2014年10月28日 · Assume x is a n by 1 column vector, then , it is easily known that xTx x T x is the sum of squares. When calculate xxT x x T, it will give you a n by n matrix. I am wondering what is the mathematical meaning of such results.

If α α is an eigenvalue of xxT x x T, then det(αI − xxT) = 0 det (α I − x x T) = 0. So it suffices to show that 1 1 is an eigenvalue of xxT x x T. But this is trivial:

I found a surprising property of the eigenvectors of the matrix A = XTX A = X T X and B = XXT B = X X T experimentally. Let X X be n × d n × d with n> d n> d.

2021年5月22日 · Hence in row echelon form of XXT X X T, there is only one row and thus, XXT X X T has n − 1 n − 1 eigenvalues as 0 0 and one eigenvalue is non-zero. With their help I was able to conclude from trace property of a matrix that the non-zero eigenvalue was (sum of square of each element in column vector X X).

2019年6月27日 · X X is a matrix of shape DxN D x N, where each column is one of the samples. XXT X X T is the multiplication of X X and the transpose, which would be the covariance if X X was zero mean.

2019年8月13日 · I'm given the matrix X = xxT ∈ Rd x d, x ∈Rd X = x x T ∈ R d x d, x ∈ R d. Does somebody know how to compute λmax(X) λ m a x (X) or λmin(X) λ m i n (X)? I only want to know these two eigenvalues, the others are not really important. I seem to …

2015年4月23日 · It is indeed quite fast, and I obtain V V, whose each column is an eigenvector for XXT X X T, and D D, a diagonal matrix holding XXT X X T 's eigenvalues. How do I go from V V and D D to the eigenvalues and eigenvectors of XTX X T X?

2016年12月2日 · Let A = I − xxT, where x ∈ Rn and I is the identity matrix of Rn We know that A is a real symmetric matrix, therefore there exists an eigenvalue decomposition of A such that A = QTΛQ

2014年9月20日 · How is differentiation of xxT with respect to x as 2xT, where x is a vector? xT means transpose of x vector.

I need a bit of help understanding a step in the derivation of the expected value of xxT, that is, E[xxT] with a Gaussian distribution. By definition, using the D-dimensional Gaussian distribution, we have: