Let  A  be nxn real matrix of rank (n-1).Show   AdjA  has  rank  1?
470202
Rank is basically the number of nonzero rows in a row-echelon reduced matrix.  So if you have a square matrix with one row of zeroes, you may be looking at something like this:\n\n| 1 0 0 |\n| 0 1 4 |\n| 0 0 0 |\n\nadj(A) is found by finding the matrix made up of the cofactors of A, and then transposing it, and changing signs in a checkerboard pattern in the matrix.  Creating the cofactor matrix is easy - for the new matrix C, Cij is found by finding the determinant of the matrix that results if you elimiate row i, column j in matrix A.  The cofactor matrix for the above example would be\n\n| 0 0 0 |\n| 0 0 0 |\n| 0 4 1 |\n\nand its transpose would be\n\n| 0 0 0 |\n| 0 0 -4 |\n| 0 0 1 |\n\nwhich is a matrix of rank 1.\n\nThe row of zeros creates multiple rows of zeros when creating the cofactor matrix.  That makes sense - when finding the determinant of a matrix with a row of zeros, the determinant always equals 0.  It turns out that the only time that you create a nonzero row in your cofactor matrix is when you eliminate the single row of zeros.
769753
Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
790751
you have to do it by induction:\nbasically:\nn=2:\nka_1 + ka_2 = k( a_1 +a_2)\n\nnow, assume it is true for n\n\nlet us prove it for n+1:\n\nsum_{i=1} ^{n+1} ka_i = sum_{i=1} ^{n} ka_i + ka_{n+1}\n= k ( sum_{i=1} ^{n} a_i )+ ka_{n+1} now you have only two summands, so:\n= k( sum_{i=1} ^{n} a_i  + a_{n+1} )\n= k sum_{i=1} ^{n+1}  a_i \n~
194771
1. Write down a couple of Jordan blocks.\n\n2. Write down their squares, cubes, maybe even a fourth power.\n\n3. Gaze at each set of identity matrix, J, J^2, J^3, ... until you see a simple linear combination of the matrices that sums to the all zero matrix.  That is your minimal polynomial.\n\n4. Generalize.\n\n5. Now it is obvious. :-)\n\nAlternative:\n\n1.  Google for some combination of:\n\nminimal polynomial Jordan block\n"minimal polynomial" "Jordan block"\nJordan block minimal polynomial\n"Jordan block" "minimal polynomial"\n\n2. Read answer.\n\n3. Too bad, it still isn't obvious. :-(
775741
I cannot prove it formally but I can mention the steps\n\na | u(n)\n\na | u(n+1)\n\nso a | u(n+1) - u(n) if a devided b and a devided c then it devides b-c\n\nso a | u(n-1)\n\nthere by we reach a | u(0) and a|u(1)\n\nu(0) does not give any infomation as u(0) =0\n\nso a | u(1) = 1\nso a= +/-1 \nu(0) = 0\nu(1)  = 1\n\nnext terms are u(2) = 1,u(3) = 2 ,u(4) = 3, then 5,8,13,
16666
it called the Cayley-Hamilton Theorem, type it in google and its proved in many places;\n\n\nhttp://delta.cs.cinvestav.mx/~mcintosh/comun/summer99/mcintosh/node9.html  <---this one is good :)\n\nLet B = x*I - A, where I is the identity matrix of the same size as A.  \nLet C be the classical adjoint of B, so that BC = (det B)I. Recall \nf(x) = det(B), by definition. We wish to show that f(A) = 0. It \nsuffices to show that f(A)*e(k) = 0 for 1 <= k <= n, where e(k) is the \nk-th unit vector, so fix a value of k in that range.\n\nNow x = A implies that for all j,\n\n   x*e(j) = A*e(j),\n\n             n\n          = SUM A(i,j)*e(i),\n            i=1\n   \n      n                     n\n   x*SUM delta(i,j)*e(i) = SUM A(i,j)*e(i),\n     i=1                   i=1\n\n    n\n   SUM [x*delta(i,j)-A(i,j)]*e(i) = 0,\n   i=1\n\n    n\n   SUM B(i,j)*e(i) = 0,\n   i=1\n\n           n\n   C(j,k)*SUM B(i,j)*e(i) = 0,\n          i=1\n\n    n\n   SUM B(i,j)*C(j,k)*e(i) = 0.\n   i=1\n\nThen, summing over all j, reversing the order of summation, and using\nthe definition of the classical adjoint,\n\n    n   n\n   SUM SUM B(i,j)*C(j,k)*e(i) = 0,\n   j=1 i=1\n\n    n    n\n   SUM [SUM B(i,j)*C(j,k)]*e(i) = 0,\n   i=1  j=1\n\n    n\n   SUM [det(B)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(x)*delta(i,k)]*e(i) = 0,\n   i=1\n\n    n\n   SUM [f(A)*delta(i,k)]*e(i) = 0,\n   i=1\n\n         n\n   f(A)*SUM [delta(i,k)]*e(i) = 0,\n        i=1\n\n   f(A)*e(k) = 0.\n\nThis proves that f(A) annihilates all the unit vectors, so must be\nthe zero matrix, and f(A) = 0.\n\n\nsecond one taken off the dr math website, but the link above might be easier to understand.
109963
The Identity Property
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
937007
If R := I - BA (:= means "equals by def.") then, by rearranging (BA = I - R, so A-1 B-1 = (I - R)-1), we get \n\nA-1 = (I - R)-1 B\n\nand the matrix (I - R)-1 can be expanded in a geometric series (just as, for numbers, 1/(1 - r) = I + r + r^2 +...), that is, \n\n(I - R)-1 = I + R + R^2 + ...\n\nbecause B approximates A-1, so R is close to 0. (More precisely, the expansion holds if each eigenvalue of R is less than 1 in absolute value.) Putting the two together, and now writing the inverse properly as A^(-1) instead of A-1 (which might be mistaken for A - I), we get \n\nA^(-1) = (I + R + R^2 + R^3 + ...)B\n\nas was required. This formula is exact, so this is not actually an iterative process, by the way (an iterative process would consist in getting successive approximations to A-1 by repeating some transformation again and again).\nPls award Best Answer :))
1065775
If a(N)=Sum(r^n, 0=>N) and -1<r<1...\n\n Is something that looks like 1/(1-r^N)
584514
Use 'Gaussian elimination' to obtain a basis for the span of matrix columns or rows.\n\nhttp://en.wikipedia.org/wiki/Gaussian_elimination
576001
By the transitive property, a=c
741630
if the matrix is squared, and it has an inverse, then the answer is YES
441742
Assume there is a pair a_i, b_j with a non zero product.\nWrite fg= (a_i x^i + other terms) (b_j x^j + other terms) =\na_i b_j x^(i+j) is non zero.\n\nContradiction.
394644
I apologize if this is too simple, but it seems this is the obvious answer:\n\nSU(2) is a collection of 2x2 matrices, so the sequence { s_j } is just a subset.  The diagonal matrices in the special unitary group consist of [[a,0],[0,conj a]] where |a| = 1.  A "diagonal subsequence" should be a subset of this group.  You should be able to identify such a subsequence with complex numbers which lie on the unit circle.
247548
You can construct one simply by interleaving a bounded sequence with an unbounded sequence. For example, the sequence 1/n with the sequence n to form \n\nS(i) = if i is odd 1/i if i is even i. \n\nThis has 1/i as a convergent subsequence.
139869
[ 0 0 ]\n[ 0 0 ] has determinate 0, thus it is not invertible.  All nxn matrices are diagnalizable, but the zero matrix is already in diagnol form, having zeros everywhere except the main diagnol.  It is just has diagnol entries (0, 0).  Thus I would say it is not non-diagnol.