A is a matrix....... how to diagonalize it?
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].
48764
I found a website where you can download a program which uses this method.\n\nHere is how it is presented:\n\nSolving a System of Linear Equations v 1.0\nThis program solves a system of linear equations by Montante's method which have the peculiarity that once the system is solve in the main diagonal is the value of the determinant of the system. It works only with integers.  \nSize: 1 Kb  Downloads: 1112 Updated: 2003-7-23
168156
get a TI 82 or above graphing calculator.  MUCH easier. program already installed, just build the type of matrix,, and plug in the numbers
608785
The same way you would measure your TV screen....diagonally
712502
You know that the diagonal forms a right triangle with the length and the width. So you can use the Pythagorean Theorem.\nx=width=a\nx+1=length=b\n4=diagonal=c\n(a^2)+(b^2)=(c^2)\n(x^2)+(x+1)^2=(4^2)\n(x^2)+(x^2)+2x+1=16\n2(x^2)+2x+1=16\n2(x^2)+2x-16=0\nIf you use the quadratic formula thing (-b plus or minus, etc.)\nthen you get:\nx=2.2838\nor\nx=-7.5\nBut you know that a dimension cannot be negative, so it must be the first one.\nSo if \nx=2.2828\nthat means that that's the width. If you add one to that, you'll get the length.\nSo, the dimensions are:\n2.2838 x 3.2838
717242
I apologize for the notation in advance.\n\nCholesky factorization assumes that a matrix is symmetric and positive definite.  Symmetric is easy to determine.  Since Cholesky is fairly efficient, it can be used to determine if in fact a symmetric matrix is positive definite.  \n\nThe gist of the algorithm is this:  let's assume that we can write a symmetric positive definite matrix A as\nA=U'*U (' means transpose), where U is upper triangular.  To think about this, let's write the matrix A in blocks:\n\nA=(a_11 A_1,A'_1 A_22)\n\nwhere a_11 is the 1,1 element of A, A_1 is the remainder of row 1 of A, and A_22 is the 2,2 cofactor of A.  Similarly write U as \n\nU=(u_11 U_1,0 U_22)\n\nThen U'*U=A gives us\n\nu_11*u_11=a_11\nu_11*U_1=A_1\nU'_1*U_1 + U'_22*U_22 = A_22\n\nThis doesn't look like it gets us that far, but notice that U'_22*U_22 is a symmetric positive definite matrix, to which the preceding steps of the algorithm can be applied.  That's the way Cholesky works; proceed as above until done.  \n\nQuick example:\nA=(1 2 3,2 8 8,3 8 11)\nThen u_11=sqrt(1)=1\nU_1=(1/u_11)*A_1=(1/1)*(2 3)=(2 3) (this is a row vector)\n\nU'_22*U_22 = A_22 - (U'_1*U_1)\n=(8 8,8 11)-(2 3)'*(2 3)=(4 2,2 2)\n\nRepeat with this new matrix:\nu_22=sqrt(4)=2\nU_2=(1/2)*(2)=(1) (writing as a row vector to emphasize similarity with above)\nU_'33*U_33 = A_33 - U'_2*U_2 = 2-1*1=1\n\nFinally, u_33=sqrt(1)=1 and the algorithm terminates.  So your matrix U is\n\nU=(1 2 3,0 2 1,0 0 1)\n\nIf at any point, your proposed value for u_jj is the square root of a negative number, the matrix you started with is not positive definite.
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)
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.
273264
Lets call this matrix A.\n[ 2 0 4 ]\n[ 3 1 5 ]\n[ -1 1 -2 ] = A\n\nNow we need to solve the equation AV = I\nthen V is the inverse matrix.  Lets write this out\n\n[ 2 0 4 ] ` ` ` ` [ 1 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 1 0 ]\n[ -1 1 -2 ] V = [ 0 0 1 ]\n\nNow just multiply elementary matrices (row operations) to the left hand side of both sides of the equation until the left-hand entity is the identity.  Multiply the following matrix on the left-hand-side both sides [[1/2 0 0][0 1 0][0 0 -1]], that is replace the first row with 1/2 times the first row, and replace the third row with -1 times the third row:\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 3 1 5 ] ` ` ` ` [ 0 ` 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\napply [[1 0 0][[-3 1 0][0 0 1]] both sides, that is replace second row with second row plus negative-three times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 1 0 ]\n[ 1 -1 2 ] V = [ 0 ` 0 -1 ]\n\nmultiply both sides on the left by the matrix [[1 0 0][0 1 0][ -1 0 1]], that replace third row with third row plus negative one times first row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 -1 ] ` ` ` ` [ -3/2 ` 1 0 ]\n[ 0 -1 0 ] V = [ -1/2 ` 0 -1 ]\n\nmultiply by [[1 0 0][0 0 -1][0 1 0]], that is place opposite of row 3 for row 2 and row 2 for row 3\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 1 -1 ] V = [ -3/2 1 0 ]\n\nnow apply [[1 0 0][0 1 0][ 0 -1 1]] both sides, that is replace third row with third row plus -1 times second row\n\n[ 1 0 2 ] ` ` ` ` [ 1/2 0 0 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 -1 ] V = [ -2 ` 1 -1 ]\n\napply [[ 1 0 2][0 1 0][0 0 -1]] both sides, replace row 1 with row 1 plus 2 times row 3, and replace row 3 with negative 1 times row 3\n\n[ 1 0 0 ] ` ` ` ` [ -3/2 2 -2 ]\n[ 0 1 0 ] ` ` ` ` [ 1/2 0 1 ]\n[ 0 0 1 ] V = [ 2 -1 1 ]\n\nNow, since the left hand item is the identity, the inverse matrix is\n\n[ -3/2 2 -2 ]\n[ 1/2 0 1 ]\n[ 2 ` -1 1 ]\n\nThe product of all of the elementary operation matrices specified is also this matrix, and the reader can verity that AV = VA = I
866887
Matrix Transpose means interchanging of rows and columns.\n\nIf A = \n[ 1 3 ]\n[ -2 -1]\n\nthen A^T =\n[ 1 -2 ]\n[ 3 -1 ]\n\nTransposing this last matrix, (A^T)^T =\n[ 1 3 ]\n[ -2 -1 ]\n= A.\n\nThis will be true no matter what the numbers are, or the number of rows or columns, so in general: (A^T)^T = A.\n\nTo show that (A+B)^T = A^T + B^T:\n\nA + B =\n[ 1+3 3+(-4) ]\n[ (-2)+2 (-1)+(-2) ] =\n\n[ 4 -1 ]\n[ 0 -3 ]\n\nso, (A+B)^T =\n[ 4 0 ]\n[ -1 -3 ]\n\nA^T =\n[ 1 -2 ]\n[ 3 -1 ]\n\nB^T =\n[ 3 2 ]\n[ -4 -2 ]\n\nAdding, A^T + B^T =\n\n[ 1+3 -2+2 ]\n[ 3+(-4) -1+(-2) ] =\n\n[ 4 0 ]\n[ -1 -3 ]\n\nSo, left side = right side. Again, the result is true for any matrices A and B, provided that they have the same number of rows and the same number of columns.
113822
A 100 by 100 matrix has 100 eigenvectors. I would use some software to do this. This is too hard to do by hand. Buy the student version of MATLAB for example. You might be able to find some free software on the internet as well.
458991
No, but if you go to List mode you can enter your matrix as a list and manipulate it.  Check out the owner's manual for help on how to do it.  I've included a link to it.  Hope this helps.
599681
It depends on exactly what you mean by a matrix. If you're talking about multi-dimensional or nested arrays in ActionScript, then here is a link to get you started...
1159782
you can use linear algebra, look up gauss jordan elimination. You will want to put each equation in its own row inside a matrix. Then you will have a coefficient matrix and an augmented matrix. Row reduce the left side and you will have your desired answers on the right. If you have a TI-83+ type in your matrix and hit RREF([A]) and it will do this for you. Go to the link I provided and scroll down a bit and it will show you.
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.
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.
268402
Consider the matrix of the real parts of a Hermitian matrix as A and the matrix of the imaginary parts as B. Use the definition of Hermitian.\nAnother hint: If a_ij = a+ib, what would a_ji have to be if the matrix A is Hermitian ? What is then true about the diagonal of a Hermitian matrix?