Is a 2*2 zero matrix non-diagonal,but diagonalizable and not invertible?
139869
[ 0 0 ]\n[ 0 0 ] has determinate 0, thus it is not invertible.  All nxn matrices are diagnalizable, but the zero matrix is already in diagnol form, having zeros everywhere except the main diagnol.  It is just has diagnol entries (0, 0).  Thus I would say it is not non-diagnol.
135158
So a hermitian matrix is defined as one that is self-adjoint, that is, when you take the conjugate transpose of the matrix you obtain the same matrix.  So let's just think about a 2D example:\n\n(Try to ignore the underscores (_), they're just there to make the matrix look more like a matrix; Answers deletes extra spaces).\n\n[ a________c+id]\n[c-id_______b]\n\nSo this is a hermitian matrix because when you take the conjugate transpose you get the matrix back (if you don't believe me, just write down the conjugate transpose and you'll see that you get that same matrix back).\n\nAnyway, so by observation we can see that we could write this as the sum of two matricies:\n\n[ a, c ]\n[ c, b ]\n\nand\n\n[0___id ]\n[-id__0 ]\n\nNotice that the first one is symmetric, because when you take the transpose, you get the matrix back.\n\nNotice also that the second one is skew symmetric (also called antisymmetric) because when you take the transpose you get the same matrix back with all of the elements inverted in sign.\n\nAdd them together, and viola, you get the original hermitian matrix back.\n\nSo we've shown by construction here that a 2x2 hermitian matrix can be written this way and we've done it in the most general case because we've considered complex off-diagonal elements and real diagonal elements (Consider: Would you have a hermitian matrix if the diagonal elements were complex rather than real?  Answer: You would not, because the conjugate transpose would not equal the original matrix.  Try it, if you don't see it.)\n\nAnyway, I've given you an example here for 2x2; 3x3 and NxN are exactly the same, except you have to write out more terms.\n\nI hope that's been helpful!  I assume you're either taking a first course in quantum or in linear algebra, so in either case good luck.  You're entering an interesting world.
874355
Using matices to do this\n\n[ 1 -2 1] * [x] = [-5]\n[ 2 -1 1]   [y]    [ 1]\n[2  1  2]   [z]    [10]\n\nAX = B\nA*A^(-1)X = A^(-1)B\nX = A^(-1)B\n\nSO we need the inverse of A this can be worked out using a long and tedious method or just use a calculator\n\nA^-1 = [ -.6 1 -.2]\n           [ -.4 0   2]   \n           [.8   -1 .6]\n\nA^-1B = [2}\n              [4}\n              [1]\nso x =2  y=4 z=1\n\n now you do the other one\nCheck\n2 - (4*2) +1 = 2 - 8 +1 = -5
802863
| A - I(lamda)| = 0 ==> eigenvalue ... \n\nlamda is an eigenvalue...\n\nsolve the matrix first
790217
Great question.\n\nMaybe you can work backwards from the characteristic polynomial.  You know the roots so:\n\np(l) = (l+2)(l-1)(l-3)  right?  Now multiply out, the coeff of the linear term is the negative of the trace.  The sum of the eigenvalues is equal to the trace.  The constant term is the determinant.  \n\nor, start with the diagonal matrix and apply a basis transformation to get a non-triangular matrix.  \n\nor, choose 3 orthogonal vectors to call the eigenvectors, then find the transformation of the ....(out of steam)
331428
Can you show that T has a real eigenvalue? If so, let v be an eigenvector corresponding to this eigenvalue. Then, the line tv for all t in Reals is invariant under T :)
819996
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
938958
No, it would have to be the square of something like ax+b, whose square is a^2x+2abx+b^2.  You would have to have a=1 or -1 and b=i or -i, but then 2ab would be imaginary, and not -2.
741630
if the matrix is squared, and it has an inverse, then the answer is YES
739088
Consider a = 1, b = 0, c = 1.  Clearly, b^2-4ac = -4 < 0.  This corresponds to the equation x^2 + 1 = 0.  But x^2 is never negative, and 1 more than that is always positive, so this has no solutions.\n\nIf you want only non-imaginaries, there are no solutions when b^2 < 4ac and only one solution when b^2 = 4ac.\n\nYou clarify: "that does not suffice because every quadratic is meant to give probable solutions because all polynomials and quotients of polynomials are rational, and rational equations don't give imaginary solutions."\n\nRational equations do indeed give complex solutions, many of them nonreal.  The example I gave above (x^2 + 1 = 0) is an example.  The algebraic closure of the reals is the complex numbers; the algebraic closure of the rationals is the algebraic numbers.  If you want all quadratics to give two solutions (counting multiplicity), you'll need to allow imaginaries.
438919
I dont know how to put this up here. I tried to put it in a nice format, but it would accept.\nBut you put the following in one matrix \n-2  6\n+2  2\n\nThen you put a * (multiply)\n\nThen you put the following in one more matrix\n\nx\ny\n\n\nThen put an ' = ' symbol\n\nThen put the following in a matrix\n\n40\n8
924063
eigenvalues: 2, 1\n\nIf you can write the eqns for the eigenvectors, surely you can solve them (the solution is unique only up to scale, so don't expect it to be unique).\n\neigenvectors (columns! i.e. right eigenvectors, not left eigenvectors which are rows) are: [1, 1] for the e.v. 2 and [3, 2] for the e.v. 1. Write them as columns and concatenate into a 2x2 matrix P whose rows are [1,3] and [1,2]. Then P^-1AP = diag(2,1).\n\nA^4 = (Pdiag(2,1)P^-1)^4 = Pdiag(2^4,1)P^-1\n= Pdiag(16,1)P^-1=....(multiply out)\n\nAns: A^4 = (written in rows) [-29,45] [-30, 46].
339198
2x0 = 0\nIt'll be the same as today... :P
414270
Add the two tops and add two bottoms.Then the sum is\n(Top)[1+0 , 3+0] and (Bottom)[2+0 , 6+0] = E\nE is any 2x2 matrix and O is the 2x2 zeromatrix.\nAlways is E+O = O+E = E.\nThis is also true for nxn matrices.
333863
Diagonalisation is the answer here.  Not sure how this didn't turn up in a web search, as I found plenty of resources for it, but the idea is that you find a matrix that diagonalises your initial matrix, and this allows powers to be done trivially.\n\nAssuming you're aiming to get a matrix A to the power n:\n\nFind P such that:\nP^-1 A P = D, where D is diagonal.  Then:\nA = (P D P^-1) and:\nA^n = (P D P^-1)(P D P^-1)(P D P^-1)....\n\nThe P^-1 P terms cancel and you're left with:\nA^n = P D^n P^-1\n\n... and this is much easier to calculate as D^n is trivial :)\n\nA good reference is in the sources.  Hope that helps!
308173
You haven't explained the question properly! These matrix can be formed any way!
1126894
This normally means that you are going to be using Gaussian (or Gauss-Jordan) reduction, where ther eare three "moves" that are allowed.\n\n1) You can switch any two rows in the matrix.\n2) You can multiply any row in the matrix by a non-zero scalar.\n3) You can add a multiple of one row to another.\n\nUsing these three "moves," the goal is to reduce the matrix into a usable form (the echelon form, for example).