7.1

Definition: Vector Space

A set \(V\) together with two operations, called addition and scalar multiplication is a vector space if the following vector space axioms are satisfied for all vectors \(\mathbf{u}, \mathbf{v}\), and \(\mathbf{w}\) in \(V\) and all scalars, \(c, d\) in \(R\).

Vector space axioms:

1. \(\mathbf{u}+\mathbf{v}\) is in \(V\)
2. \(c \mathbf{u}\) is in \(V\)
3. \(\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}\)
4. \((\mathbf{u}+\mathbf{v})+\mathbf{w}=\mathbf{u}+(\mathbf{v}+\mathbf{w})\)
5. There is a vector, denoted by \(\mathbf{0}\), in \(V\) such that \(\mathbf{u}+\mathbf{0}=\mathbf{u}\) for all \(\mathbf{u}\) in \(V\)
6. For each \(\mathbf{u}\) in \(V\), there is an element, denoted by \(-\mathbf{u}\), in \(V\) such that \(\mathbf{u}+(-\mathbf{u})=\mathbf{0}\)
7. \((c d) \mathbf{u}=c(d \mathbf{u})\)
8. \((c+d) \mathbf{u}=c \mathbf{u}+d \mathbf{u}\)
9. \(c(\mathbf{u}+\mathbf{v})=c \mathbf{u}+c \mathbf{v}\)
10. \(1 \mathbf{u}=\mathbf{u}\)

Examples:
    \(\mathbb{R}^k\) with the usual operations of addition and scalar multiplication is a vector space.
    The set \(\mathbb{M}^{k,n}\), the set of all \(k \times n\) matrices with the usual operations of addition and scalar multiplication, is a vector space.

 

Linear Algebra Review

Definition: Eigenvalues and Eigenvectors

We say \(\lambda\) is an eigenvalue of the linear transformation \(T: V \rightarrow V\) if there exists a nonzero vector \(\mathbf{x}\) in \(V\) such that \(T(\mathbf{x})=\lambda \mathbf{x}\). The vector \(\mathbf{x}\) is said to be an eigenvector corresponding to the eigenvalue \(\lambda\).

Example \(\PageIndex{1}\)

Find the eigenvalues and eigenvectors for \(T(\mathbf{x})=\left[\begin{array}{ll}
4 & 1 \\
5 & 0
\end{array}\right] \mathbf{x}.\) 

Solution

Note \(\left[\begin{array}{ll}4 & 1 \\ 5 & 0\end{array}\right]\left[\begin{array}{r}-1 \\ 5\end{array}\right]=\left[\begin{array}{r}1 \\ -5\end{array}\right]=-1\left[\begin{array}{r}-1 \\ 5\end{array}\right].\) Thus -1 is an eigenvalue of \(A\) and \(\left[\begin{array}{r}-1 \\ 5\end{array}\right]\) is a corresponding eigenvector of \(A\).

Note \(\left[\begin{array}{ll}4 & 1 \\ 5 & 0\end{array}\right]\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}5 \\ 5\end{array}\right]=5\left[\begin{array}{l}1 \\ 1\end{array}\right].\) Thus 5 is an eigenvalue of \(A\) and \(\left[\begin{array}{l}1 \\ 1\end{array}\right]\) is a corresponding eigenvector of \(A\).

Note \(\left[\begin{array}{ll}4 & 1 \\ 5 & 0\end{array}\right]\left[\begin{array}{l}2 \\ 8\end{array}\right]=\left[\begin{array}{l}16 \\ 10\end{array}\right] \neq k\left[\begin{array}{l}2 \\ 8\end{array}\right]\) for any \(k.\) Thus \(\left[\begin{array}{l}2 \\ 8\end{array}\right]\) is NOT an eigenvector of \(A\).

The motivation for this last example is the following: \(\left[\begin{array}{l}2 \\ 8\end{array}\right]=\left[\begin{array}{r}-1 \\ 5\end{array}\right]+3\left[\begin{array}{l}1 \\ 1\end{array}\right]\)
Thus \(A\left[\begin{array}{l}2 \\ 8\end{array}\right]=A\left(\left[\begin{array}{r}-1 \\ 5\end{array}\right]+3\left[\begin{array}{l}1 \\ 1\end{array}\right]\right)=A\left[\begin{array}{r}-1 \\ 5\end{array}\right]+3 A\left[\begin{array}{l}1 \\ 1\end{array}\right]\) \(=-1\left[\begin{array}{r}-1 \\ 5\end{array}\right]+3 \cdot 5\left[\begin{array}{l}1 \\ 1\end{array}\right]=\left[\begin{array}{l}16 \\ 10\end{array}\right]\)

 

 

How to find eigenvalues

Suppose \(A \mathbf{x}=\lambda \mathbf{x} \quad\) (Note \(A\) is a SQUARE matrix).
Then \(A \mathbf{x}=\lambda I \mathrm{x}\) where \(I\) is the identity matrix.
Thus \(\lambda I \mathbf{x}-A \mathbf{x}=(\lambda I-A) \mathbf{x}=\mathbf{0}\)
Thus if \(A \mathbf{x}=\lambda \mathbf{x}\) for a nonzero \(\mathbf{x}\), then \((\lambda I-A) \mathbf{x}=\mathbf{0}\) has a nonzero solution.

Thus \(\operatorname{det}(\lambda I-A) \mathbf{x}=0\).
Note that the eigenvectors corresponding to \(\lambda\) are the nonzero solutions of \((\lambda I-A) \mathbf{x}=\mathbf{0}\).

Thus to find the eigenvalues of \(A\) and their corresponding eigenvectors: \(\square\)

Step 1: Find eigenvalues: Solve the equation
\[ \notag
\operatorname{det}(\lambda I-A)=0 \text { for } \lambda .
\]

Step 2: For each eigenvalue \(\lambda_0\), find its corresponding eigenvectors by solving the homogeneous system of equations
\[ \notag
\left(\lambda_0 I-A\right) \mathbf{x}=0 \text { for } \mathbf{x} \text {. }
\]

Definition: Characteristic equation

We say \(\operatorname{det}(\lambda I-A)=0\) is the characteristic equation of \(A\).

Theorem \(\PageIndex{1}\)

The eigenvalues of an upper triangular or lower triangular matrix (including diagonal matrices) are identical to its diagonal entries.

Definition: Eigenspace

The eigenspace corresponding to an eigenvalue \(\lambda_0\) of a matrix \(A\) is the set of all solutions of \(\left(\lambda_0 I-A\right) \mathbf{x}=\mathbf{0}\).

Note

An eigenspace is a vector space as:

1. The vector \(\mathbf{0}\) is always in the eigenspace.
2. The vector \(\mathbf{0}\) is never an eigenvector.
3. The number 0 can be an eigenvalue.

Theorem \(\PageIndex{1}\)

A square matrix is invertible if and only if \(\lambda=0\) is not an eigenvalue of \(A\).

 

\(n\)th order LINEAR differential equation:

Theorem (from CH2)

If \(p\) and \(g\) are continuous on \((a, b)\) and the point \(t_0 \in(a, b)\), then there exists a unique function \(y=\phi(t)\) defined on \((a, b)\) that satisfies the following initial value problem:
\[ \notag
y^{\prime}+p(t) y=g(t), \quad y\left(t_0\right)=y_0 \text {. }
\]

Theorem (from CH3)

If \(p:(a, b) \rightarrow R, q:(a, b) \rightarrow R\), and \(g:(a, b) \rightarrow R\) are continuous and \(a<t_0<b\), then there exists a unique function \(y=\phi(t), \phi:(a, b) \rightarrow R\) that satisfies the initial value problem
\[ \notag
\begin{array}{c}
y^{\prime \prime}+p(t) y^{\prime}+q(t) y=g(t) \\
y\left(t_0\right)=y_0, \quad y^{\prime}\left(t_0\right)=y_1
\end{array}
\]

Theorem \(\PageIndex{1}\)

If \(p_i:(a, b) \rightarrow R, i=1, \ldots, n\) and \(g:(a, b) \rightarrow R\) are continuous and \(a<t_0<b\), then there exists a unique function \(y=\phi(t), \phi:(a, b) \rightarrow R\) that satisfies the initial value problem
\[ \notag
\begin{array}{c}
y^{(n)}+p_1(t) y^{(n-1)}+\ldots+p_{n-1}(t) y^{\prime}+p_n(t) y=g(t) \\
y\left(t_0\right)=y_0, \quad y^{\prime}\left(t_0\right)=y_1, \ldots, \quad y^{(n-1)}\left(t_0\right)=y_{n-1}
\end{array}
\]

Proof

We proved the case \(n=1\) using an integrating factor. When \(n>1\), see more advanced textbook.

Example \(\PageIndex{2}\)

Transform the \(n\)-th order linear  differential equation into a first order differential equation: \(y^{\prime \prime}-5 y^{\prime}+6 y=0, \quad y(0)=1, \quad y^{\prime}(0)=2 \)

Solution

We are going to solve for the highest order derivative and we are going to relabel \(y'\) and \(y\) as \(x_2\) and \(x_1\) respectfully. we then have the following:

\[ \notag
\begin{aligned}
& y=x_1 \\
& y^{\prime}=x_2=x_1^{\prime} \\
& y^{\prime \prime}=x_2' = -6x_1 + 5x_2 \\
\end{aligned}
\]

This written in matrix form will give us:

\[ \notag
\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]'=\left[\begin{array}{cc}
0 & 1 \\
-6 & 5
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_2
\end{array}\right]
\]

Example \(\PageIndex{3}\)

Transform the \(n\)-th order linear  differential equation into a first order differential equation:\(y''''-5y''+6y=\sin(t)\)

Solution

This is a fourth order differential equation that we can treat as first order linear differential equation with four unknowns, say \(x_1,x_2,x_3,\) and \(x_4\). We then have 

\[ \notag
\begin{aligned}
& y=x_1 \\
& y^{\prime}=x_2=x_1^{\prime} \\
& y^{\prime \prime}=x_3=x_2' \\
& y^{\prime \prime \prime}=x_{4}=x_3'
\end{aligned}
\]

This allows us to rewrite the original as \(y''''=x_4^{\prime}=-6 x_1+5 x_3+\sin (t)\) which we can then write in matrix form as follows:

\[ \notag
\left[\begin{array}{l}
x_1 \\
x_2 \\
x_3 \\
x_4
\end{array}\right]'=\left[\begin{array}{llll}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
-6 & 0 & 5 & 0
\end{array}\right]\left[\begin{array}{l}
x_1 \\
x_1 \\
x_3 \\
x_4
\end{array}\right]+\left[\begin{array}{c}
0 \\
0 \\
0 \\
\sin (t)
\end{array}\right]
\]