Impérialiste
10-02-2010, 10:09 AM
If any person uses any of my proofs, credit my work with a link. This is my original work, derivation, and proof.
The interest of this topic is in linear systems such that they can be written as matrices. The key issue to consider is, like in Linear Algebra, R^2 and R^3 are all elements of R^n and work in the exact same manner. There is nothing different about adding another dimensional variable. And note too that in R^4, though time is the fourth dimension, it's still an element of the fourth dimension.
There is another consideration too. Only homogeneous solutions have to be proven. There is a reason why.
Let X(t) = X,c + X,p such that X,c is the homogeneous solution and X,p is the non-homogeneous solution. Also, note that X(t) = C*Φ(t). Then, X,c and X,p are subsets of C*Φ(t). Since, X,c has already been solved for, then X,p must be solved for. Consider. . .
X' = AX in general. Then X',p = AX,p. If X = CΦ, then X,p = CΦ as well. Hence
d/dt[X,p] = d/dt[C*Φ] = C*Φ' + Φ*C'.
A non-homogeneous solution is in the form of X'(t) = AX + F(t). Then, due to the superposition principle, X',p = C*Φ' + Φ*C' such that either C*Φ' or Φ*C' is AX or F(t). Let CΦ' = AX, so CΦ' = ACΦ, then Φ' = AΦ. Hence F = Φ*C'. Then. . .
Let X' = X',P, then. . .
X' = Ax + F
X',p = C*Φ' + Φ*C';
X' = ACΦ + F
X' = C*AΦ + Φ*C';
X' = ACΦ + F
X' = ACΦ + Φ*C';
X' = ACΦ + F
-1*X' = -1*(ACΦ + Φ*C');
X' = ACΦ + F
-X' = -ACΦ - ΦC'
X' - X' = ACΦ - ACΦ + F - Φ*C';
0 = 0 + F - Φ*C'
F = Φ*C'
F = C' / (Φ^-1);
F*Φ^(-1) = C'.
Revisit X(t) = X,c + X,p such that X,c and X,p are subsets of CΦ(t). Since the homogeneous component, Ax has already been solved for, then X,c as an element of CΦ has been solved for. X,p is also an element of CΦ, but as an element of CΦ, it must be either of a different root, r, or multiplied by the next order up (e.g. A to At + B, or At + b to At^2 + bt + c, such that each order up is the integral of the previous order with respect to time such that the final integral is At^n + bt^(n - 1) + ct^(n - 2) + . . . + z,n*t^(n - n)). Due to this, and the fact that F = Φ*C', then consider:
X'(t) = Ax + F = C*Φ + Φ*C'; then integrating X'(t) and integrating C' after a substitution, then
X(t) = C*Φ + Φ*integral of C' = C*Φ + Φ*integral of F*Φ^(-1). Since C*Φ(t), then integrate F*Φ^(-1) with respect to a dummy variable, s, in terms of [a, b] being [to, t].
Regardless, homogeneous solutions are subsets of non-homogeneous solutions, and proving the homogeneous case proves the general case.
Hence x'(t) = A(t)x + b(t)y + e(t)
y'(t) = C(t)x + d(t)y + f(t).
That's the general form. A homogeneous equation is one such that e(t) and f(t) are equal to 0. Since 0 is an element of F(t), then homogeneous solutions are subsets of non-homogeneous solutions. Hence. . .
x'(t) = Ax + by
y'(t) = Cx + dy.
x' = Ax + by
y' = Cx + dy.
d/dt(x') = d/dt(Ax + by); x'' = Ax' + by'; x'' = Ax' + b(Cx + dy); x'' = Ax' + bcx + bdy; x'' = ax' + bcx + bd[(x' - ax)/b]; x'' = ax' + bcx + d(x' - ax); x'' = ax' + dx' + bcx - adx; x'' = x'(a + d) + x(bc - ad); x'' - x'(a + d) - x(bc - ad) = 0; x'' - x'(a + d) + x(ad - bc) = 0.
Hence, x'' - x'(a + d) + x(ad - bc) = 0 which turns to r^2 - r(a + d) + ad - bc = 0 such that x^(n) = r^n and x^(0) = 1.
Let K = ad - bc, then. . .
K + d^2 = ad + d^2 - bc; k + d^2 = d(a + d) - bc; (k + d^2 + bc) = d(a + d); [(k + d^2 + bc) / d] = (a + d); -[(k + d^2 + bc) / d] = -(a + d) = M.
Hence, r^2 - r(a + d) + (ad - bc) = r^2 + M*r + K = 0.
This is now in the form of the Quadratic Equation: Let A = 1, B = M, and C = K.
r = [-B ± square root(B^2 - 4AC)] / 2A = [-M ± square root of (M^2 - 4K)] / 2. Only the square root of (M^2 - 4K) is of any real interest.
Let D: square root of (U) ≥ 0 for any real number. M^2 - 4K is a subset of U, so the square root of (M^2 - 4k) ≥ 0.
If the square root of (M^2 - 4k) is greater than 0, then two real solutions exist such that X(t) = C,1*e^(r,1*t) + C,2*e^(r,2*t).
If the square root of (M^2 - 4k) is equal to 0, then a repeating eigenvalue case exists. More on this solution in a bit.
If the square root of (M^2 - 4k) is less than 0, then a complex case must be considered. More on this solution as well.
x' = ax + by
y' = cx + dy.
If the determinant of the above system is not equal to 0, then a unique solution is guaranteed. The determinant is ad - bc. If it is equal to 0, then the above system either has no solution (geometrically, we're talking about parallel lines; in three-dimensions, we're talking about planes) or an infinite number of solutions due to an intersection between an arbitrary number of constants.
x' = ax + by
y' = cx + dy
fits the form of x' = Ax such that A is a matrix of [A B, C D] such that the comma represents going down one level. The point being is that the three subsets of M^2 - 4K and its relation to 0. The determinant of A is always not equal to 0. Moreover, there must be n eigenvectors for an n x n matrix.
If the square root of (M^2 - 4K) is greater than 0, then the solution works as follows:
There is a matrix A = [a b, c d] such that the determinant is not equal to 0. A = λI such that I = [1 0, 0 1]. Hence, λI = [λ 0, 0 λ]. So A - λI = 0. Following matrix addition, then [a-λ b, c d-λ]. The determinant of [A - λI] will yield two real numbers known as eigenvalues. Let λ = {λ,1 and λ,2} so λ,1 and λ,2 are elements of λ such that λ,1 and λ,2 are both real numbers.
To find the solution of A in terms of λ,1, then A*V1 = λ,1*I*V1 such that V1 [V1, V2]. Since I is just an identity matrix and hence functions as 1, then λ,1*I = λ,1. So A*V1 = λ,1*V1; A*V1 - λ,1*V1 = (A - λ,1)*V1 = 0. A - λ,1 = [a-λ,1 b, c d-λ,1]*[V1, V2] = [0, 0] such that a-λ,1 and d-λ,1 are elements of the real number set. This will yield two equations with variables V1 and V2 as elements of V1:
(a - λ,1)*V1 + b*V2 = 0
C*V1 + (d - λ,1)*V2 = 0.
Solving for V1 and V2 will provide X1(t) = C1*[V1, V2]*e^(λ,1t).
Using λ,2 instead of λ,1 and using V2 for [V3, V4], then
(a - λ,2)*V3 + b*V4 = 0
C*V3 + (d - λ,2)*V4 = 0.
Solving for V3 and V4 will provide X2(t) = C2*[V3, V4)*e^(λ,2t)
So X(t) = C1*V1*e^(λ,1t) + C2*V3*e^(λ,2t)
and X(t) = C1*V2*e^(λ,1t) + C2*V4*e^(λ,2t) such that both solutions are elements of X(t).
Case II: If the square root of (M^2 - 4K) is equal to 0, then the solution works as follows:
There is a matrix A = [a b, c d] such that the determinant is not equal to 0. A = λI such that I = [1 0, 0 1]. Hence, λI = [λ 0, 0 λ]. So A - λI = 0. Following matrix addition, then [a-λ b, c d-λ]. The determinant of [A - λI] will yield one real numbered eigenvalue.
A*V1 = λ*I*V1; A*V1 = λ*V1; A*V1 - λ*V1 = 0; (A - λ)*V1 = 0 such that A - λ = [a-λ b, c d-λ], V1 = [V1, V2], and 0 = [0, 0]. Hence,
(a - λ)*V1 + b*V2 = 0
c*V1 + (d - λ)*V2 = 0.
Solving for V1 and V2 will yield two real numbers. However, two unique eigenvectors are in demand for the two x two matrix (or n unique eigenvectors for the n x n matrix), so. . .
(A - λ)*V2 = V1 such that V2 = [V3, V4].
(a - λ)*V3 + b*V4 = V1
c*V3 + (d - λ)*V4 = V2.
That will yield a solution for the second eigenvector.
However, X1(t) = C1*V1*e^(λt) = C1*[V1, V2]*e^(λt).
The integral of X1(t) = C1*e(λt)*the integral of V1*dt;
X2(t) = C2e^(λt)*(V1*t + C);
X2(t) = C2e^(λt)*(V1*t + V2);
X2(t) = C2e^(λt)*([V1, V2]t + [V3, V4]);
X2(t) = C2*e^(λt)*([V1*t, V2*t] + [V3, V4]);
X2(t) = C2*e^(λt)*[V1*t + V3, V2*t + V4].
So X(t) = X1(t) + X2(t) = e^(λt)*(C1*V1 + C2*(V1*t + V3)) or e^(λt)*(C1*V2 + C2*(V2*t + V4)).
If this were an n x n matrix, under the aforementioned conditions, then X(t) = X1(t) + X2(t) + . . . Xn(t) such that Xn(t) = Cn*e^(λt)*{[V1*t^(n - 1)/(n - 1)!] + [V2*t^(n - 2)/(n - 2)!] + [V3*t^(n - 3)/(n - 3)!] + . . . + [Vn^(n - n)/(n - n)!]}.
What if the square root of (M^2 - 4K) is less than 0?
Before this is dealt with, let's provide a lemma:
Taylor's Theorem dictates that a function can be approximated via F(t) + [F'(t)*(t - to)^1]/1! + [F''(t)*(t - to)^2]/2! + . . . + [F(n)*(X - Xo)^n]/n!.
F(t) = Cos(t). Let u = t, so du = dt, hence du/dt = 1.
F(t) = Cos(t); d/dt[F(t)] = d/dt[Cos(u)] = -sin(u)du/dt = -sin(t)*1 = -sin(t) = F'(t);
F''(t) = -d/dt[sin(t)] = -cos(t); F^(3)(t) = -d/dt[cos(t)] = -(-sin(t) = sin(t);
F^(4)(t) = d/dt[sin(t)] = cos(t); F^(5)(t) = d/dt [cos(t)] = -sin(t). This is a repeating cycle.
Cos(T) = Cos(To) - [sin(To)*(t - to)]/1! - [Cos(To)*(t - to)^2]/2! + [Sin(To)*(t - to)^3]/3! + [Cos(To)*(t - to)^4]/4! + . . . +[Cos(To)*(T - to)^n]/n!
Letting To = 0, then
Cos(t) = Cos(0) - [sin(0)*(t - 0)]/1! - [Cos(0)*(t - 0)^2]/2! + [Sin(0)*(t - 0)^3]/3! + [Cos(0)*(t - 0)^4]/4! + . . . +[Cos(0)*(T - 0)^n]/n! = 1 - 0 - (t^2 / 2!) + 0 + (T^4 / 4!) - . . . - t^(2k)/(2k)! + t^(2k + 2)/(2k + 2)! = 1 - (t^2 / 2!) + (T^4 / 4!) - . . . - t^(2k)/(2k)! + t^(2k + 2)/(2k + 2)!
Using this standard for Sin(t) such that F(t) = Sin(t), F'(t) = Cos(t), F''(t) = -sin(t), F^(3)(t) = -cos(t), F^(4)(t) = sin(t), creating a cycle, using the same to value of 0, then. . .
Sin(t) = t - (t^3 / 3!) + (t^5 / 5!) - . . . - t^(2k + 1) / (2k + 1)! + t^(2k + 3) / (2k + 3)!.
Since i is the only element of the complex number set that differentiates its elements from the real number set and i^2 yields a real number such that the complex number plane can map onto the real number plane, then this will have direct implications for Cos(t) and Sin(t).
Cos(t) has nothing but even, whole number exponents, hence f(t) = f(-t), or cos(t) = cos(-t).
Sin(t) has nothing but odd, whole number exponents, hence f(t) is not equal to f(-t), as f(-t) = -f(t).
Now, this will prove the next case:
A = λI; A = λ; A - λ = 0; [a-λ b, c d-λ] = [0, 0] such that the determinant of A - λ will be a complex number a ± bi such that a is an element of the real number set and b, being an element of the real number set, functions as a scalar multiple. i is an element of the complex number set. Either a + bi or a - bi will work, but using both is superfluous and counterproductive.
A*V1 = λV1; (A - λ)V1 = 0; [a-λ b, c d-λ]*[V1, V2] = [0, 0] such that λ is an element of the complex number set.
(A - λ)V1 + b*V2 = 0
C*V1 + (d - λ)*V2 = 0.
(A - λ) will be an element of the complex number set and b is an element of the real number set. V1 or V2 can be any valid solution such that (A - λ)*V1 = 0. The important factor is differentiating between numbers as elements of either the complex set or the real set such that any real number is part of the real number set, and any complex number is part of the complex number set.
Let λ = a + bi even though λ could just as easily be a - bi. One has the freedom to choose.
Then X(t) = V1*e^(λt) = V1*e^[(a + bi)t) = V1*e^(at + bit) = V1*e^(at)*e^(bit) = e^at*V1*e^(bit) = e^at*[V1, V2]*e^bit = e^at*[V1, V2]*Cos(t) + i(sint) = e^at*[V1*(Cos(t) + i(sin(t)), V2*(Cos(t) + i*sin(t)). If V1 is an element of the real number set, then V1 = C1*Cos(t) + C2*sin(t). If V1 is an element of the complex number set, then V1 = -C1sin(t) [due to i^2 = -1] + C2*cos(t). Generalizing this, C1 denotes an element of the real number set and C2 denotes an element of the complex number set. This can be generalized to V2 as well.
{X1(t), X2(t)} are elements of X(t). X1(t) = e^(at)V1*(cos(t) + i(sin(t)) and X2(t) = e^(at)*V2(cos(t) + isin(t)) such that i = C2 and 1 = C1 since 1 is an element of R and i is an element of C where what differentiates C and R is as the square root of any negative real number is i*the square root of the real number.
What this shows is linear algebra is merely a lemma for matrix algebra.
Now, there are a few more considerations. Linear algebra is a lemma to matrix algebra such that matrix algebra and linear algebra are consistent with one another (and this can only be done by creating an algorithm of matrix algebra's operators). Though the case and its subsets dealt with here were 2 x 2 matrices, with two solutions, this is only a matter of R^2 being an element of R^n, and this would hold for any n since matrix operators wouldn't change with added dimensions to n.
Let's consider a few more issues, that of a space.
If it's true that an n x n matrix works just as well as a 2 x 2 matrix (and it does, for operators will not change by adding dimensions), then consider this:
Let V be a vector space; let A, B, and C be matrices; let m and n be scalar quantities.
I. If A is an element of V and B is an element of V, then (A + B) is an element of V.
II. If A is an element of V and m is a scalar multiple, then mA is an element of V.
III. A + (B + C) = (A + B) + C such that both are elements of V (Transitivity).
IV. A + (-A) = A - A = 0 as elements of V.
V. A + 0 = 0 + A = A as an element of V.
VI. A + B = B + A as elements of V (Symmetry).
VII. m(A + B) = mA + mB as elements of V.
VIII. (m + n)A = mA + nA as elements of V.
IX. (mn)A = m(nA) as elements of V.
X. I*A = A as an element of V.
That proves this entire system is a vector space.
Having proven that, then it's also demonstrative to note that it's now a subspace of R^n = V.
QED.
The interest of this topic is in linear systems such that they can be written as matrices. The key issue to consider is, like in Linear Algebra, R^2 and R^3 are all elements of R^n and work in the exact same manner. There is nothing different about adding another dimensional variable. And note too that in R^4, though time is the fourth dimension, it's still an element of the fourth dimension.
There is another consideration too. Only homogeneous solutions have to be proven. There is a reason why.
Let X(t) = X,c + X,p such that X,c is the homogeneous solution and X,p is the non-homogeneous solution. Also, note that X(t) = C*Φ(t). Then, X,c and X,p are subsets of C*Φ(t). Since, X,c has already been solved for, then X,p must be solved for. Consider. . .
X' = AX in general. Then X',p = AX,p. If X = CΦ, then X,p = CΦ as well. Hence
d/dt[X,p] = d/dt[C*Φ] = C*Φ' + Φ*C'.
A non-homogeneous solution is in the form of X'(t) = AX + F(t). Then, due to the superposition principle, X',p = C*Φ' + Φ*C' such that either C*Φ' or Φ*C' is AX or F(t). Let CΦ' = AX, so CΦ' = ACΦ, then Φ' = AΦ. Hence F = Φ*C'. Then. . .
Let X' = X',P, then. . .
X' = Ax + F
X',p = C*Φ' + Φ*C';
X' = ACΦ + F
X' = C*AΦ + Φ*C';
X' = ACΦ + F
X' = ACΦ + Φ*C';
X' = ACΦ + F
-1*X' = -1*(ACΦ + Φ*C');
X' = ACΦ + F
-X' = -ACΦ - ΦC'
X' - X' = ACΦ - ACΦ + F - Φ*C';
0 = 0 + F - Φ*C'
F = Φ*C'
F = C' / (Φ^-1);
F*Φ^(-1) = C'.
Revisit X(t) = X,c + X,p such that X,c and X,p are subsets of CΦ(t). Since the homogeneous component, Ax has already been solved for, then X,c as an element of CΦ has been solved for. X,p is also an element of CΦ, but as an element of CΦ, it must be either of a different root, r, or multiplied by the next order up (e.g. A to At + B, or At + b to At^2 + bt + c, such that each order up is the integral of the previous order with respect to time such that the final integral is At^n + bt^(n - 1) + ct^(n - 2) + . . . + z,n*t^(n - n)). Due to this, and the fact that F = Φ*C', then consider:
X'(t) = Ax + F = C*Φ + Φ*C'; then integrating X'(t) and integrating C' after a substitution, then
X(t) = C*Φ + Φ*integral of C' = C*Φ + Φ*integral of F*Φ^(-1). Since C*Φ(t), then integrate F*Φ^(-1) with respect to a dummy variable, s, in terms of [a, b] being [to, t].
Regardless, homogeneous solutions are subsets of non-homogeneous solutions, and proving the homogeneous case proves the general case.
Hence x'(t) = A(t)x + b(t)y + e(t)
y'(t) = C(t)x + d(t)y + f(t).
That's the general form. A homogeneous equation is one such that e(t) and f(t) are equal to 0. Since 0 is an element of F(t), then homogeneous solutions are subsets of non-homogeneous solutions. Hence. . .
x'(t) = Ax + by
y'(t) = Cx + dy.
x' = Ax + by
y' = Cx + dy.
d/dt(x') = d/dt(Ax + by); x'' = Ax' + by'; x'' = Ax' + b(Cx + dy); x'' = Ax' + bcx + bdy; x'' = ax' + bcx + bd[(x' - ax)/b]; x'' = ax' + bcx + d(x' - ax); x'' = ax' + dx' + bcx - adx; x'' = x'(a + d) + x(bc - ad); x'' - x'(a + d) - x(bc - ad) = 0; x'' - x'(a + d) + x(ad - bc) = 0.
Hence, x'' - x'(a + d) + x(ad - bc) = 0 which turns to r^2 - r(a + d) + ad - bc = 0 such that x^(n) = r^n and x^(0) = 1.
Let K = ad - bc, then. . .
K + d^2 = ad + d^2 - bc; k + d^2 = d(a + d) - bc; (k + d^2 + bc) = d(a + d); [(k + d^2 + bc) / d] = (a + d); -[(k + d^2 + bc) / d] = -(a + d) = M.
Hence, r^2 - r(a + d) + (ad - bc) = r^2 + M*r + K = 0.
This is now in the form of the Quadratic Equation: Let A = 1, B = M, and C = K.
r = [-B ± square root(B^2 - 4AC)] / 2A = [-M ± square root of (M^2 - 4K)] / 2. Only the square root of (M^2 - 4K) is of any real interest.
Let D: square root of (U) ≥ 0 for any real number. M^2 - 4K is a subset of U, so the square root of (M^2 - 4k) ≥ 0.
If the square root of (M^2 - 4k) is greater than 0, then two real solutions exist such that X(t) = C,1*e^(r,1*t) + C,2*e^(r,2*t).
If the square root of (M^2 - 4k) is equal to 0, then a repeating eigenvalue case exists. More on this solution in a bit.
If the square root of (M^2 - 4k) is less than 0, then a complex case must be considered. More on this solution as well.
x' = ax + by
y' = cx + dy.
If the determinant of the above system is not equal to 0, then a unique solution is guaranteed. The determinant is ad - bc. If it is equal to 0, then the above system either has no solution (geometrically, we're talking about parallel lines; in three-dimensions, we're talking about planes) or an infinite number of solutions due to an intersection between an arbitrary number of constants.
x' = ax + by
y' = cx + dy
fits the form of x' = Ax such that A is a matrix of [A B, C D] such that the comma represents going down one level. The point being is that the three subsets of M^2 - 4K and its relation to 0. The determinant of A is always not equal to 0. Moreover, there must be n eigenvectors for an n x n matrix.
If the square root of (M^2 - 4K) is greater than 0, then the solution works as follows:
There is a matrix A = [a b, c d] such that the determinant is not equal to 0. A = λI such that I = [1 0, 0 1]. Hence, λI = [λ 0, 0 λ]. So A - λI = 0. Following matrix addition, then [a-λ b, c d-λ]. The determinant of [A - λI] will yield two real numbers known as eigenvalues. Let λ = {λ,1 and λ,2} so λ,1 and λ,2 are elements of λ such that λ,1 and λ,2 are both real numbers.
To find the solution of A in terms of λ,1, then A*V1 = λ,1*I*V1 such that V1 [V1, V2]. Since I is just an identity matrix and hence functions as 1, then λ,1*I = λ,1. So A*V1 = λ,1*V1; A*V1 - λ,1*V1 = (A - λ,1)*V1 = 0. A - λ,1 = [a-λ,1 b, c d-λ,1]*[V1, V2] = [0, 0] such that a-λ,1 and d-λ,1 are elements of the real number set. This will yield two equations with variables V1 and V2 as elements of V1:
(a - λ,1)*V1 + b*V2 = 0
C*V1 + (d - λ,1)*V2 = 0.
Solving for V1 and V2 will provide X1(t) = C1*[V1, V2]*e^(λ,1t).
Using λ,2 instead of λ,1 and using V2 for [V3, V4], then
(a - λ,2)*V3 + b*V4 = 0
C*V3 + (d - λ,2)*V4 = 0.
Solving for V3 and V4 will provide X2(t) = C2*[V3, V4)*e^(λ,2t)
So X(t) = C1*V1*e^(λ,1t) + C2*V3*e^(λ,2t)
and X(t) = C1*V2*e^(λ,1t) + C2*V4*e^(λ,2t) such that both solutions are elements of X(t).
Case II: If the square root of (M^2 - 4K) is equal to 0, then the solution works as follows:
There is a matrix A = [a b, c d] such that the determinant is not equal to 0. A = λI such that I = [1 0, 0 1]. Hence, λI = [λ 0, 0 λ]. So A - λI = 0. Following matrix addition, then [a-λ b, c d-λ]. The determinant of [A - λI] will yield one real numbered eigenvalue.
A*V1 = λ*I*V1; A*V1 = λ*V1; A*V1 - λ*V1 = 0; (A - λ)*V1 = 0 such that A - λ = [a-λ b, c d-λ], V1 = [V1, V2], and 0 = [0, 0]. Hence,
(a - λ)*V1 + b*V2 = 0
c*V1 + (d - λ)*V2 = 0.
Solving for V1 and V2 will yield two real numbers. However, two unique eigenvectors are in demand for the two x two matrix (or n unique eigenvectors for the n x n matrix), so. . .
(A - λ)*V2 = V1 such that V2 = [V3, V4].
(a - λ)*V3 + b*V4 = V1
c*V3 + (d - λ)*V4 = V2.
That will yield a solution for the second eigenvector.
However, X1(t) = C1*V1*e^(λt) = C1*[V1, V2]*e^(λt).
The integral of X1(t) = C1*e(λt)*the integral of V1*dt;
X2(t) = C2e^(λt)*(V1*t + C);
X2(t) = C2e^(λt)*(V1*t + V2);
X2(t) = C2e^(λt)*([V1, V2]t + [V3, V4]);
X2(t) = C2*e^(λt)*([V1*t, V2*t] + [V3, V4]);
X2(t) = C2*e^(λt)*[V1*t + V3, V2*t + V4].
So X(t) = X1(t) + X2(t) = e^(λt)*(C1*V1 + C2*(V1*t + V3)) or e^(λt)*(C1*V2 + C2*(V2*t + V4)).
If this were an n x n matrix, under the aforementioned conditions, then X(t) = X1(t) + X2(t) + . . . Xn(t) such that Xn(t) = Cn*e^(λt)*{[V1*t^(n - 1)/(n - 1)!] + [V2*t^(n - 2)/(n - 2)!] + [V3*t^(n - 3)/(n - 3)!] + . . . + [Vn^(n - n)/(n - n)!]}.
What if the square root of (M^2 - 4K) is less than 0?
Before this is dealt with, let's provide a lemma:
Taylor's Theorem dictates that a function can be approximated via F(t) + [F'(t)*(t - to)^1]/1! + [F''(t)*(t - to)^2]/2! + . . . + [F(n)*(X - Xo)^n]/n!.
F(t) = Cos(t). Let u = t, so du = dt, hence du/dt = 1.
F(t) = Cos(t); d/dt[F(t)] = d/dt[Cos(u)] = -sin(u)du/dt = -sin(t)*1 = -sin(t) = F'(t);
F''(t) = -d/dt[sin(t)] = -cos(t); F^(3)(t) = -d/dt[cos(t)] = -(-sin(t) = sin(t);
F^(4)(t) = d/dt[sin(t)] = cos(t); F^(5)(t) = d/dt [cos(t)] = -sin(t). This is a repeating cycle.
Cos(T) = Cos(To) - [sin(To)*(t - to)]/1! - [Cos(To)*(t - to)^2]/2! + [Sin(To)*(t - to)^3]/3! + [Cos(To)*(t - to)^4]/4! + . . . +[Cos(To)*(T - to)^n]/n!
Letting To = 0, then
Cos(t) = Cos(0) - [sin(0)*(t - 0)]/1! - [Cos(0)*(t - 0)^2]/2! + [Sin(0)*(t - 0)^3]/3! + [Cos(0)*(t - 0)^4]/4! + . . . +[Cos(0)*(T - 0)^n]/n! = 1 - 0 - (t^2 / 2!) + 0 + (T^4 / 4!) - . . . - t^(2k)/(2k)! + t^(2k + 2)/(2k + 2)! = 1 - (t^2 / 2!) + (T^4 / 4!) - . . . - t^(2k)/(2k)! + t^(2k + 2)/(2k + 2)!
Using this standard for Sin(t) such that F(t) = Sin(t), F'(t) = Cos(t), F''(t) = -sin(t), F^(3)(t) = -cos(t), F^(4)(t) = sin(t), creating a cycle, using the same to value of 0, then. . .
Sin(t) = t - (t^3 / 3!) + (t^5 / 5!) - . . . - t^(2k + 1) / (2k + 1)! + t^(2k + 3) / (2k + 3)!.
Since i is the only element of the complex number set that differentiates its elements from the real number set and i^2 yields a real number such that the complex number plane can map onto the real number plane, then this will have direct implications for Cos(t) and Sin(t).
Cos(t) has nothing but even, whole number exponents, hence f(t) = f(-t), or cos(t) = cos(-t).
Sin(t) has nothing but odd, whole number exponents, hence f(t) is not equal to f(-t), as f(-t) = -f(t).
Now, this will prove the next case:
A = λI; A = λ; A - λ = 0; [a-λ b, c d-λ] = [0, 0] such that the determinant of A - λ will be a complex number a ± bi such that a is an element of the real number set and b, being an element of the real number set, functions as a scalar multiple. i is an element of the complex number set. Either a + bi or a - bi will work, but using both is superfluous and counterproductive.
A*V1 = λV1; (A - λ)V1 = 0; [a-λ b, c d-λ]*[V1, V2] = [0, 0] such that λ is an element of the complex number set.
(A - λ)V1 + b*V2 = 0
C*V1 + (d - λ)*V2 = 0.
(A - λ) will be an element of the complex number set and b is an element of the real number set. V1 or V2 can be any valid solution such that (A - λ)*V1 = 0. The important factor is differentiating between numbers as elements of either the complex set or the real set such that any real number is part of the real number set, and any complex number is part of the complex number set.
Let λ = a + bi even though λ could just as easily be a - bi. One has the freedom to choose.
Then X(t) = V1*e^(λt) = V1*e^[(a + bi)t) = V1*e^(at + bit) = V1*e^(at)*e^(bit) = e^at*V1*e^(bit) = e^at*[V1, V2]*e^bit = e^at*[V1, V2]*Cos(t) + i(sint) = e^at*[V1*(Cos(t) + i(sin(t)), V2*(Cos(t) + i*sin(t)). If V1 is an element of the real number set, then V1 = C1*Cos(t) + C2*sin(t). If V1 is an element of the complex number set, then V1 = -C1sin(t) [due to i^2 = -1] + C2*cos(t). Generalizing this, C1 denotes an element of the real number set and C2 denotes an element of the complex number set. This can be generalized to V2 as well.
{X1(t), X2(t)} are elements of X(t). X1(t) = e^(at)V1*(cos(t) + i(sin(t)) and X2(t) = e^(at)*V2(cos(t) + isin(t)) such that i = C2 and 1 = C1 since 1 is an element of R and i is an element of C where what differentiates C and R is as the square root of any negative real number is i*the square root of the real number.
What this shows is linear algebra is merely a lemma for matrix algebra.
Now, there are a few more considerations. Linear algebra is a lemma to matrix algebra such that matrix algebra and linear algebra are consistent with one another (and this can only be done by creating an algorithm of matrix algebra's operators). Though the case and its subsets dealt with here were 2 x 2 matrices, with two solutions, this is only a matter of R^2 being an element of R^n, and this would hold for any n since matrix operators wouldn't change with added dimensions to n.
Let's consider a few more issues, that of a space.
If it's true that an n x n matrix works just as well as a 2 x 2 matrix (and it does, for operators will not change by adding dimensions), then consider this:
Let V be a vector space; let A, B, and C be matrices; let m and n be scalar quantities.
I. If A is an element of V and B is an element of V, then (A + B) is an element of V.
II. If A is an element of V and m is a scalar multiple, then mA is an element of V.
III. A + (B + C) = (A + B) + C such that both are elements of V (Transitivity).
IV. A + (-A) = A - A = 0 as elements of V.
V. A + 0 = 0 + A = A as an element of V.
VI. A + B = B + A as elements of V (Symmetry).
VII. m(A + B) = mA + mB as elements of V.
VIII. (m + n)A = mA + nA as elements of V.
IX. (mn)A = m(nA) as elements of V.
X. I*A = A as an element of V.
That proves this entire system is a vector space.
Having proven that, then it's also demonstrative to note that it's now a subspace of R^n = V.
QED.