One who asks a question is a fool for a minute; one who does not remains a fool forever, Chinese proverb

Any fool can criticize, complain, and condemn —and most fools do. But it takes character and self-control to be understanding and forgiving, Dale Carnegie

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:

- y is the dependent variable (a function of the independent variable t),
- y′ and y′′ are the first and second derivatives of y with respect to t,
- t is the independent variable,
- A and B are constants.

This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.

A linear first-order system has the form:

$\begin{cases} x’ = a(t)x + b(t)y + r_1(t) \\ y’ = c(t)x + d(t)y + r_2(t) \end{cases}$ where:

- a(t), b(t), c(t), and d(t) are functions of time t (they can be constants in simpler cases).
- r
_{1}(t) and r_{2}(t) represent external forcing functions or inputs that depend on t. - x(t
_{0}) = x_{0}and y(t_{0}) = y_{0}are the initial conditions.

Let’s explore homogeneous linear systems of first-order ordinary differential equations (ODEs) with constant coefficients. Specifically, we will analyze the following system:

$\begin{cases} x’ = -2x + 2y \\ y’ = 2x - 5y \end{cases}$ where:

- x(t) and y(t) are the dependent variables,
- t is the independent variable (typically representing time),
- x’ = $\frac{dx}{dt}$ and y’ = $\frac{dy}{dt}$ denote the first derivatives of x(t) and y(t) with respect to t.

We will solve this system using the eigenvalue and eigenvector method, also known as the exponential ansatz. This approach transforms the system into an algebraic problem, allowing us to find the general solution by identifying the system’s eigenvalues and corresponding eigenvectors.

The given system of ODE’s is a **homogeneous linear system** because:

- The equations are linear in x and y.
- There are no constant or forcing terms (i.e., terms that do not depend on x or y).

In a previous solution approach, we used substitution to reduce the system to a second-order differential equation. The solution was found to be:

$\begin{cases} x = C_1e^{-t} + C_2e^{-6t} \\ y = \frac{1}{2}C_1e^{-t} - 2C_2e^{-6t} \end{cases}$

$\begin{cases} x’ = -2x + 2y \\ y’ = 2x - 5y \end{cases}$

To streamline the solution process, we can express the system in matrix form. This facilitates the application of linear algebra techniques, particularly those involving eigenvalues and eigenvectors.

$(\begin{smallmatrix}x\\ y\end{smallmatrix})’ = (\begin{smallmatrix}-2 &2\\2 &-5\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix})$ (🚀) where the matrix A = $(\begin{smallmatrix}-2 &2\\2 &-5\end{smallmatrix})$ contains the coefficients of the system. The vector X = $(\begin{smallmatrix}x\\ y\end{smallmatrix})$represents the dependent variables.

Thus, the system can be succinctly written as: X’ = AX. The goal is to find the general solution to this system.

The eigenvalue and eigenvector method involves finding solutions of the form: $X(t) = ve^{λt}$ where:

- λ is an eigenvalue of matrix A
- v is the corresponding eigenvector.

Assume a solution of the form: $X(t) = (\begin{smallmatrix}x\\ y\end{smallmatrix}) = ve^{λt} = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})e^{λt}$

Substituting into the original system X’ = AX: (🚀) $X’ = λve^{λt} = Ave^{λt}$

Cancelling the exponential term e^{λt} (since it’s never zero): λv = Av. This equation is the matrix eigenvalue equation.

**Setting Up the Characteristic Equation**. To find the eigenvalues λ, we solve the characteristic equation: det(A -λI) = 0 where I is the identity matrix.

Compute A -λI = $(\begin{smallmatrix}-2-λ & 2\\2 &-5-λ\end{smallmatrix})$

Calculate the determinant: $det(A -λI) = (-2-λ)(-5-λ)-4 = 0 ↭[\text{Simplify}] (2+λ)(5+λ)-4 = 0 ↭ λ^2+7λ + 6 = 0$.

Solving for λ using the quadratic formula, $λ = \frac{-7±\sqrt{7^2-4·1·6}}{2·1} = \frac{-7±\sqrt{25}}{2}$.

**Finding the Corresponding Eigenvectors**

The roots are λ_{1} = -1, λ_{2} = -6.

For λ_{1} = -1. Substitute λ = 1 into the (A−λI)v=0:

$(\begin{smallmatrix}-2-(-1) & 2\\2 &-5-(-1)\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}-1 & 2\\2 &-4\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$

This yields the system:

$\begin{cases} -a_1 + 2a_2 = 0\\ 2a_1 -4a_2 = 0 \end{cases}$

The second equation is a multiple of the first, so we can choose a_{2} = 1, which gives a_{1} = 2. The eigenvector corresponding to λ = 1 is: $(\begin{smallmatrix}a_1\\a_2\end{smallmatrix})=(\begin{smallmatrix}2\\ 1\end{smallmatrix})$. The solution associated with λ = 1 is: $(\begin{smallmatrix}2\\ 1\end{smallmatrix})e^{-t}$

For λ_{2} = 6. Substitute λ = 6 into (A-λI)v = 0, we aim to calculate a_{1}, a_{2}.

$(\begin{smallmatrix}-2-(-6) & 2\\2 &-5-(-6)\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}4 & 2\\2 & 1\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$

This yields the system:

$\begin{cases} 4a_1 + 2a_2 = 0\\ 2a_1 +a_2 = 0 \end{cases}$

The second equation is a constant multiple (2) of the first one. Typically, we make a_{1} = 1, then a_{1} = -2. The eigenvector corresponding to λ = -6 is: $(\begin{smallmatrix}a_1\\a_2\end{smallmatrix})=(\begin{smallmatrix}1\\ -2\end{smallmatrix})$. The solution associated with λ = -6 is: $(\begin{smallmatrix}1\\ -2\end{smallmatrix})e^{-6t}$

Using the eigenvalues and eigenvectors, the general solution to the system is a linear combination of the solutions corresponding to each eigenvalue: X(t) = $C_1v_1e^{dλ_1t} + C_2v_2e^{dλ_2t}$.

In other words, by the principle of superposition, the general solution to the system is: $(\begin{smallmatrix}x\\y\end{smallmatrix}) = \tilde{c_1}(\begin{smallmatrix}2\\ 1\end{smallmatrix})e^{-t}+ \tilde{c_2}(\begin{smallmatrix}1\\ -2\end{smallmatrix})e^{-6t}$. This matches the solution obtained via substitution, but presented in a clearer vector form, the difference is that $\frac{c_1}{2} = \tilde{c_1}, c_2 = \tilde{c_2}$. The constants C_{1} and C_{2} are determined by initial conditions.

In the study of ordinary differential equations (ODEs), understanding systems of linear differential equations is crucial for modeling and analyzing a wide range of physical, biological, and engineering phenomena.

Consider the following homogeneous system of two coupled first-order linear differential equations with constant coefficients: $\begin{cases} x’ = ax + by \\ y’ = cx + dy \end{cases}$

where:

- x(t) and y(t) are the dependent variables,
- t is the independent variable (typically representing time),
- x′ = $\frac{dx}{dt}$ and y′ = $\frac{dy}{dt}$ denote the first derivatives of x(t) and y(t) with respect to t,
- a, b, c, and d are constant coefficients that characterize the system’s behavior.

**Matrix Representation**

$(\begin{smallmatrix}x\\ y\end{smallmatrix})’ = (\begin{smallmatrix}a & b\\c & d\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix})$ (🚀) where the matrix A = $(\begin{smallmatrix}a & b\\c &d\end{smallmatrix})$ contains the coefficients from the original system of equations. The vector X = $(\begin{smallmatrix}x\\ y\end{smallmatrix})$represents the system’s state at any time t.

X′ = AX is the system in matrix form. Our goal is to find the general solution for x(t) and y(t) ↭ $\vec{x’} = A\vec{x}$

We begin by assuming that the solutions for x(t) and y(t) take the form of exponential functions: $X(t) = ve^{λt} = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})e^{λt}$ where

- λ is an eigenvalue of matrix A,
- v = $(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$ is the corresponding eigenvector,
- $e^{λt}$ represents the exponential growth or decay rate.

**Substituting the Trial Solution $X(t) = ve^{λt}$ into the System** X’ = AX.
leads to: $λve^{λt} = Ave^{λt}$. Since e^{λt} is never zero, we can divide both sides by e^{λt}. This equation is the eigenvalue equation: $Av = λv$.

This trial solution assumes that the solution grows (or decays) exponentially with time at a rate determined by λ.

**Formulating and Solving the Eigenvalue Problem**

To find the eigenvalues λ and eigenvectors v, we solve the characteristic equation derived from the eigenvalue equation.

The characteristic equation is obtained by ensuring that the system Av = λv has non-trivial solutions (i.e., v ≠ 0): det(A -λI) = 0 where I is the identity matrix.

The determinant is: det(A -λI) = $(\begin{smallmatrix}a-λ & b\\c & d-λ\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (a-λ)(d-λ) -bc = 0 ↭ λ^2 +(a+d)λ + (ad -bc) = 0$. This is a quadratic equation in λ, known as the characteristic equation.

The eigenvalues λ_{1} and λ_{2} are the roots of the characteristic equation: $λ^2 -(a+d)λ + (ad -bc) = 0$

Using the quadratic formula: $λ = \frac{(a+d)±\sqrt{(a+d)^2-4·1·(ad-bc)}}{2·1}$. These eigenvalues determine the nature of the system’s solutions (e.g., exponential growth, decay, oscillations).

For each eigenvalue λ_{i}, we find the corresponding eigenvector v_{i} by solving:

$(A-λ_iI)v_i = 0 ↭ (A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

$(\begin{smallmatrix}a-λ_i & b\\c &d-λ_i\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$.

$\begin{cases} (a -λ_i)a_1 + ba_2 \\ ca_1 + (d-λ_i)a_2 = 0 \end{cases}$

This gives us a linear system for a_{1} and a_{2} that we can solve to find the components of the corresponding eigenvector v_{i} associated with each eigenvalue λ_{i}.

Once we have the eigenvalues and eigenvectors, the general solution to the system is a linear combination of the solutions corresponding to each eigenvalue: $\vec{x(t)} = c_1\vec{v_1}e^{λ_1} + c_2\vec{v_2}e^{λ_2}$ where $\vec{v_1} = (\begin{smallmatrix}a_{11}\\a_{12}\end{smallmatrix})$ and $\vec{v_2} = (\begin{smallmatrix}a_{21}\\a_{22}\end{smallmatrix})$ are the eigenvectors corresponding to the eigenvalues λ_{1} and λ_{2} respectively, and c_{1} and c_{2} are arbitrary constants determined by initial conditions.

- Solve the system of differential equations:

$\begin{cases} x’_1 = 2x_1 + 3x_2 \\ x’_2 = 4x_1 -2x_2 \end{cases}$

We can rewrite the system in matrix form as X′= AX, where: $X = (\begin{smallmatrix}x_1\\ x_2\end{smallmatrix}), A = (\begin{smallmatrix}2 & 3\\ 4 & -2\end{smallmatrix})$. The general solution will involve finding the eigenvalues and eigenvectors of matrix A.

To find the eigenvalues, we compute the characteristic polynomial, which requires calculating det(A − λI).

Compute A -λI = $(\begin{smallmatrix}2-λ & 3\\ 4 & -2-λ\end{smallmatrix})$

Calculate the determinant: det(A -λI) = $det(\begin{smallmatrix}2-λ & 3\\ 4 & -2-λ\end{smallmatrix}) = (2-λ)(-2-λ)-12 = λ^2-2λ+2λ-4-12 = λ^2-16$.

Thus, we obtain the characteristic equation: $λ^2-16 = 0$

Solving for λ: $λ^2-16 = 0 ↭ (λ-4)(λ+4) = 0$. The roots (eigenvalues) are λ_{1} = 4, λ_{2} = -4.

For λ_{1} = 4. Substitute λ_{1} into (A -λI)v = 0

$(\begin{smallmatrix}-2 & 3\\ 4 & -6\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$

This yields the system:

$\begin{cases} -2a_1 + 3a_2 = 0 \\ 4a_1 -6a_2 = 0 \end{cases}$

The second equation is a multiple of the first so we can choose a_{1} = 3 and a_{2} = 2. The eigenvector corresponding to λ_{1} = 4 is $v_1 = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}3\\ 2\end{smallmatrix})$. The solution associated with λ_{1} = 4 is: $(\begin{smallmatrix}3\\ 2\end{smallmatrix})e^{4t}$

For λ_{2} = -4. Substitute λ_{2} into (A -λI)v = 0

$(\begin{smallmatrix}6 & 3\\ 4 & 2\end{smallmatrix})(\begin{smallmatrix}a_1\\ a_2\end{smallmatrix})$

This yields the system:

$\begin{cases} 6a_1 + 3a_2 = 0 \\ 4a_1 +2a_2 = 0 \end{cases}$

The second equation is a multiple (^{2}⁄_{3}) of the first so we can choose a_{1} = 1 and a_{2} = -2. The eigenvector corresponding to λ_{1} = -4 is $v_1 = (\begin{smallmatrix}a_1\\ a_2\end{smallmatrix}) = (\begin{smallmatrix}1\\ 2\end{smallmatrix})$. The solution associated with λ_{1} = 4 is: $(\begin{smallmatrix}1\\ 2\end{smallmatrix})e^{-4t}$

The general solution to the system is a linear combination of the solutions corresponding to each eigenvalue: X(t) = $C_1v_1e^{dλ_1t} + C_2v_2e^{dλ_2t} = C_1(\begin{smallmatrix}3\\ 2\end{smallmatrix})e^{4t} + C_2(\begin{smallmatrix}1\\ 2\end{smallmatrix})e^{-4t}$ where C_{1} and C_{2} are constants determined by initial conditions.

Consider a model where a fish tank is divided into three compartments, each maintained at different temperatures:

- t
_{1}for the tropical region. - t
_{2}for the ice region, and - t
_{3}for the normal region.

The compartments are well insulated from the outside environment, and heat flows only between these compartments. Over time, these compartments exchange heat, and our goal is to determine how the temperatures in each compartment evolve over time (Refer to Figure i for a visual representation and aid in understanding it).

To represent this mathematically, let x_{i}(t) represent the temperature in the i-th compartment at time t. The rate of change of temperature in each compartment depends on the temperature differences between neighboring compartments.

**Setting Up the Differential System**

We aim to calculate the variables x_{i}, which represent the temperatures in each compartment as functions of time.

We assume that the heat flow from compartment i to compartment j is proportional to the difference in temperature between the compartments, following Newton's law of cooling. This means the rate at which heat flows is proportional to x_{j} - x_{i}.

Thus, the system of equations governing the temperature changes is.

Thus, for compartment 1, the net rate of heat flow into it is proportional to (x_{2} - x_{1}) + (x_{3} - 1_{1}). Therefore, the differential equation for x_{1} is:

$x_1’ = a(x_3-x_1)+a(x_2-x_1) ↭[\text{This can be simplified to: }] x_1’ = -2ax_1+ax_2+ax_3$

There is symmetry in our differential system regarding the heat flow between compartments (each compartment interacts with the others in a similar way). For simplicity, we assume that a = 1, which leads to the following system of differential equations:

$\begin{cases} x_1’ = -2x_1+x_2+x_3 \\ x_2’ = x_1-2x_2+x_3 \\ x_2’ = x_1+x_2-2x_3 \end{cases}$

This system can be written in matrix form as: x’ = Ax where x = $(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix})$ and A = $\Bigl( \begin{smallmatrix}-2 & 1 & 1\\ 1 & -2 & 1\\ 1 & 1 & -2\end{smallmatrix} \Bigr)$

We now aim to find the general solution to this system, which requires solving for the eigenvalues and eigenvectors of the matrix A.

To find the eigenvalues, we solve the characteristic equation: |A - λI| = $\Bigl \vert\begin{smallmatrix}-2-λ & 1 & 1\\ 1 & -2-λ & 1\\ 1 & 1 & -2-λ\end{smallmatrix}\Bigr \vert = -(λ+2)^3-3(-2-λ)+2 = 0 ↭−(8+12λ+6λ^2 +λ^3)−3(−2−λ)+2 = 0 ↭ (−8+6+2)+(−12λ+3λ)+(−6λ^2) −λ^3 =0 ↭\text{Simplify and multiply both sides by -1} λ^3 +6λ^2 +9λ = 0$

Factoring it: λ(λ +3)^{2} = 0. Eigenvalues: λ_{1} = 0, λ_{2, 3} = -3 (repeated root).

- Eigenvalue λ
_{1}= 0

We solve $(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

$(\begin{smallmatrix}-2 & 1 & 1\\1 &-2 & 1 \\1 & 1 & -2\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix}) = 0$.

This leads to:

$\begin{cases} -2a_1 + a_2 + a_3 = 0 \\ a_1 -2a_2 + a_3 = 0 \\ a_1 + a_2 -2a_3 = 0 \end{cases}$

This is a non-independent solution, the determinant of the coefficients is zero. This system has infinitely many solutions.

Solution: $(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix})e^{0t} = (\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix})$. This is a constant solution, and intuitively there is a trivial solution, all compartments share the same temperature, hence there is no reason why they should change their temperatures as time goes on. So, the problem itself suggest a solution $v_1 =(\begin{smallmatrix}1\\1\\1\end{smallmatrix})$

We can check that this solution is indeed a solution of the system of equations.

- Eigenvalue λ
_{2, 3}= 3 (repeated)

We solve $(A-λ_iI)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0 ↭ (A+3I)(\begin{smallmatrix}a_1\\a_2\end{smallmatrix}) = 0$

$(\begin{smallmatrix}1 & 1 & 1\\1 &1 & 1 \\1 & 1 & 1\end{smallmatrix})(\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix}) = 0$.

$\begin{cases} a_1 + a_2 + a_3 = 0 \\ a_1 + a_2 + a_3 = 0 \\ a_1 + a_2 + a_3 = 0 \end{cases}$

This a dependent system of equation, meaning they describe the same relationship between the variables. Futhermore, I just have one equation, but repeated two times, namely $a_1 + a_2 + a_3 = 0 ⇒a_3 = -(a_1+a_2)$

- If a
_{1}= 1, a_{2}= 0 ⇒ a_{3}= -1, $v_2 = (\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})$ - If a
_{1}= 1, a_{2}= -1 ⇒ a_{3}= 0, $v_3 = (\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})$

Two independent solutions are: $(\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})e^{-3t}$ and $(\begin{smallmatrix}1\\ -1 \\ 0\end{smallmatrix})e^{-3t}$

The reader should be cautions, e.g., $(\begin{smallmatrix}0 \\ 1 \\ -1\end{smallmatrix})$ is not independent of the previous both: $(\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix}) - (\begin{smallmatrix}1\\ -1 \\ 0\end{smallmatrix}) = (\begin{smallmatrix}0 \\ 1 \\ -1\end{smallmatrix})$

The general solution is a linear combination of the eigenvectors: $(\begin{smallmatrix}x_1\\x_2\\x_3\end{smallmatrix}) = c_1(\begin{smallmatrix}1 \\ 0 \\ -1\end{smallmatrix})e^{-3t} + c_2(\begin{smallmatrix}1\\ -1 \\ 0\end{smallmatrix})e^{-3t}+c_3(\begin{smallmatrix}1\\1\\1\end{smallmatrix})$

**Long-Term Behavior**. As t → ∞, $(\begin{smallmatrix}x_1\\x_2\\x_3\end{smallmatrix}) → c_3(\begin{smallmatrix}1\\1\\1\end{smallmatrix})$. All solutions tend to the solution where all compartments reach the same temperature, corresponding to the eigenvalue λ = 0.

A complete eigenvalue (or a non-defective eigenvalue) is a repeated eigenvalue for which there exists a sufficient number of linearly independent eigenvectors to form a complete basis. This means that if λ is an eigenvalue with algebraic multiplicity m (the number of times it is repeated), there are m linearly independent eigenvectors associated with λ. As a result, you can construct a full set of independent solutions.

In our problem, λ = −3 is a complete eigenvalue because we found two linearly independent eigenvectors corresponding to its algebraic multiplicity of 2.

A defective eigenvalue (or a non-complete eigenvalue) is also a repeated eigenvalue, but in this case, there are not enough linearly independent eigenvectors to match its algebraic multiplicity.

The Spectral or Principal Axis Theorem is a fundamental result in linear algebra and has significant implications in various fields such as physics, engineering, and mathematics. It states that for a real symmetric matrix, the eigenvalues are real, and there exists an orthogonal basis of eigenvectors.

The Spectral or Principal Axis Theorem states that if A is a real n × n symmetric matrix (i.e., A = A^{T}), then:

- Real Eigenvalues: All eigenvalues of the matrix A are real numbers. This means that when we solve the characteristic equation det(A −λI) = 0, all solutions λ (the eigenvalues) are real numbers.
- Complete Eigenvalues: All eigenvalues are complete, meaning that for each eigenvalue, there exists a full set of n linearly independent eigenvectors for A. Moreover, these eigenvectors can be chosen to be orthogonal.
For each eigenvalue, there exists a corresponding eigenvector, and the collection of all such eigenvectors forms a basis for ℝ

^{n}. Since the eigenvectors can be chosen to be orthogonal, we can simplify the matrix A to a diagonal form using an orthogonal transformation.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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