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How many $4$-digit palindromes are divisible by $3$? How many 4 4 -digit palindromes are divisible by 3 3? I'm trying to figure this one out I know that if a number is divisible by 3 3, then the sum of its digits is divisible by 3 3 All I have done is listed out lots of numbers that work I haven't developed a nice technique for this yet
matrices - When will $AB=BA$? - Mathematics Stack Exchange Given two square matrices A, B A, B with same dimension, what conditions will lead to this result? Or what result will this condition lead to? I thought this is a quite simple question, but I can find little information about it Thanks
Proofs of determinants of block matrices [duplicate] I know that there are three important results when taking the Determinants of Block matrices $$\\begin{align}\\det \\begin{bmatrix} A amp; B \\\\ 0 amp; D \\end
The commutator of two matrices - Mathematics Stack Exchange The commutator [X, Y] of two matrices is defined by the equation $$\begin {align} [X, Y] = XY − YX \end {align}$$ Two anti-commuting matrices A and B satisfy $$\begin {align} A^2=I \qu
If eigenvalues are positive, is the matrix positive definite? This question does a great job of illustrating the problem with thinking about these things in terms of coordinates The thing that is positive-definite is not a matrix M M but the quadratic form x ↦ xTMx x ↦ x T M x, which is a very different beast from the linear transformation x ↦ Mx x ↦ M x For one thing, the quadratic form does not depend on the antisymmetric part of M M, so
Trace of AB = Trace of BA - Mathematics Stack Exchange We can define trace if A =∑i ei, Aei A = ∑ i e i, A e i where ei e i 's are standard column vectors, and x, y =xty x, y = x t y for suitable column vectors x, y x, y With this set up, I want to prove trace of AB and BA are same, so it's enough to prove that
Show that $ e^{A+B}=e^A e^B$ - e^ {A+B}=e^A e^B$ - Mathematics Stack . . . As a remark, it is actually legitimate to assume that A A and B B are simultaneously diagonalisable (surprise, surprise!), so the proposition is trivial But obviously, the reason why we can make such an assumption is way beyond the scope of undergraduate (or even graduate) linear algebra courses