Section 5.3. System of linear equations

Objective:

1. Basic terms of a system of linear equations

2. Linearly independent, eigenvalues, eigenvectors

Definition: (a) A system of $n$ linear equations in $n$ variables,

$$\begin{array}{cccc}{a}_{11}{x}_1+& a_{12}{x}_2+& \cdots & +a_{1n}{x}_n=b_1\\ a_{21}{x}_1+& a_{22}{x}_2+& \cdots & +a_{2n}{x}_n=b_2\\ ⋮& ⋮& ⋮\\ a_{n1}{x}_1+& a_{n2}{x}_2+& \cdots & +a_{nn}{x}_n=b_n\end{array}$$

 can be expressed as a matrix equation $A\overrightarrow{x}=\overrightarrow{b}$:

$$\left[\begin{array}{cccc}{a}_{11}& a_{12}& \cdots & a_{1n}\\ a_{21}& a_{22}\\ ⋮& ⋮\\ a_{n1}& a_{n2}& \cdots & a_{nn}\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right]=\left[\begin{array}{c}{b}_1\\ b_2\\ ⋮\\ b_n\end{array}\right].$$

If $\overrightarrow{b}=0$, then system is homogeneous; otherwise it is nonhomogeneous.

(b) A set of vectors $\overrightarrow{v_1}$,…,$\overrightarrow{v_p}$ is linearly dependent if there exists scalars $c_1,c_2,...,c_p$, not all zero, such that 

$$c_1\overrightarrow{v_1}+c_2\overrightarrow{v_2}+\dots .+c_p\overrightarrow{v_p}=\overrightarrow{0},$$

 otherwise it is linearly independent.

(c) A set of vector functions $\overrightarrow{v_1(t)}$,…,$\overrightarrow{v_p(t)}$ is linearly dependent if there exists scalars $c_1,c_2,...,c_p$, not all zero, such that 

$$c_1\overrightarrow{v_1(t)}+c_2\overrightarrow{v_2(t)}+\dots .+c_p\overrightarrow{v_p(t)}=\overrightarrow{0},$$

 otherwise it is linearly independent.

Example 1: Determinate if the set of the vector functions is linearly independent or linearly dependent. 

$\overrightarrow{v_1}=\left[\begin{array}{c}3e^t\\ e^{2t}\end{array}\right]$,  $\overrightarrow{v_2}=\left[\begin{array}{c}{e}^t\\ 4e^{2t}\end{array}\right]$, and $\overrightarrow{v_3}=\left[\begin{array}{c}2e^t\\ -3e^{2t}\end{array}\right]$.

Exercise 1: Determinate if the set of the vector functions is linearly independent or linearly dependent.  $\overrightarrow{v_1}=\left[\begin{array}{c}2e^{3t}\\ e^{4t}\end{array}\right]$, $\overrightarrow{v_2}=\left[\begin{array}{c}{e}^{3t}\\ -2e^{4t}\end{array}\right]$, and $\overrightarrow{v_3}=\left[\begin{array}{c}7e^{3t}\\ e^{4t}\end{array}\right]$.

Fact: (a) If the coefficient matrix $A$ is nonsingular, then it is invertible and we can solve $A\overrightarrow{x}=\overrightarrow{b}$ and the solution is $\overrightarrow{x}=A^{-1}\overrightarrow{b}$. The only solution to $A\overrightarrow{x}=\overrightarrow{0}$is the trivial solution $\overrightarrow{x}=\overrightarrow{0}$. 

(b) The columns (or rows) of $A$ are linearly independent iff $A$ is nonsingular iff  $A^{-1}$ exists.

(c) A is nonsingular iff $\text{det}A\ne 0$. 

Example 2:  Show $\overrightarrow{v_1}=\left[\begin{array}{c}3e^t\\ e^{2t}\end{array}\right]$,  $\overrightarrow{v_2}=\left[\begin{array}{c}{e}^t\\ 4e^{2t}\end{array}\right]$ are linearly independent. 

Exercise 2: Show$\overrightarrow{v_1}=\left[\begin{array}{c}2e^{3t}\\ e^{4t}\end{array}\right]$, $\overrightarrow{v_2}=\left[\begin{array}{c}{e}^{3t}\\ -2e^{4t}\end{array}\right]$ are linearly independent. 

Definition:  Let $\lambda$ be a value such that $A\overrightarrow{x}=\lambda \overrightarrow{x}$ for some $\overrightarrow{x}\ne 0$, then $(A-\lambda I)\overrightarrow{x}=\overrightarrow{0}$ has a nontrivial solution, $\overrightarrow{x}\ne \overrightarrow{0}$. We must have $\text{det}(A-\lambda I)=0$. We call $\lambda$ an eigenvalue of $A$} and $\overrightarrow{x}$ is called the eigenvector of $A$} corresponding to the eigenvalue $\lambda$. If an eigenvalue $\lambda$ is repeated $m$ times, i.e. it is a repeated root of $\text{det}(A-\lambda I)=0$ for $m$ times, then its algebraic multiplicity is $m$.

Example 3:  Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$ and $B=\left[\begin{array}{cc}3& 4\\ 0& 3\end{array}\right]$.

Exercise 3: Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}4& 0\\ 0& 4\end{array}\right]$ and $B=\left[\begin{array}{cc}4& -3\\ 0& 4\end{array}\right]$. 

Example 4: Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}5& -1\\ 3& 1\end{array}\right]$. 

Exercise 4:  Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}4& 1\\ -2& 1\end{array}\right]$. 

Group Work: 

1. Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}3& -1\\ 5& -3\end{array}\right]$. 

2. Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}5& 3\\ 3& 5\end{array}\right]$. 

3. Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}2& 1\\ -1& 4\end{array}\right]$. 

4. Find the eigenvalues and their eigenvectors of $A=\left[\begin{array}{cc}2& 7\\ 7& 2\end{array}\right]$.