　　Original | https://mp.weixin.qq.com/s/TDj3aCEHjaKHATZ7uviQMA

Rectangular matrices with positive definite matrix

　　Before we have been discussing the square, but a large number of practical problems applied to a rectangular matrix, such as used in the least squares in A ^T A .

　　If A is a rectangular matrix of m × n, then A ^T A is a symmetric matrix, the square of course, we are interested in A ^T A positive qualitative. For A ^T A , our feature vectors and its determinant ignorance need to X ^T ( A ^T A ) X to determine their qualitative n> 0:

　　If and only if Ax = 0, the above equation is equal to zero, so only need to look at when Ax = 0.

　　We discussed in the null space matrix, for one of the m × n rectangular matrix, if the matrix is a full column rank is, m> n, then the null space of the matrix only the zero vector. Thus, when A is a matrix of full column rank, the only if X = 0 when Ax of = 0, this time for any nonzero vector, must have X ^T ( A ^T A ) X > 0, A is positive definite.

Similarity matrix

　　A and B are an n × n matrix, if there is invertible matrix M , so that B = M ^-1AM , called A and B mutually similar matrix, denoted as A ~ B .

Similarity matrix eigenvalues

　　In fact, we would have seen a similar matrix. If A has n linearly independent eigenvectors, then A can be diagonalized as A = SΛS ^-1 , corresponding to S ^-1the AS = Lambda , A , and eigenvalues matrix Lambda mutual similarity matrix, where M = S , it is a feature vector matrix. Indeed A similarity matrix has a lot, we can use an arbitrary invertible matrix M instead of S , thereby to obtain other similarity matrix, Lambda is similar to many of the most simple of a matrix.

　　Summons a matrix:

　　Λ and A mutually similar matrix. If we take another set of invertible matrices can be obtained A further similarity matrix:

　　Observation B will find that it is the trace of 4 (feature vector sum), 3 determinant (the product of the feature vector), which implies that we B feature vectors of A same. In fact this is the similarity matrix characteristics: has the same similarity matrix eigenvalues. Characterized in that virtually all of the 1 and 3 are second order matrix A similarity matrix.

　　Why similarity matrix exhibit the same characteristic value it? Now set A and B mutually similar matrix, B = M ^-1AM , according to the characteristic equation:

　　Now there is a new characteristic equation, B eigenvectors are M ^-1X , characteristic value λ, and A feature values match. Of course, do not expect the same feature vectors, if the feature vectors are the same, the same becomes a matrix exactly equal.

Similar Matrices

For B = M ^-1AM

　　Set A , B , and C is an arbitrary matrix of the same order, there are:

　　(1) anti-wear properties: A～A

　　(2) symmetry: if A ~ B , then B ~ A

　　(3) is transitive: if A ~ B , B ~ C , then A ~ C

　　（4）若A~B，则二者的特征值相同、行列式相同、秩相同、迹相同。

　　（5）若A~B，且A可逆，则B也可逆，A^-1~B^-1。

特征值相等的情况

　　当A的所有特征值互不相同时，A必然存在n个线性无关的特征向量，此时A能够对角化；如果存在完全相等的特征值，是否能够对角化就不好说了，需要另行判断，我们对这类矩阵的相似矩阵同样感兴趣。

　　上面的对角矩阵有两个相同的特征值：λ₁=λ₂=4，如果A有相似矩阵，我们看看这个相似矩阵是什么：

　　此时A的相似矩阵是A本身，类似A这种特征值重复的对角矩阵，它们只和自己相似。

　　另一种特征值相同的矩阵则可能有很多相似矩阵，两个特征值都是4的这类矩阵中最简洁的是：

　　这个矩阵无法对角化，如果它能对角化，那么：

　　这显然是不成立的。类似A的矩阵虽然有完全相同的特征向量，但无法对角化，比如把右上角的元素1改成其他值。其中A是这类矩阵中最简单的一个，称为诺尔当标准型。

诺尔当标准型

　　诺尔当指出，对于特征值完全相同的方阵A，就算不能对角化，也一定能够通过变换得到与对角矩阵很接近的诺尔当标准型。具体来说，对于方阵A，一定有同样规模的可逆矩阵P，使得P^-1AP=J，J是诺尔当标准型。

　　诺尔当标准型到底是个啥？举个例子：

　　上面的矩阵就是诺尔当标准型，其中空白区域的元素全是0，每一个红色方块是一个诺尔当块。每个诺尔当块都要满足两个性质：主对角线元素完全相同（特征值完全相同），主对角线上方的次对角线元素全为1（如果有次对角线的话）。上面的矩阵是5个诺尔当块构成的，其中[4]比较特别，它只有主对角线，没有次对角线，是大小为1的诺尔当块。

　　若尔当标准型是由若干个若尔当块按对角排列组成的准对角矩阵。

　　有时候，诺尔当标准型不是那么容易辨别。来看几个诺尔当标准型：