What is Rank of a Matrix?
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The RANK of matrix A, an m x n matrix, is the dimension of the row spaces or the column space. Or, the number of basic variables in Ax = 0. \n\nUsually written as rank A.\n\nAlso, for m x n matrix A, the row space and the column space have the same dimension, \n\nwritten dim R(A) = dim C(A).
201316
This website may help you.  Good luck.\n\nhttp://www.sosmath.com/matrix/matrix.html
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.
199628
Assuming, of course, that each of the vectors has 3 entries...\n\nDo you know how to put a matrix formed from those four vectors in row-echelon form?\n\nI'm trying to decide the best way to teach this to you, which is why I need to know if you know how to get the matrices in REF, but the final "Answer" is to see if three of those vectors are linearly independent. If three of them are linearly independent, then they must span R^3.\n\nFor your second question, no. Remember that a singular matrix is a square matrix that is not invertible. Therefore, a singular matrix will not have a pivot in every column, which means that the columns are linearly dependent, which means that the rank is less than the dimension. Any square matrix with a column that is all 0's will will be singular.
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Depends on what you are analyzing.  Matrix algebra is used extensively in communications and signal processing applications.  Not very much, though, in financial applications.
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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.
475802
Basically, Cramer's Rule lets you solve for a variable in a system of equations via matrix operations without solving the whole matrix.  The process requires some knowledge of finding determinants.  The site below walks through an example of how this works.
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I am assuming that you mean A^2 = A, meaning that A is idempotent.\n\nI'm not really sure.  I know that rank(A) + rank(I-A) = n in the case that A is idempotent, since \n\nrank(A) + rank(I-A) = tr(A) + tr(I-A) (since I-A is also idempotent)\n= tr(A + I - A)\n= tr(I)\n= n\n\nBut since I+A is not idempotent, you can't do the same thing.  Sorry.\n\n\nUPDATE:  I just found a case that contradicts that rank(I+A) + rank(I-A) = n\n\nI is 3 by 3.\nA =\n| 10/11 3/11 -1/11 |\n| 3/11 2/11 3/11 |\n| -1/11 3/11 10/11 |\n\nA turns out to be idempotent.  rank(I-A) = 1, rank(I+A) = 3, so rank(I+A) + rank(I-A) = 4, not 3.\n\nAre you sure you copied the problem correctly?
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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].
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Say a[i,j] are the elements of A. The definition of the determinant is:\n\ndetA = Sum over P: (-1)^P * a[1,P(1)] * a[2,P(2)]*...* a[n,P(n)]\n\nwhere the sum is over all permutations P of (1,2,...,n) and (-1)^P is -1,+1 for odd and even permutations respectively. (P(i) is the image of i under P).\n\nNow let's say (for example) that matrix B is created multiplying the second row by a constant c and adding it to the first row:\n\na[1,i] --> a[1,i]+c*a[2,i]\n\nThen the determinant of this new matrix will be:\n\ndetB = Sum over P:\n(-1)^P * (a[1,P(1)]+c*a[2,P(1)]) * a[2,P(2)]*...* a[n,P(n)]\n\n= det A +\nc * Sum over P: (-1)^P * a[2,P(1)] * a[2,P(2)]*...* a[n,P(n)]\n\nbut the last sum is zero, because for each permutation P there is exactly one other permutation differing only by a transposition T of (1,2)\n\nP' = PT , P = P'T\n\nand P and P' have opposite parity so the two terms generated by each of these n!/2 pairs cancel each other --> detB = detA
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| . x . x |\n| -x . -x |\n\nwhere x is any real number and x ≠ 0 (The last part is only needed so that A will be non-zero). The transpose also works.\n\nMultiply it out yourself. The result has nothing to do with triangular matrices, so I also suggest that you do not merely cite special characteristics of such matrices in your answer.
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I assume you mean invertible, not convertible.\n\nIt depends on how detailed you want to get. If you do not want to dig in extremely deeply, you can just do it by construction. \n\n1. A symmetric matrix M has an orthogonal decomposition\n\nM=O'DO\n\nwhere O is an orthogonal matrix, O'O=I, the identity, and D is a diagonal matrix with eigenvalues of M down the diagonal.\n\n2. If M is positive definite, then the diagonal elements of D are >0\n\n3. Produce the matrix \n\nA=O'(D^(-1))O\n\n\n4. Verify that A is the inverse of M\n\ni.e., \n\nA*M=M*A=I
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The size of a matrix multiplication is as follows:\n\n(nxm) (mxp)= (nxp)\n\n(1x2) (1x1) = 1x1\n\n(3x1) (1x3) = 3x3
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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.
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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.
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divide the first row by 2\n1 1/2\n1  -1\n\nSubtract the first row from the second row:\n1  1/2\n0  -3/2\n\nMutiply the second row by -2/3\n1  1/2\n0   1\n\nSubtract 1/2 of the second row from the first row:\n1  0\n0  1\n\nAnd you have the identity matrix
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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).