Recent content by charlies1902

But previous I had shown ##PA=A## so ##y=Ax=PAx=P(Ax)## Why is that not correct?

Homework Statement ##P=A(A^*A)^{-1}A^*## where A is a mxn real/complex matrix and ##A^*A## is invertible. ##A^*## means the conjugate transpose of A. Homework EquationsThe Attempt at a Solution Let y be in the range(A), such that ##y = Ax## for some ##x##. We can see that ##PA =...

Hmmm. I might be missing something here: So if I square both sides I obtain: ##||x||_1= (\sum |x_i|)^2## and ##||x||_2=(\sum |x_i|^2)## Since, the square of sums of absolute values is always greater or equal to the sum of squares, we have proved that ##||x||_1 \geq ||x||_2##. Is what I just...

Thanks. But isn't that just a different way of saying what I said above?

Homework Statement Show that ##||x||_1\geq ||x||_2## 2. Homework Equations ##||x||_1 = \sum_{i=1}^n |x_i|## ##||x||_2 = (\sum_{i=1}^n |x_i|^2)^.5## The Attempt at a Solution I am having a hard time with this, because the question just seems so trivial, that I don't even know how to prove...

The denominator is a vector norm, so are we saying that the numerator is a vector norm for all square A matrices?

The induced matrix norm for a square matrix ##A## is defined as: ##\lVert A \rVert= sup\frac{\lVert Ax \rVert}{\lVert x \rVert}## where ##\lVert x \rVert## is a vector norm. sup = supremum My question is: is the numerator ##\lVert Ax \rVert## a vector norm?

It is a typo I mean "in the matrix." I believe that is a definition of orthogonal matrix, along with other variations with the same meaning.

Thanks for the answer. To find out if a matrix is orthogonal (I know there are various ways), is it sufficient to show that the dot product of any given 2 column vectors in the vector is zero?

Can an orthogonal matrix involve complex/imaginary values?

Hello. I have a question regarding curvature and second derivatives. I have always been confused regarding what is concave/convex and what corresponds to negative/positive curvature, negative/positive second derivative. If we consider the profile shown in the following picture...

When looking at Elliptic PDEs that describe a physical system, do these typically not involve a time term? I have yet to see an elliptic PDE involving a time term, which seem to be associated with parabolic/hyperbolic PDEs rather than elliptic. Can anyone confirm?

But even if the viscosity varies with temperature, it doesn't necessarily mean it is a non-Newtonian fluid right? As long as the viscosity does not vary with the strain rate, the shear stress is still a linear function of the strain rate, even if the viscosity is varying with other parameters?

I was reading this: http://www.creatis.insa-lyon.fr/~dsarrut/bib/Archive/others/phys/www.mas.ncl.ac.uk/%257Esbrooks/book/nish.mit.edu/2006/Textbook/Nodes/chap06/node29.html Under the first figure it states "Figure 6.20: The boundary layer at a stagnation point on an airfoil has a constant...

Oh, so if the viscosity becomes a function of the strain rate. In that case, it would violate Newtonian fluid?