-36" type="math/tex">A</script> 的阶数 n<script id="MathJax-Element-37" type="math/tex">n</script> 进行归纳，来证明 Cholesky 分解的存在性和唯一性. n=1<script id="MathJax-Element-38" type="math/tex">n=1</script> 时, 结论显然成立. 设阶数 <n<script id="MathJax-Element-39" type="math/tex">A<script id="MathJax-Element-40" type="math/tex">A</script> 的第 n-1<script id="MathJax-Element-41" type="math/tex">n-1</script> 个顺序主子阵 A_{n-1}<script id="MathJax-Element-42" type="math/tex">A_{n-1}</script> 也是正定实对称阵, 故由归纳假设，A_{n-1}<script id="MathJax-Element-43" type="math/tex">A_{n-1}</script> 存在唯一的 Cholesky 分解：A_{n-1}=C_{n-1}‘C_{n-1},
<script id="MathJax-Element-44" type="math/tex; mode=display">A_{n-1}=C_{n-1}‘C_{n-1},</script> 其中 C_{n-1}<script id="MathJax-Element-45" type="math/tex">C_{n-1}</script> 是主对角元全大于零的 n-1<script id="MathJax-Element-46" type="math/tex">n-1</script> 阶上三角阵. 设 A=\begin{bmatrix} A_{n-1} & \alpha \\ \alpha‘ & a_{nn} \end{bmatrix}=\begin{bmatrix} C‘_{n-1} & 0 \\ x‘ & y \end{bmatrix}\begin{bmatrix} C_{n-1} & x \\ 0 & y \end{bmatrix}=\begin{bmatrix} C_{n-1}‘C_{n-1} & C_{n-1}‘x \\ x‘C_{n-1} & x‘x+y^2 \end{bmatrix},
<script id="MathJax-Element-47" type="math/tex; mode=display">A=\begin{bmatrix} A_{n-1} & \alpha \\ \alpha‘ & a_{nn} \end{bmatrix}=\begin{bmatrix} C‘_{n-1} & 0 \\ x‘ & y \end{bmatrix}\begin{bmatrix} C_{n-1} & x \\ 0 & y \end{bmatrix}=\begin{bmatrix} C_{n-1}‘C_{n-1} & C_{n-1}‘x \\ x‘C_{n-1} & x‘x+y^2 \end{bmatrix},</script> 其中 \alpha,\beta,x<script id="MathJax-Element-48" type="math/tex">\alpha,\beta,x</script> 为 n-1<script id="MathJax-Element-49" type="math/tex">n-1</script> 维列向量, y<script id="MathJax-Element-50" type="math/tex">y</script> 为数. 由此可得： \alpha=C_{n-1}‘x,\,\,a_{nn}=x‘x+y^2.
<script id="MathJax-Element-51" type="math/tex; mode=display"> \alpha=C_{n-1}‘x,\,\,a_{nn}=x‘x+y^2.</script> 由上式可唯一解得：x=(C_{n-1}‘)^{-1}\alpha,
<script id="MathJax-Element-52" type="math/tex; mode=display">x=(C_{n-1}‘)^{-1}\alpha,</script>y^2=a_{nn}-\alpha‘C_{n-1}^{-1}(C_{n-1}‘)^{-1}\alpha=a_{nn}-\alpha‘A_{n-1}^{-1}\alpha=\frac{|A|}{|A_{n-1}|}>0,\,\,y=\sqrt{\frac{|A|}{|A_{n-1}|}}.
<script id="MathJax-Element-53" type="math/tex; mode=display">y^2=a_{nn}-\alpha‘C_{n-1}^{-1}(C_{n-1}‘)^{-1}\alpha=a_{nn}-\alpha‘A_{n-1}^{-1}\alpha=\frac{|A|}{|A_{n-1}|}>0,\,\,y=\sqrt{\frac{|A|}{|A_{n-1}|}}.</script> 令 C=\begin{bmatrix} C_{n-1} & (C_{n-1}‘)^{-1}\alpha \\ 0 & \sqrt{\frac{|A|}{|A_{n-1}|}} \end{bmatrix},
<script id="MathJax-Element-54" type="math/tex; mode=display">C=\begin{bmatrix} C_{n-1} & (C_{n-1}‘)^{-1}\alpha \\ 0 & \sqrt{\frac{|A|}{|A_{n-1}|}} \end{bmatrix},</script> 则 A=C‘C<script id="MathJax-Element-55" type="math/tex">A=C‘C</script> 即为 A<script id="MathJax-Element-56" type="math/tex">A</script> 的唯一的 Cholesky 分解. \Box<script id="MathJax-Element-57" type="math/tex">\Box</script>
