linear algebra - How to find the matrix of a quadratic form ...
3. It suffices to note that if A A is the matrix of your quadratic form, then it is also the matrix of your bilinear form f(x, y) = 14[q(x + y) − q(x − y))] f (x, y) = 1 4 [q (x + y) − q (x − y))], so that. aij = f(ei,ej) = 1 4(q(ei +ej) − q(ei −ej)) a i j = f (e i, e j) = 1 4 (q (e i + e j) − q (e i − e j))
Derivative of Quadratic Form - Mathematics Stack Exchange
I'm trying to get to the $\mu_0$ of Gaussian Discriminant Analysis by maximizing the log likelihood and I need to take the derivative of a Quadratic form. Either the result I mentioned above is wrong (shouldn't be because I went over my arithmetic several times) or the form I arrived at above is not the terribly useful to my problem (because I ...
linear algebra - How to find max and min of this quadratic form ...
For the quadratic form $\mathbb q (x)= (Ax, x)$ find the maximal and minimal value of $\mathbb q (x)$ on the unit spher $\mathbb S = \ {x| (x,x) = 1\}$. For this I'm stumped, I know it clearly must involve what I've done and I've read questions about quadratic forms which say there is diagonalizable matrix with eigenvectors on the diagonal such ...
diagonalize quadratic form - Mathematics Stack Exchange
Q. So to solve Q(v) = 0 Q (v) = 0, you only need the eigenvectors for the eigenvalue 0 0; however, for the diagonalization, all eigenvectors are going to be useful (in an orthonormal basis of eigenvectors, the linear form Q Q will be diagonal with the eigenvalues on its diagonal; and the transform matrix to get from the canonical basis to this ...
linear algebra - Symmetry in the quadratic form matrix
Feb 21, 2021 · 1. Therein, the author states the matrix in quadratic forms must be symmetric. No, he didn't. And he didn't say the ij i j -th component in A A is aij a i j. Here the quadratic form can be written as x⊤Ax x ⊤ A x such that A A is a symmetric matrix. The representation is not unique without posing any restriction on A A.
norm of a quadratic form - Mathematics Stack Exchange
May 3, 2017 · Thus, the quantity you are asking about is just the maximum eigenvalue's norm. More generally, given A A symmetric (no loss of generality here) and B B positive definite, you can prove that. max{xTAx: xTBx = 1} m a x {x T A x: x T B x = 1} is the maximum eigenvalue of B−1A B − 1 A at x x the corresponding eigenvector. Share.
Why do we assume that a matrix in quadratic form is symmetric?
I am looking to the review document for linear algebra (Zico Kolter (updated by Chuong Do), Linear Algebra Review and Reference), and the part of the quadratic form (pg17) mentions about an assumption of being symmetric for a matrix in quadratic form. It also includes some declarative equality for that proposed argument.
metric spaces - Quadratic form as generalized distance?
Jul 24, 2015 · A simple example of a generalisation of distance encoded by a quadratic form is given by the Minkowski metric for space-time; for the squared length of a vector from the origin: d2 =x2 +y2 +z2– (ct)2 d 2 = x 2 + y 2 + z 2 – (c t) 2. … which I’m sure you can see can be expressed as XTPX X T P X where P P is a 4 × 4 4 × 4 matrix with ...
Quadratic form equals zero - Mathematics Stack Exchange
Derivative of Quadratic Form as a Linear Approximation. 3. derivative of a symmetric bilinear form ...
linear algebra - Skew-symmetric matrix and quadratic form
For the controverse, let S:= At + A S:= A t + A. Since for all x ∈Rn x ∈ R n we have qA(x) = xtAx = 0 = −xtAx q A (x) = x t A x = 0 = − x t A x, we get xtSx = 0 x t S x = 0 for all x x. We denote by ej e j the vector whose entries are 0 0, except the j j which is 1 1, and sij s i j the entries of S S.