Recent content by rs1n

Rotating the object by an angle of \theta is the same as rotating the coordinate system by -\theta -- it's all relative. For all intents and purposes, you can just consider only rotating the object (and changing the angle to its negation where needed). The second version is basically the same...

To get back to the problem, though... over the complex numbers, the inner product is presumably a Hermitian inner product. So ##\begin{align*} \| u + v \|^2 & = \langle u + v, u+v \rangle = \langle u,u \rangle + \langle u,v \rangle + \langle v,u \rangle + \langle v, v \rangle\\ & = \langle u,u...

You are absolutely right! My eyes failed me, somehow.

To plot by hand, it is generally much easier to plot where the plane intersects one of the 8 octants. For example, given ax+by+cz=d, set x=0 and you get by+cz=d. This produces a line that you can graph on the yz-plane. Similarly, set y=0 to get ax+cz=d -- a line in the xz-plane. Lastly, ax+by=d...

By definition, \langle v_1, v_2 \rangle = \| v_1 \| \cdot \| v_2 \| \cdot \cos(\theta) where \theta is the angle between vectors v_1 and v_2. If you also additionally know that \langle v_1, v_2 \rangle = \| v_1 \| \cdot \| v_2 \| , then the angle between the two vectors must either be 0 or 180...

The source (of the original problem) probably uses a different convention for mutiple integration. It looks like Cauchy's formula. From wikipedia: https://upload.wikimedia.org/math/4/9/9/4996ac5454f0b1e6a964f1ec572ba146.png

If all n eigenvalues are distinct, then the matrix is diagonalizable. However, the converse is not true. There are matrices that are diagonalizable even if their eigenvalues are not distinct, as your example clearly shows.

Would the SVD be helpful? It would presumably still require you to compute the entire surface, though.

If you look at the picture here: http://etc.usf.edu/clipart/43200/43216/unit-circle8_43216_lg.gif you will see that only the first quadrant is needed (and even then, only the angles between 0 and 45 degrees are needed (everything else can be derived by symmetry of the circle. As for the...

If f(-x) = f(x) then the function is even. If f(-x) = -f(x) then the function is odd. Example: f(x)=2x^4+4x^2-1 is even since f(-x) = 2(-x)^4+4(-x)^2-1 = 2x^4+4x^2-1 = f(x). Example: f(x)=x^3-3x is odd since f(-x) = (-x)^3 - 3(-x) = -x^3 + 3x = -(x^3-3x) = -f(x) Exaple: f(x)=x^2-x is neither...

If you write the summations as dot products, you can readily see the answer. Let \vec{x} = \langle x_1, x_2, \dotsm, x_n \rangle and \vec{y} = \langle y_1, y_2, \dotsm, y_n \rangle. Then the Cauchy-Schwartz inequality can be restated as: \underbrace{(\vec{x} \cdot \vec{x})}_{\text{dot product}}...

It seems perhaps your issue is with the definitions. Then span of a set of n vectors, written as span(\{ v_1, v_2, \dotsm , v_n \}) is the set of all possible linear combinations of those n vectors. A basis is a collection of linearly independent vectors whose span is a vector space. To verify...

The determinant can be interpreted geometrically. If A is an n\times n matrix, then let r_1, r_2, \dotsm, r_n be the n rows of A. The absolute value of the determinant of A would be the n-dimensional voume of the parallelotope corresponding to these n vectors. (Imagine one corner of the...

Yes, this is mathematically sound. A and R would have the same singular values.

Did you meant to write that the singular values of A are the square root of the eigenvalues of A^HA ?