how to calculate eigen values of a N X N matrix.?
550410
There are a lot of software libraries that do that:\n\nhttp://tinyurl.com/k9zah\n\nincluding some for Excel:\n\nhttp://tinyurl.com/hrda7
945015
Modal analysis is one application of Eigenvalue analysis but Eigenvalue analysis can be used for other things.\n\nIn the context of modal analysis (using FEM), the eigenvalue=eigenfrequency=natural frequency (mode).\n\nThe eigenvector is the diplacement vector or mode shape of the mode.\n\nI haven't seen the term eigensystem, but it probably refers to the matrix and vector for the problem (i.e. everything as a whole).
346932
well, in case I don't answer sufficiently, I provided a link for Wikipedia to define the ranks of matrices.\n\nBasically, ranks are when the matrices are reduced and you want to know the number of independent rows and columns.\n\nSince A^2 = I, this confirms A is nxn, so we're off to a good start.\n\nThe product of each row i x column i = 1, and row i x column j = 0, when i <> j.\n\nI think rank(I+A) + rank(I-A) somewhere is the solution by comparing the number of independent rows left of I+A and noticing there's only so many independent rows left in I-A.\n\nOne might wonder what does rank(I+A) + rank(I-A) suggest about the equation... (I+A)x(I-A)\n\n(I+A)x(I-A) = matrix[I(i,j)+A(i,j)] x matrix[I(k,l)-A(k,l)]\n            = matrix[sigma(row(i)xcolum(l))]\n            = matrix[sigma(j)(I(i,j)*I(i,l)-I(i,j)*A(j,l)+I(i,j)*A(j,l) -A(i,j)*A(j,l))]\n            = matrix[sigma(j)(I(i,j)*I(i,l)-A(i,j)*A(j,l))]\n            = I^2-A^2\n            = I - I\n            = 0\n\nSorry I couldn't answer your question completely.
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 :)
805688
U can use either Kruskal's algorithm or prims algorithm. More information about these algorithm u can find on internet.\n\nPrims algorithm is based on adjacency matrix.
954567
Check out Markov Chain's.  You will find many uses for matrices.
656507
Use the law of sines\n\nn/sin(N)=m/sin(M)\n\nso \n\nm = n*sin(M)/sin(N)\n\nPlug in the values and use your calculator.
100903
If A is an n-by-n symmetric positive definite matrix with real entries, that means ((v transpose)Av) > 0 for nonzero real n-by-1 column vectors v.\n\nConsider the 3x3 case with A equal to\n\n[ a b c ]\n[ b d e ]\n[ c e f ]\n\nLet u, v, and w be the nonzero column vectors\n\n[ 1 ]\n[ 0 ]\n[ 0 ]\n\n[ 0 ]\n[ 1 ]\n[ 0 ]\n\n[ 0 ]\n[ 0 ]\n[ 1 ]\n\nrespectively.  Then\n\n((u transpose)Au) = a > 0\n((v transpose)Av) = d > 0\n((w transpose)Aw) = f > 0\n\nCan you see how to generalize this for other values of n besides three and/or for matrices and vectors with complex entries?
304446
hey!! chek out this website...\nBUT Y DO U WANT TO USE MATLAB FOR SEPARATING EVEN AND ODD?? THERE ARE OTHER METHODS...
949909
So Im guessing your have a transformation matrix.  \n\nYou have an initial state specified by matrix [A] and the linear transform specified by matrix [T].\n\nTo find the state after iteration 1 you multiply\n[A]x[T]=[y1]\n\nIf [y1] was your initial matrix you would multiply it by [T] to get the state after iteration.\n[y1]x[T]=[y2]\nby substitution\n([A] x [T]) x [T] = [y2]\nor \n[A] x [T]^2 = [y2]\n\nso what is T at the nth iteration?\n[A] x [T]^n = [yn]\n\nso what is the limit as n goes to inifinity?\n[A] x lim([T]^n,n->inf) = [y_inf]\n\nNumerical results break down, consider a rotation.  The limit as it goes to infinity is undefined.\nT(1,1)=sin(x)\nT(1,2)=cos(x)\nT(2,1)=-1*cos(x)\nT(2,2)=sin(x)\n\nWhen you square it you get\nT2(1,1)=sin(x)^2-cos(x)^2 = -cos(2*x)\nT2(1,2)=2*cos(x)*sin(x) = sin(2*x)\nT2(2,1)=-2*cos(x)*sin(x) = -sin(2*x)\nT2(2,2)=sin(x)^2-cos(x)^2 = -cos(2*x)\n\nWhen you cube it you get\nT3(1,1)=-sin(3*x)\nT3(1,2)=-cos(3*x)\nT3(2,1)=cos(3*x)\nT3(2,2)=-sin(3*x)\n\nWhen you raise it to the power of 100 (come on MatLab, dont die on me now, darnit) ... ahem.  When you raise it to the power of 10 (?)  aha\nT10(1,1)=-cos(10*x)\nT10(1,2)=sin(10*x)\nT10(2,1)=-sin(10*x)\nT10(2,2)=-cos(10*x)\n\nSee how that works?  Not bad huh?\n\nFor the pure rotation it never ends.\nWhat about a shrinking, a scaling by a matrix whose norm is less than 1? (note if you use 0.9, the norm is 1.8.)\n\nT(i,j)=0.45\nyields\nT2(i,j)=0.4050\nT3(i,j)=0.3645\nT10(i,j)=0.1743\nT1000(i,j)=0.1328*10^-4\n\nIt disappears to zero.\n\nintuitively you can also figure out that for a scaling by a matrix whose norm is larger than one the transform eventually goes to infinity as the limit is taken to infinity.\n\nyou could make a non-trigonometric rotation, like the sin(1/x)  Thats a fun expression.\n\nT(i,i)=sin(1/x) \nT(1,2)=cos(1/x)\nT(2,1)=-1*cos(1/x)\n\nAs x goes to infinity what happens?\nit becomes a reflection\nT_inf(i,i)=0\nT_inf(1,2)=1\nT_inf(2,1)=-1\n\nIm digressing.  Its a fun field with interesting applications.  Weather forecasting, deformation of solids, heat transfer, optimization, all use linear transformations to do their work.\n\nHave fun.
629797
if u have \n\nt=1:2:1000\n\nS =10sin30t\n\nplot(S)\n\nFs=fft(S,512)\n\nplot(Fs) \n\n\nso if u have any signal it will be stored in vector on the matlab .. like .. X= 155x1 matrix size\n\njust convert the X by\n\nXftt= fft(X)\nor\nXfft2=ffft2(X)\nplot(Xfft)\npause\nplot (Xfft)\n\n\nand anything u want u can use matlab help it is v cool \n\n\ni wish i could help u
964615
Please type smith's algorithm for polynomial matrices at yahoo search you get links hope fully one of them should provide you with a solution.
199213
You probably mean eigen-values. We say that x is an eigen value for the square matrix A if there is a vector v such that Av=xv. Eigen-values are usually found by solving det(A-xI)=0\nwhere I is the identity matrix.
1132678
Directions how to do this can be found in the text book.\n\nDo your own homework!
491342
A square matrix is n by n, so it would have n^2 entries.  Each entry has 2 possible values: 0 or 1.  So, by the counting principle, the number of matrices with entries that are either 0 or 1 is 2^(n^2).
823237
There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
458222
0 1 0 0\n0 0 1 0\n1 0 0 0\n0 0 0 1\nPut the identity matrix behind your matrix.\nTurn you matrix into the identity matrix using row operations.  The matrix in the back will be the inverse.\n0 1 0 0  1 0 0 0 \n0 0 1 0  0 1 0 0 \n1 0 0 0  0 0 1 0 \n0 0 0 1  0 0 0 1\nSwap row 1 with row 3\n1 0 0 0  0 0 1 0 \n0 0 1 0  0 1 0 0 \n0 1 0 0  1 0 0 0 \n0 0 0 1  0 0 0 1\nSwap row 2 with row 3\n1 0 0 0  0 0 1 0 \n0 1 0 0  1 0 0 0\n0 0 1 0  0 1 0 0 \n0 0 0 1  0 0 0 1\nThus, the inverse is\n0 0 1 0\n1 0 0 0\n0 1 0 0 \n0 0 0 1