How can I find the Jordan form and basis of this 5*5 matrix?
830322
I just noticed that two rows are the same, right?  This is big, but I'll have to study up what it means.  Intuitively, I think that it means that the matrix has a null space, and thus and eigenvalue of zero.  \n\nor is it a type?\n\n\nRead up on it:\n\nhttp://en.wikipedia.org/wiki/Jordan_normal_form
1183592
If your matrix is not square, then you can compute a generalized inverse matrix.  There are some links below.  There are many generalized inverses for matrices (other than those that are square and invertible; those only have one generalized inverse, which is its inverse).  But there is only one Moore-Penrose inverse for any given matrix.\n\nedit: reetz, I have already reported you for copying.  I suggest you back off.
480863
There's something missing in that hint, there must still be an 'n' in there somewhere.\n\nTo work out the general formula:\n\nThe number of 2*2 matrices is (n-1)(m-1). The number of 2*3 is (n-1)(m-2). The number of 2*4 is (n-1)(m-3)... and so on until you reach (n-1)(1) = (n-1)(m-(m-1))\n\nSo the total number of 2 * j matrices is\nSigma(j = 1) { (n-1)(m-j) }\n\nThen the number of 3 * j matrices is\nSigma(j = 1) { (n-2)(m-j) }\n\nAnd so on, up to\nSigma(j = 1) { (1)(m-j) }\n\nThe sum of all these gives the number of i * j matrices as:\n\nSigma(i=1 to n-1)Sigma(j=1 to m-1) {(n-i)(m-j)}
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)
738212
you need to set up a matrix equation.  \nthe rows should look like this:\n\n1      ,   -2    ,     1    ,      / -8\n3      ,   -3    ,    -2    ,     /-5\n-2     ,   -3     ,    3      ,    /5\n\nOnce you set this up follow the rules of matrices to get this to a unit matrix.  Once you have done this the three numbers on the right will become your answers for x, y and z.  x being which ever row has the 1 in the first column.  y being which ever row has the 1 in the second column.  z being which ever row has the 1 in the third column.
590873
You have a 3*2 matrix, into which each cell can have either a 0 or a 1.  You have two choices - 0 or 1 - and you have six different places to put your choices (2*3 = 6).  To find the total number of matrices, you multiply each cell by every other cell.  In this case,\n\n2*2*2*2*2*2 = 64, which can also be written as 2^6 (2 to the 6th power) = 64\n\nFollowing this logic, to find the number of n*m matrices that have an entry of 0 or 1, you would again multiply every cell by every other cell.  In this case, using exponential notation only, since we don't know what n*m equals,\n\n2^(n*m) or 2 raised to power of n*m
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.
542644
Here are a few websites which explain it fairly well...\n\nhttp://www.mai.liu.se/~halun/matrix/\n\nhttp://www.easycalculation.com/matrix/matrix-multiplication.php\n\nThat one has a matrix calculator online. Good luck!
889697
well gauss jordan is a easy method for matrix manipulations. if u use reverse triangulation also in conjunction with it . it wud be much more simpler.\n\n2x-y+3z=8\n-x+2y+z=4\n3x+y-z=0\n\nthis is set of equations given.write them in matrix form\n\n|| 2  -1  3 ||  ||x||     ||8||\n|| -1  2  1 ||  ||y||  = ||4|| \n|| 3   1  -1||  ||z||     ||0||\n\nA    X= B\nnow the equation is in the form of AX=B\n\nnow our concentration will be on matrices A and B\n\nour aim is to make the lower triangle elements equal to 0.\n\nnow follow the steps one by one. apply the calculations on both A and B  matrices.\n\nwe have 3 rows in matrix A and let us name them as R1 , R2 and R3.\n\nreplace R2 by R2/-1 and R3 by R3/3\n\nnow the matrices will be\n\n||2   -1    3    || ||x||   ||8 ||\n||1   -2   -1    || ||y||= ||-4||\n||1   1/3  -1/3 || ||z||   ||0 ||\nreplace R2=R1-2xR2\n||2   -1    3     || ||x||    ||8 ||\n||0    3    5     || ||y||=  ||16||\n||1   1/3  -1/3 ||  ||z||   ||0 ||\ncheck once what we have done here. \nevery element of R2 is calculated from the preceding element in that colum by the formula R1-2xR2.\n\nthe elelment of colum one is\n2-2x1=0\nthe element of colum 2 is -1-(2x-2)=3\nthe element of colum 3 is 3-(2x-1)=5\n\nlike wise \nreplace R3=R1-2xR3\n||2   -1    3      || ||x||     ||8 ||\n||0    3    5      || ||y|| =  ||16||\n||0   -5/3  11/3 || ||z||     ||8 ||\n\nnow we got the first colum zero in the lower triangle.\n\nnow the second element in the third row needs to be zeroed.\nreplace R3=R3x(-3/5) or  R3=R3/(-5/3)\n||2   -1    3      || ||x||   ||  8   ||\n||0    3    5      || ||y||= ||  16  ||\n||0    1  11/5   || ||z||   ||-24/5 ||\nreplace R2=R2/3\n||2   -1    3      || ||x||   ||  8   ||\n||0    1    5/3   || ||y||= || 16/3 ||\n||0    1  11/5   || ||z||   ||-24/5 ||\n\nnow replace R3=R3-R2\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\n\nnow apply the final steps of reverse triangulation technique.\nfrom the matrix equation, if again converted in to normal equations \n\n2x-y+3z   =8\n  y+5z/3  =16/3\n    8z/15 =-152/15\nnow z= (-152/15) divided by (8/15)\n= (-152/15)/(8/15) =-152/8=-19\nthere fore z=-19\nnow substitute z=-19 in equation 2\n\ny+5x(-19)/3=16/3\ny=(16/3)+(95/3)=111/3=37\nsubstitute the values of y and Z in equation 1\n\n2x-37+3x-19=8\n2x=8+37+57=102\nX=102/2=51\nfinally we got the values of x = 51, y=37 and z=-19\n\nactually u can continue the guass jordan method without applying the reverse triangulation method to get the values of x, y \n\nand z directy from the matrix for that after making the lower triangle zero we have to make the upper triangle elements also \n\nequal to zero.\n\nlet us see how to do that.\n||2   -1    3      || ||x||   ||  8     ||\n||0    1    5/3   || ||y||= || 16/3   ||\n||0    0    8/15 || ||z||   ||-152/15 ||\nreplace R3=R2/(8/15)\n||2   -1    3     || ||x||   ||  8     ||\n||0    1    5/3  || ||y||= || 16/3   ||\n||0    0    1     || ||z||   || -19    ||\nR2=R2-5xR3/3\n||2   -1    3    || ||x||  ||  8     ||\n||0    1    0    || ||y||= ||  37    ||\n||0    0    1    || ||z||  || -19    ||\n\nR1=R1/3\n||2/3   -1/3    1    || ||x||   ||  8/3   ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\n\nR1=R1-R3\n||2/3   -1/3    0    || ||x||   ||  65/3  ||\n||0       1     0     || ||y||= ||  37    ||\n||0       0     1     || ||z||   || -19    ||\nR1=R1x-3\n||-2    1     0    || ||x||   ||  -65   ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1-R2\n||-2    0     0    || ||x||   ||  -102  ||\n||0       1     0    || ||y||= ||  37    ||\n||0       0     1    || ||z||   || -19    ||\nR1=R1/(-2)\n||1    0     0    || ||x||  ||  51    ||\n||0       1     0    || ||y||=||  37    ||\n||0       0     1    || ||z||  || -19    ||\nnow we can diretly read the matrix into equations as\nx=51, y=37 and z=-19\n\n\nhope und
487838
1) Assuming an aritmetic sequence\n\nbn = b1 + (n-1)d \nletting b3=a1 = 20, then b7=a5 = 32, and b25 = b23\na5 = a1 + (5-1)d\n32 = 20 + (5-1)d\n32 = 20 + 4d\n12 = 4d\n3 = d\n\nb25 = a23 = 20 + (23-1)*3\nb25 = 20 + 22*3\nb25 = 20 + 66\nb25 = 86\n=====================================================\n2)\na75 = a1 + (75-1)d; d=-3\n\na75 = 5 + (75-1)*-3\na75 = 5 + 74*-3\na75 = 5 + -222\na75 = -217\n\nSn = n/2(a1+an)\nS75 = 75/2(5 + -217)\nS75 = 75/2(-212)\nS75 = -7950\n===================================================\n3) ????\n\na5 = 5*5 - 4 = 21\na15 = 5*15 - 4 = 71\na25 = 5*25 - 4 = 121\n\nOR MAYBE\n\na1 = 5*1 - 4\na1 = 5 - 4\na1 = 1\na30 = 5*30 - 4\na30 = 150 - 4\na30 = 146\n\nSn = n/2(a1+an)\nS30 = 30/2(1 + 146)\nS30 = 30/2(147)\nS30 = 2205\n================================================\n4)\n\nrth term is nCr-1m^n-(r-1)k^r-1\n\n5th term =\n4C4m^4-(5-1)k^5-1\n4C4 = 1\n\n5th term = k^4
339354
(2x + b) * (a + 3c)\n\n\nsteps:\n2ax + 6xc + ba + 3bc \n1) group first two, and last two\n(2ax + 6xc) + (ba + 3bc)\n\n2) factor out 2x from the first, and then b from the second group\n2x*(a + 3c) +  b*(a + 3c)\n\n3) since (a + 3c) is common, combine 2x and b\n(2x + b)*(a + 3c)
978893
The last link was the most informative.\nThe function is as follows:\n(x' y')T = [(1 1)T (1 -1)T] (x y)T\n\nIf you multiply the basis vectors by this matrix you get:\n(0 2)T for the (1 -1)T vector and\n(-1 -5)T for the (-3 2)T vector.\n\n(0, 2) = c1 * (1, -1) + c2 * (-3, 2)\nso:\n0 = c1 - 3 * c2\n2 = - c1 + 2 * c2\nc1 = -6\nc2 = -2\n\nlikewise:\n(-1, -5) = c3 * (1, -1) + c4 * (-3, 2)\nc3 = 17 and c4 = 6\n\nM = [(-6 -2)T (17 6)T]
551932
Assuming your matrix is of order nxn and you want submatrix of order mxm.\n\nThen there is (n-m+1)^2 submatrices.\n\nSo you can run this loop(Java code)\n\nfor(i=0;i==n-m;i++)\n{\nint k;\nfor(k=0;k==n-m;k++)\n{\nint p,q;\nfor(p=0;p<m;p++)\n{\nfor(q=0;q<m;q++)\n{\nprint("  ");\nprint(a[p][q]);\n}\nprintln();\n}\n}\n\n}\n\nTo print all the submatrices iterate over this procedure canging m from 1 to n.
718910
ill use this as a key for a standard matrix:\n\na b\nc d\n\nin your coeff. matrix:\n\na=equation 1 x coeff\nb=eq.1 y coeff.\nc=eq.2 x coeff.\nd=eq.2 y coeff.\n\nso the coeff. matrix looks like\n\n3   11\n-1  4\n\nthe value of the determinant is always "ad-bc" for a 2 by 2, so yours youls be\n\n(3)(4)-(-1)(11) = 12+11 = 23\n\n\nand i dont know what you mean by adjoint, sorry
219213
So you need to find a lower diagonal matrix L and an upper diagonal matrix U such that L*U equals the given matrix. The most straightforward way of doing this is to perform Gaussian elimination while keeping track of all the row operations you perform. The row operations are used to construct the matrix L, and the end product of the Gaussian elimination will be U.\n\nStart with the 1 in the top left corner and use it to eliminate all the non-zero terms below it. So take twice the first row and subtract it from the second row, giving you:\n1 1 1\n0 0 0\n3 3 3\n\nWhile you are doing this you should keep track of a matrix that will undo all the operations you have done (this will be your L matrix in the end). Start with the identity matrix and add a 2 in the first column of the second row (it's a +2 since we subtracted twice the first row but we are looking for a matrix that will undo all the operations we are performing and get us back to the original matrix)\n1 0 0\n2 1 0\n0 0 1\n\nNow eliminate the 3 in the bottom left corner by taking the third row and subtracting 3 times the first row, giving you:\n\n1 1 1\n0 0 0\n0 0 0\nwhich is upper triangular (so it'll be your U) and an L matrix:\n1 0 0\n2 1 0\n3 0 1\nwhich is lower triangular. You can confirm that L*U does in fact equal your original matrix, so you're done.
985452
Firstly, there is a formula. However, its quite easy to forget - so why not learn how to actually come up with it?\nSuppose you are given coordinates (a,b) in the first system. That really corresponds to the point a[2,1] + b[2,0]. Writing that in matrix form, we have: (// denoting next line)\n[2 2 // 1 0][a // b]\nNow, we want to express it as c[-2,-1]+d[0,1]. Writing that in matrix form we have:\n[-2 0 // -1 1][c // d].\nSo how do we work out [c // d]? Just multiply on the left by the inverse of that second matrix.\nSo it would be [-2 0 // -1 1]^(-1) * [2 2 // 1 0].\nThat turns out to be [-1 -1 // 0 -1].
920734
2xy(-15x^2y^4)+8x^2)-5/-4xy^3\n(-1/2)[-15x^2y^4+8x^2-5)/y^2\n(-1/2y^2)[-15x^2y^4+8x^2-5]