Reputation
Next tag badge:
75/100 score
37/20 answers
Badges
4 33 69
Newest
 Yearling
Impact
~530k people reached

1h
comment A question about product of three positive definite matrices
Obviously this is false in the scalar case. Also, when the sizes of $D$ is odd, it must have at least one real eigenvalue and hence the vector of spectrum cannot be any arbitrary point in $\mathbb C^3$. If you are asking whether every complex number $z$ is realisable as an eigenvalue of some $D$, you should edit your question.
3h
comment A question on matrix norm
A duplicate of A question about matrix norm .
3h
comment Vector perpendicular to timelike vector must be spacelike?
It's very nice. +1
1d
comment The number of $2\times 2$ complex matrices satisfying $A^{3}=A$
@user118494 Yes. In particular, $\pmatrix{1&z\\ 0&-1}$ is a solution for every complex number $z$.
2d
answered Matrix with all 1's diagonalizable or not?
2d
revised Finding an explicit eigenvector
added 138 characters in body
2d
answered Exist basis, simultaneously upper-triangular?
2d
comment About a matrix identity.
As a remark, the Matrix Cookbook is all about real or complex matrices. The two expressions can be meaningless (not to say true) over some finite fields (such as $GF(2)$).
Aug
29
comment The relationship between diagonal entries and eigenvalues of a diagonalizable matrix
This is nothing strange. Multiply out the relevant expressions and compare them, you will see that they are equal.
Aug
29
comment The Calculation Process
Do you know how to calculate the dot product of two vectors?
Aug
29
comment Eigen vectors of a matrix multiplied with its transpose
What do you mean by a subspace of a matrix?
Aug
29
comment Why do polynomial regressions have larger variance at the end?
Secondly, the OP asks why the deviation of the data points from the interpolating curve are larger near the end of the curve, but in fact this is data dependent. E.g. if the data points are concentrated at (20,50) and (80,50) and there are a few data points with age between 30 and 70 but large variance in wages, the regression curve will likely pass through somewhere near (20,50) and (80,50) and the phenomenon mentioned by the OP may not occur.
Aug
29
comment Why do polynomial regressions have larger variance at the end?
I don't think you are answering the OP's question. First of all, regression is not interpolation. In interpolation, we try to find a function (usually a polynomial) that approximates another function. In regression, however, there may be data points with identical abscissa (age) but different ordinates (wages) and hence the data set is not presented as a sample of some function.
Aug
28
awarded  Yearling
Aug
27
revised Prove that $\det(I-CD)=\det(I-DC) $
edited tags
Aug
27
comment Prove that $\det(I-CD)=\det(I-DC) $
The question inside the first box is a special case of Sylvester's determinant identity or Show that $|I_m-AB|=|I_n-BA|$, but I am not going to close this as duplicate because the OP is also interested to know if his proof attempt is correct.
Aug
27
comment Equivalence of positivity
What do you mean by $\pm\tilde{L}\ge0$? if both $\tilde{L}$ and $-\tilde{L}$ are positive semidefinite, $\tilde{L}$ must be zero, but I think you don't mean this, do you?
Aug
27
comment A question about matrix kernels and Kronecker products
@Jacquard The answer is still no if $d_A=d_B\ge3$. E.g. suppose $d_A=d_B=3$ and $v_A=(1,0,0)^T$. Take a random real symmetric $C$ whose first three rows/columns are zero. It's almost always not a Kronecker product of two $3\times3$ matrices because its nonzero $3\times3$ sub-blocks are not scalar multiples of each other.
Aug
27
comment proof-similar matrices have the same characteristic polynomial
What Marc says is also what I thought when I first saw the question. Attributing the equality $M^{-1}xIM=xI$ to the fact that $I$ commutes with every square matrix is just wrong. To prove this equality, one can use the fact that $xI$ commutes with $M$, but not the fact that $I$ commutes with $xM$. At any rate, proving the equality using commutativity is unnatural. What matters here is actually --- as Marc says --- another fact, namely, $(xA)B=A(xB)$.
Aug
27
answered A question about matrix kernels and Kronecker products