Boost Your Understanding of Linear Equation Forms

I'm having a presentation on Gauss-Seidel iterative method, and although it isn't mandatory , I would like to have some practical examples for this method (a system of linear equations with $n\ge <!--\; \ge \; -->1000$, preferrably in .txt form), as well as some implementation details (maybe block GS or something, since I haven't looked that up).
I would love it if you can give me examples from real-world calculations, like in a scientific paper or something.

Consider a linear matrix differential equation of the form
$\frac{\mathrm{d}C}{\mathrm{d}t}=AC+C{A}^{\mathrm{T}}$
where $C$ is a symmetric $n\times <!--\; \times \; -->n$ matrix and $A$ is a $n\times <!--\; \times \; -->n$ matrix. Find $C(t)$.
Is there a formal solution for the above equation? This is in principle linear equation if we treat the matrix $C$ and $A$ as a $n^2$ vector. However, it does not seem to be practical way to solve the problem.
This kind of differential equations for matrices is quite new to me. Besides the formal solution let me know some books considering similar topic.

The problem is this:
The impulse response of a system is the output from this system when excited by an input signal $\mathrm{\delta <!--\; \delta \; -->}(k)$ that is zero everywhere, except at $k=0$, where it is equal to 1. Using this definition and the general form of the solution of a difference equation, write the output of a linear system described by:
$y(k)–3y(k–1)–4y(k–2)=\mathrm{\delta <!--\; \delta \; -->}(k)+2\mathrm{\delta <!--\; \delta \; -->}(k–1)$
The initial conditions are: $y(–2)=y(–1)=0$.
My question is: How can the particular solution be found using the method of undetermined coefficients if the non-homogeneous equation is also a difference equation?

The problem reads:
The solution of a certain differential equation is of the form
$y(t)=a\exp <!--\; \; -->(5t)+b\exp <!--\; \; -->(8t)$
where $a$ and $b$ are constants.
The solution has initial conditions $y(0)=5$ and $y^{\prime }(0)=5$
Find the solution by using the initial conditions to get linear equations for $a$ and $b$. ....................
What I did was solve using the initial conditions and I found that
$a+b=5$
and $5a+8b=5.$
Am I totally on the wrong track? I don't know what it means to find a linear equation for $a$ and $b$. I'd appreciate it if you could solve it step by step.

The homogenous representation of a circle is given by $x^2+y^2+2gxz+2fyz+c{z}^2=0$ (or, equivalently, if we set $z=1$, $x^2+y^2+2gx+2fy+c=0$). Now, given 3 points (in a homogenous form), we can solve a system of linear equations and retrieve the unknowns $f$, $g$ and $c$.
This is all very nice (because of linear algebra), but what do these unknowns actually represent with respect to the circle? Which of these numbers represent the x and y coordinates of a circle and which one represents the radius?
Apparently, $-<!--\; -\; -->f$ and $-<!--\; -\; -->g$ would be the $x$ and $y$ coordinate of the center of the circle? Why is that the case? I would like to see a proof/derivation of it. Also, what is the radius then?

I've noticed that in Statistics, when the topic of "linearity" comes up, what is usually meant is that a fit can be made with a function of the form
$\mathbb{R}\mapsto <!--\; \mapsto \; -->\mathbb{R}:y(x)=mx+q$
Now in linear algebra this is obviously not true (equations of the form y=mx+q are not necessarily linear).
Am i mistaking something or is this just a point where statisticans and mathematicians use the same words for different things?

So, I can model growth and decay if I start with assuming that the growth rate is constant:
$\frac{p^{\prime }(t)}{p(t)}=\mathrm{\alpha <!--\; \alpha \; -->}$
and then I have
$p^{\prime }(t)-<!--\; -\; -->\mathrm{\alpha <!--\; \alpha \; -->}p(t)=0$
A general linear differential equation, however, would have the form
$p^{\prime }(t)-<!--\; -\; -->g(t)p(t)=h(t)$
So the growth rate is g(t). What is h(t)?
And what type of thing is this form of equation used to model?

The problem is as follows:
The general form of a linear, homogeneous, second order equation with constant coefficients is $d^{2}y/d{t}^2+p(dy/dt)+qy=0$
(a) Show that if $q$ does not equal 0, then the origin is the only equilibrium point of the system.
(b) Show that if $q$ does not equal 0, then the only solution of the second-order equation with $y$ constant is $y(t)=0$ for all $t$.
I solved part a by creating a system of equations:
$dy/dt=v$
$dv/dt=-<!--\; -\; -->pv-<!--\; -\; -->qy$
I then set each equation equal to zero. Since $v$ must equal zero, I plugged $v=0$ into the second equation: $-<!--\; -\; -->p(0)-<!--\; -\; -->qy=0$ $\Rightarrow <!--\; \Rightarrow \; -->$ $-<!--\; -\; -->qy=0$. Therefore, $y$ must equal zero to satisfy the equation since $q$ cannot be zero.
I'm not sure how to approach part (b) because I don't understand how the two questions are any different.

There is a famous quote on linear algebra:
We share a philosophy about linear algebra: we think basis-free, we write basis-free, but when the chips are down we close the office door and compute with matrices like fury.
What does it mean to think and write basis free? Are there any "basis free" approaches to linear algebra?
Update: The comments below raise another question related to the above quote. How can one interpret studying systems of linear equations in a basis free way.
For an $m\times <!--\; \times \; -->n$ matrix its reduced row echelon form (used in Gauss Jordan elimination) yields a unique basis of ${\mathbb{R}}^n$ (take the basis of corresponding row space and adjoin to it the basis of the corresponding null space). So each matrix $A$ has a special basis associated with it which does something nice: it solves $Ax=0$. How can one talk of these things in Halmos' viewpoint?

Let A be a real, invertible $n\times <!--\; \times \; -->n$ matrix. I am interested in finding the vectors $\mathbf{x}\in <!--\; \in \; -->{\mathbb{R}}^n$ that solve the following equation:
$\mathbf{x}=A\tanh <!--\; \; -->(\mathbf{x})$
where the tanh is applied element-wise. More generally, we can consider other kinds of non-linearities instead of the tanh (but always applied element-wise).
Is there a generic approach to studying the solutions of these type of equations? Probably exploiting the eigen decomposition of $A$?
I added the tag "reference-request" in case someone can suggest relevant references to the literature.

Consider an incoming beam of particles with $E>U$ scattering from the potential
$V(x)=\{\begin{array}{ll}0& \text{ for }x<-<!--\; -\; -->a\\ U& \text{ for }|x|<a\\ 0& \text{ for }x>a\end{array}$
So Schrodinger's (time-independent) equation reads
${\mathrm{\psi <!--\; \psi \; -->}}^{″}=\{\begin{array}{ll}-<!--\; -\; -->\frac{2mE}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}\mathrm{\psi <!--\; \psi \; -->}& \text{ for }|x|>a\\ -<!--\; -\; -->\frac{2m}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}(E-<!--\; -\; -->U)\mathrm{\psi <!--\; \psi \; -->}& \text{ for }|x|<a\end{array}$
Then the scattering states are given by
$\mathrm{\psi <!--\; \psi \; -->}(x)=\{\begin{array}{ll}\exp <!--\; \; -->(ikx)+R\exp <!--\; \; -->(-<!--\; -\; -->ikx)& \text{ for }x<-<!--\; -\; -->a\\ A\exp <!--\; \; -->(-<!--\; -\; -->lx)+B\exp <!--\; \; -->(lx)& \text{ for }|x|<a\\ T\exp <!--\; \; -->(ikx)& \text{ for }x>a\end{array}$
where $k=\sqrt{\frac{2mE}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}}$, $l=\sqrt{-<!--\; -\; -->\frac{2m}{{\mathrm{\hslash <!--\; \hslash \; -->}}^2}(E-<!--\; -\; -->U)}$.
[The above is from my lecture notes].
Now my questions are:
1. From the form of the solution in $|x|<a$ it looks like we are assuming that $l\in <!--\; \in \; -->\mathbb{R}$, i.e. that $E<U$. Why is that? Why can't we have a combination of sines and cosines?
2. There's a theorem that says that
Any solution of the time-independent Schrodinger equation with a symmetric potential is a linear combination of solutions of definite parity.
We clearly have a symmetric potential here, and for real l, $\exp <!--\; \; -->(\pm <!--\; \pm \; -->lx)$ doesn't have definite parity (and as far as I know can't be written as a (finite) linear combination of functions of definite parity). So given this theorem, the 'assumption' that $l\in <!--\; \in \; -->\mathbb{R}$ seems even stranger/less justified. Do we just throw this theorem away because we are dealing with scattering solutions?
What's going on?

I am trying to fit data to an equation of this form using non-linear regression:
$y=1-<!--\; -\; -->\frac{a}{\exp <!--\; \; -->\left(n\left(x-<!--\; -\; -->x_0\right)\right)}$
I would like to select three sets of $(x,y)$ data and solve the above equation for $a$, $n$ and $x_0$ in order to improve the initial estimates of these parameters. I eliminated $a$ using the following expression:
$a=\left(1-<!--\; -\; -->y_1\right)\exp <!--\; \; -->\left(n\left(x_1-<!--\; -\; -->x_0\right)\right)$
where $(x_1,y_1)$ is one of the selected data points and then back-substituted it to give:
$y_2=1-<!--\; -\; -->\frac{\left(1-<!--\; -\; -->y_1\right)\exp <!--\; \; -->\left(n\left(x_1-<!--\; -\; -->x_0\right)\right)}{\exp <!--\; \; -->\left(n\left(x_2-<!--\; -\; -->x_0\right)\right)}$
where $(x_2,y_2)$ in another of the selected data points. This becomes
$y_2=1-<!--\; -\; -->\left(1-<!--\; -\; -->y_1\right)\exp <!--\; \; -->\left(n\left(x_1-<!--\; -\; -->x_2\right)\right)$
and thus
$n=\frac{1}{x_1-<!--\; -\; -->x_2}\ln <!--\; \; -->\left(\frac{1-<!--\; -\; -->y_2}{1-<!--\; -\; -->y_1}\right)$
If I use point $(x_3,y_3)$ instead I get:
$n=\frac{1}{x_1-<!--\; -\; -->x_3}\ln <!--\; \; -->\left(\frac{1-<!--\; -\; -->y_3}{1-<!--\; -\; -->y_1}\right)$
The problem is that I have two different expressions for n but none for $x_0$ because it was eliminated along the way. I get a similar result if I eliminate $x_0$ first.
Is it possible to solve for $a$, $n$ and $x_0$ in this manner?

Given an arbitrary system of equations, why is direction in space "stored" in the variables when considering the system as linear equations, but "stored" in vectors when considering the system as a vector equation? For example suppose we have a system of three equations in three variables where each equation is of the form $(a_i)x+(b_i)y+(c_i)z=d_i$. Lets also suppose they represent three distinct planes in $R^3$. In the context of the system representing planes in space, it seems to me that dimension/direction is sort of "stored" in the variables $x,y$, and $z$. Considering the system as a linear combination of vectors, the coefficients associated with any one variable make a column vector; for example, for variable $x$ lets call the vector $v_1=$. The vector equation associated with the system would then be $x{v}_1+y{v}_2+z{v}_3=$. In this context it seems as though dimension/direction is stored in the vectors, and the variables $x,y$, and $z$ now just scale them. I realize that both contexts have the same solution set, and both take place in $R^3$, but is there a more intuitive explanation for this relatedness?

I wish to find the functional whose minimisation yields the follwoing equation on the vector function u $(\mathrm{\lambda <!--\; \lambda \; -->}+\mathrm{\mu <!--\; \mu \; -->})\mathrm{\nabla <!--\; \nabla \; -->}(\mathrm{\nabla <!--\; \nabla \; -->}\cdot <!--\; \cdot \; -->u)+\mathrm{\mu <!--\; \mu \; -->}{\mathrm{\nabla <!--\; \nabla \; -->}}^{2}u=0$, the Navier equation of linear elasticity. I know that this equation has a vaiational principle from physical reasons, but struggle to find it. The term containing the laplacian is easier to handle, I can not even prove simmetry of the weak form for the first one: how to integrate by parts (grad div u) v to obtain a term symmetric in u e v?

I have to convert the equation $y^2+xy+y=x^3$ by a change of linear variables to the form $Y^2=X^3+aX+b$ where $a$ and $b$ are rational numbers. So far, by completing the square method I've reduced it to
$Y^2=x^3+x^2/4+x/2+1/4$
where $Y=y+(x+1)/2$. However, I can't figure out how to reduce it to the form asked in the question. Any help would be appreciated thanks.

If we have a system of linear equations $Ax=b$, then in case $A$ is invertible it easy to say that the solution is $x=A^{-<!--\; -\; -->1}b$. In all other cases, even if $A$ is invertible, the solution if it exist will be in the form $x=A^{+}b+(I-<!--\; -\; -->A^{+}A)w$, where $A^{+}$ is called pseudoinvers of $A$, and w is a vector of free parameters. My question here is how to compute $A^{+}$ for any matrix $A$? what is the way for doing that? It is easy to compute $A^{+}$ if the column are linearly independent (so that $m\ge <!--\; \ge \; -->n$, $A^{+}=(A^{T}A{)}^{-<!--\; -\; -->1}A$, and also if the rows are linearly independent (so that $m\le <!--\; \le \; -->n$), $A^{+}=A^T(A{A}^T{)}^{-<!--\; -\; -->1}$, but I dont know how to compute $A^{+}$ if $A$ is sequare non invertible matrix or if the columns or rows are not linearly independents. The above does not works. If anyone have any idea about computing pseudoinverse or if there is easier method for finding the solution $x$ for system of linear equations $Ax=b$, please help me in all cases, and please keep in mind that I want to find the general form for the solution if it exist not just for a particular $b$.

I am working with a linear equation of the following form.
$x=q+Px,$
where x,q are vectors of size $K$ and $P$ is a square matrix of size $K$. The variables satisfy: (1) $q>0$, (2) $0\le <!--\; \le \; -->P_{i,j}\le <!--\; \le \; -->1$ for all $i,j$ and (3) $\underset{j}{\sum <!--\; \sum \; -->}{P}_{i,j}<1$ for each $i$.
If $(I-<!--\; -\; -->P)$ is nonsingular, then the equation can be solved by $x=(I-<!--\; -\; -->P{)}^{-<!--\; -\; -->1}q$. My question is whether or not $(I-<!--\; -\; -->P)$ is always nonsingular under the above conditions, and if not, what is the condition to add on to make sure the invertibility. I conjecture that it has something to do with the eigenvalues of the matrix but cannot figure out an answer.

I'm currently reading Pugh's Analysis. He makes the statement that the line between two points x and y is the set of linear combinations $sx+ty$ where $s+t=1$. I'm satisfied that this is true, as the line between two points $x$ and $y$ has the equation $y-<!--\; -\; -->x_2=\frac{x_2-<!--\; -\; -->y_2}{x_1-<!--\; -\; -->y_1}(x-<!--\; -\; -->x_1)$ and the points of form $(s{x}_1+t{y}_1,s{x}_2+t{y}_2)$ are solutions of the equation. I still don't have any reasonable geometric intuition for why this is true though.

I have a simple question which is troubling me.
I have seen this theorem in Linear Algebra which I quote here :
The Row Echelon Form of an Inconsistent System: An augmented matrix corresponds to an inconsistent system of equations if and only if the last column (i.e., the augmented column) is a pivot column.
Suppose I have a RREF augmented matrix with the last row containing 0s in the coefficient and 1 in the augmented column ( I mean $00...0|1$ as the last row). Surely, the presence of this row would make the whole system inconsistent, as it would imply that $0=1$, which is not possible.
But, what I don't understand is that for this to happen, why is there a need for all the other entries of the last column to be zero. This doubt stems from the definition of pivot column which says it is a column containing pivot, and hence we know that all the entries in a column containing pivot must be zero.
But I don't see why other entries should be zero. In all the examples of inconsistent system RREF matrices also the last column is always having all the entries except the pivot to be zero.
I hope I made myself clear. I would be grateful to you if you can help me with this.