问下：哪本书里面证明Jordan Normal Form存在性，最容易懂

[YWY]

AA’和A’A是同秩的吧？也就是6X6和2X2是同秩的？都是两个eigenvalue，两个eigenvector。

Theorem. For any m x n matrix A, we have rank(A'A) = rank(A). Hence rank(AA') = rank(A).

Proof. For a column vector x in R^n, Ax = 0 iff x'A'Ax = (Ax)'(Ax) = 0 iff A'Ax = 0. Thus A and A'A have the same null space, hence they have the same rank. Thus rank(AA') = rank(A') = rank(A).

[TheMatrix]

Theorem. For any m x n matrix A, we have rank(A'A) = rank(A). Hence rank(AA') = rank(A).

Proof. For a column vector x in R^n, Ax = 0 iff x'A'Ax = (Ax)'(Ax) = 0 iff A'Ax = 0. Thus A and A'A have the same null space, hence they have the same rank. Thus rank(AA') = rank(A') = rank(A).

嗯。对。这个6X6和2X2应该有比较密切的关系。也就是AA'和A'A，可能都有意义。

[FoxMe]

Theorem. For any m x n matrix A, we have rank(A'A) = rank(A). Hence rank(AA') = rank(A).

Proof. For a column vector x in R^n, Ax = 0 iff x'A'Ax = (Ax)'(Ax) = 0 iff A'Ax = 0. Thus A and A'A have the same null space, hence they have the same rank. Thus rank(AA') = rank(A') = rank(A).

这个证明好。如果用SVD，也能看出rank(AA') = rank(A)，并且AA'和A'A的非零特征值是相同的（其实AB和BA的非零特征值是相同的）。

能解释为啥x'A'Ax = (Ax)'(Ax) = 0 iff A'Ax = 0吗？

[FoxMe]

回答一下TheMatrix的问题。和前面rgg发的帖的字母选取有稍许不同。先写出来有4点不同，
1，他说A矩阵是mxn，（n是subjects，m是features），我这里用nxp，n是subjects, p是features。我为啥要这样用？因为“n by p”矩阵是统计系里面的标准表述；他肯定是别的出身比如EE，不是统计系出身）

rgg应该是数学系科班出身，水平非常高。

[YWY]

这个证明好。如果用SVD，也能看出rank(AA') = rank(A)，并且AA'和A'A的非零特征值是相同的（其实AB和BA的非零特征值是相同的）。

能解释为啥x'A'Ax = (Ax)'(Ax) = 0 iff A'Ax = 0吗？

Ax = 0 -> A'Ax = 0 -> x'A'Ax = (Ax)'(Ax) = 0 -> Ax = 0, so they are all equivalent (as they form a circle).

[FoxMe]

噢