Recent content by Vaedoris

What you need is regression analysis and someone with good knowledge of Statistics and experiment design.

One way to find the solution is to express the LTI system in State Space model (SS). Then Laplace-transform the two equations of SS, make an algebraic substitution, and take the inverse Laplace transform of the substitution. Voila... This is a great note (one that I now treasure) on how to...

which means {red, green, blue} is a basis if mathematically (disregarding physical sense) we assume each color is independent

This is informal but you'll get the idea... ## \begin{align*} \begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\left(k_1\begin{pmatrix} a\\ b\\ c \end{pmatrix}+k_2\begin{pmatrix} e\\ f\\ g \end{pmatrix} \right )&=\begin{pmatrix} 0 & 1 & 0 \end{pmatrix}\begin{pmatrix} k_1a+k_2e\\ k_1b+k_2f\\...

Reflecting twice off the same "surface" gets you the original vector back.

##PP^T## is an n-by-n matrix. (prove: ##P## is n-by-1 and ##P^T## is 1-by-n). So ##PP^T\neq P^TP## unless if n = 1. ##H## is symmetric. ##H^TH=I_{n}## which has dimension n-by-n.

Don't you realize that Cauchy-Schwarz inequality is at the very root of that "cosine rule"?

Didn't I say dot product is inner product already? (edit: note I didn't say inner product is restricted to dot product only) The point here is not the dot product but rather the Cauchy-Schwarz inequality itself which applies to R^n if you take the inner product to be dot product. Besides...

Many textbooks define an overdetermined system as simply a system that has more equations than unknowns. I have yet seen one definition that restricts it to being "irreducible" or that it must have full column rank! For example: You can have a matrix that has more rows than columns with its...

what you on about? Cauchy-Schwarz inequality applies to any inner product space including ##\mathbb{R}^n##!

It is a consequence of Cauchy-Schwarz inequality: ##\left |\left \langle a,b \right \rangle \right | \leq \left\|a\right\|\left\|b\right\|## Hence the ratio: ##cos\theta = \frac{\left \langle a,b \right \rangle}{\left\|a\right\|\left\|b\right\|}## Dot product is a inner product.

Well, that's what HallsofIvy claimed (probably just some unintentional mistake). Not me.P.S.: You could create unlimited number of counter-examples to that claim by just inspecting Ax=b. "Intersection of straight lines" though may not be obvious to some people.

I don't think you are talking about the same thing. Besides, your example of intersection of lines is just one case of overdetermined system (where rank = number of columns - if you write the problem in matrix form). What I explained above is the other case: For overdetermined system...

You could find the reduced row echelon form of the matrix (of an overdetermined system) to determine which rows are dependent. Then, take them out, but you must also erase the corresponding components of the vector b. Edit: However, this does not guarantee that the reduced matrix is square as...

Edit: another way to solve this is to multiply both sides by ##A^{-1}## from the right. Sorry, it seems that I almost forgot one property of matrix inverse :)