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 [latex]n[/latex] linear equations in [latex]n[/latex] variables,

\[\begin{array}{cccc} a_{11} x_1+ & a_{12}x_2+ &  \cdots & +a_{1n}x_n =b_1\\ a_{21}x_1+ & a_{22}x_2+& \cdots & +a_{2n}x_n=b_2\\ \vdots & \vdots & \vdots\\ a_{n1}x_1+ & a_{n2}x_2+ & \cdots &+ a_{nn}x_n=b_n  \end{array} \]

 can be expressed as a matrix equation [latex]A\overrightarrow{x}=\overrightarrow{b}[/latex]:

\[\left[\begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22}\\ \vdots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array}\right]\left[\begin{array}{c} x_{1}\\ x_{2}\\ \vdots\\ x_{n} \end{array}\right]=\left[\begin{array}{c} b_{1}\\ b_{2}\\ \vdots\\ b_{n} \end{array}\right]. \]

If [latex]\overrightarrow{b}=0[/latex], then system is homogeneous; otherwise it is nonhomogeneous.

(b) A set of vectors [latex]\overrightarrow{v_{1}}[/latex],…,[latex]\overrightarrow{v_{p}}[/latex] is linearly dependent if there exists scalars [latex]c_{1},c_{2},...,c_{p}[/latex], not all zero, such that 

\[ c_{1}\overrightarrow{v_{1}}+c_{2}\overrightarrow{v_{2}}+….+c_{p}\overrightarrow{v_{p}}=\overrightarrow{0}, \]

 otherwise it is linearly independent.

(c) A set of vector functions [latex]\overrightarrow{v_{1}(t)}[/latex],…,[latex]\overrightarrow{v_{p}(t)}[/latex] is linearly dependent if there exists scalars [latex]c_{1},c_{2},...,c_{p}[/latex], not all zero, such that 

\[ c_{1}\overrightarrow{v_{1}(t)}+c_{2}\overrightarrow{v_{2}(t)}+….+c_{p}\overrightarrow{v_{p}(t)}=\overrightarrow{0},\]

 otherwise it is linearly independent.

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

[latex]\overrightarrow{v_{1}}=\left[\begin{array}{c} 3e^{t}\\ e^{2t} \end{array}\right][/latex],  [latex]\overrightarrow{v_{2}}=\left[\begin{array}{c} e^{t}\\ 4e^{2t} \end{array}\right][/latex], and [latex]\overrightarrow{v_{3}}=\left[\begin{array}{c} 2e^{t}\\ -3e^{2t} \end{array}\right][/latex].

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

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

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

(c) A is nonsingular iff [latex]\text{det}A\neq0[/latex]. 

Example 2:  Show [latex]\overrightarrow{v_{1}}=\left[\begin{array}{c} 3e^{t}\\ e^{2t} \end{array}\right][/latex],  [latex]\overrightarrow{v_{2}}=\left[\begin{array}{c} e^{t}\\ 4e^{2t} \end{array}\right][/latex] are linearly independent. 

Exercise 2: Show[latex]\overrightarrow{v_{1}}=\left[\begin{array}{c} 2e^{3t}\\ e^{4t} \end{array}\right][/latex], [latex]\overrightarrow{v_{2}}=\left[\begin{array}{c} e^{3t}\\ -2e^{4t} \end{array}\right][/latex] are linearly independent. 

Definition:  Let [latex]\lambda[/latex] be a value such that [latex]A\overrightarrow{x}=\lambda\overrightarrow{x}[/latex] for some [latex]\overrightarrow{x}\neq0[/latex], then [latex](A-\lambda I)\overrightarrow{x}=\overrightarrow{0}[/latex] has a nontrivial solution, [latex]\overrightarrow{x}\neq\overrightarrow{0}[/latex]. We must have [latex]\text{det}(A-\lambda I)=0[/latex]. We call [latex]\lambda[/latex] an eigenvalue of [latex]A[/latex]} and [latex]\overrightarrow{x}[/latex] is called the eigenvector of [latex]A[/latex]} corresponding to the eigenvalue [latex]\lambda[/latex]. If an eigenvalue [latex]\lambda[/latex] is repeated [latex]m[/latex] times, i.e. it is a repeated root of [latex]\text{det}(A-\lambda I)=0[/latex] for [latex]m[/latex] times, then its algebraic multiplicity is [latex]m[/latex].

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

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

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

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

Group Work: 

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

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

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

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