What is a matrix norm?
818178
Think of a matrix as defining a linear transformation. Now imagine what happens to the unit sphere when that transformation is applied: it will get stretched and rotated into some type of ellipsoid. The norm of the matrix is the distance from the origin to the furthest point on that ellipsoid.
732131
ha! I did this today in Algebra!!\n\nOkay, basically you take each of the coordinates and plug them into this fomula\n\n\nplusorminus 1/2  det| x1 y1 1|\n                                 | x2 y2 1|\n                                 | x3 y3 1|         solve for 1/2 of this detemanant just like you would solve for a 3 x 3 matrix.  You answer of course must always be positive.\n\nfor some reason, the two lines under plusorminus... got moved over, they should fall directly under |x1 y1 1|
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.
211287
here you find a free download from a book (from the American Mathematical Society):\nhttp://www.ams.org/online_bks/coll17/\n\nThe wikipedia is also a good source\nhttp://en.wikipedia.org/wiki/Matrix_(mathematics)\n\nAnd here's an on-line calculator for matrix operations (it seems cool... I didn't try it)\nhttp://www.quickmath.com/www02/pages/modules/matrices/index.shtml\n\nAnd... mathworld\nhttp://mathworld.wolfram.com/Matrix.html\n\n\nGood work!\n(I'll download the online book myself: I didn't gave a look at it before giving you the link; the book is a little old (1937) but matrices haven't changed so much from that time on...)
288230
It should work on any norm, whether the norm is the infinity norm, |[a,b]| = max(a,b) or the 2 norm, |[a,b]| = sqrt(a^2 + b^2).  In the former case you have a rectangle.  In the latter case you have a disk.  All that is necessary is that the norm satisfies the properties of being a norm, which is a generalization of a distance function:\n\nPositive definiteness: |[a,b]| > 0 for all [a,b] != [0,0]\nLinearity: |[ka,kb]| = k|[a,b]| for all [a,b] and real k.   \nTriangle inequality: |[a,b]| + |[b,c]| >= |[a,c]|.
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.
95774
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
813865
The singular value decomposition for a matrix A writes A as a product (hanger)(stretcher)(aligner). \n \n\nIt's an amazing and useful fact that every m x n matrix has a singular value decomposition. \n\n-------------------\nIn linear algebra, the singular value decomposition (SVD) is an important factorization of a rectangular real or complex matrix, with several applications in signal processing and statistics.\n\nThe SVD can be seen as a generalization of the spectral theorem to arbitrary, not necessarily square, matrices. The spectral theorem says that normal matrices can be unitarily diagonalized using a basis of eigenvectors.\n\n---------------------------------\nSingular value decomposition takes a rectangular matrix of gene expression data (defined as A, where A is a n x p matrix) in which the n rows represents the genes, and the p columns represents the experimental conditions.\n\n---------------------------------\nA very powerful set of techniques dealing with sets of equations or matrices that are either singular or numerically very close to singular is the so-called singular value decomposition (SVD). SVD allows one to diagnose the problems in a given matrix and provides numerical answer as well.\n\nAny m x n matrix a (m >= n) can be written as the product of an m x n column-orthogonal matrix u, an n x n diagonal matrix with positive or zero elements, and the transpose of an n x n orthogonal matrix v:\n\n\nI hope that all these are useful, good luck!!!!!!!!!!!!
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
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.
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).
252498
block matrices are those with equal number of rows and columns.\n\nex:\n\n1 1 1\n1 2 2 \n3 4 5\n\nthat's a 3x3 matrix.  same number of rows and columns.\n\nthey're the only matrices that can be multiplied. \n\nmatrices can never be divided by another matrix.
734225
i just learned this in class... \nafter you enter your mass matix M,\nyour Dampening Matrix C\nand your Stiffness Matrix K\nyou take [K] - lamda [M]= (b) \nyou can type \n[Evals]=eigs(new)\nThese will also be the square of your natural frequencies.\nyour mode shapes are the eigvectors! fyi
49374
Wow this question rocks because I found out that search engines like Yahoo and Google use this math trick to figure out which web sites are number one etc in page rank.\n\nIt is also cool for doing stretchy graphics manipulations in graphics programs and  video game designs.\n\nIn so-called "linear algebra" you can make a table or array of numbers (sort of like a spreadsheet) called a matrix. Hey, not "The Matrix" (bullet time rocks!), just "a" matrix.\n\nA matrix can be used for example to stretch, blur, focus, distort or otherwise transform an image. Some programs like PaintShop Pro even let you stuff numbers into a matrix to create a special-effects filter.\n\nIt is very useful to know the quantities called "eigenvectors" and "eigenvalues" for a matrix, because "eigen" is German for"Nuthing happening here dude". This means we can toss them away and make a complicated (big) matrix smaller. That's what "deflation" means. For computer programming you want fast, fast results so it helps to deflate big matrices BEFORE you start computing.\n\nThe Power method is a way to approximately find those pesky eigens using simple arithmetic. The math is not really too hard, you can manually do it with a $1 calculator, just multiplying and adding mostly. But it is best to use a computer program because it can be tedious.\n\nSo get out there and wipe out those eigens!!!
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
415924
A matrix is an array of numbers that is usually studied in relation to solving systems of equations and more general problems in linear algebra. Familiarity with linear algebra is the gateway to higher mathematics; the matrix part of this is little more than advanced arithmetic.