In this appendix, we state basic facts of linear algebra concerning matrices, eigenvalues and eigenvectors. No proofs are given and the reader should consult linear algebra texts for more details. The brief review presented below, although far from being complete, should however provide sufficient information for a reader to follow most of the linear stability arguments made in the previous chapters.

Vector spaces

Definitions

• $\mathcal{S}$ is a real (resp. complex) vector space if and only if it is closed under addition and under multiplication by a scalar. In other words,

$\begin{array}{}\text{(1)}& & \mathrm{\forall }x,y\in \mathcal{S},x+y\in \mathcal{S}\text{(2)}& & \mathrm{\forall }x\in \mathcal{S},\mathrm{\forall }\alpha \in \mathbb{R}\text{ (resp. }\mathbb{C}),\alpha x\in \mathcal{S}.\end{array}$

• The vectors in $\{u_i\in S,i=1\dots n\}$ are linearly independent if and only if any linear combination equal to zero must have all of its coefficients equal to zero. In other words,

$\begin{array}{}\text{(3)}& \mathrm{\forall }& \{{\alpha }_i,i=1,\dots n\}\subset \mathbb{R}\text{ (or }\mathbb{C}),\text{(4)}& & \sum _{i=1}^n{\alpha }_i{u}_i=0⟹{\alpha }_i=0,\mathrm{\forall }i=1,\dots n.\end{array}$

• $\mathcal{S}$ is finite dimensional if there exists a finite set of linearly independent vectors that span $\mathcal{S}$.
• Such a set is called a basis of $\mathcal{S}$. In what follows, we are only concerned with finite dimensional vector spaces.
• The dimension of a finite dimensional vector space $\mathcal{S}$ is the number of vectors in any basis of $\mathcal{S}$.

Linear mappings

We say that the mapping $\mathcal{T}:\mathcal{S}\to \mathcal{U}$ from a vector space $\mathcal{S}$ to a vector space $\mathcal{U}$ is linear if for every $x,y\in \mathcal{S}$,

$\begin{array}{}\text{(5)}& & \mathcal{T}(x+y)=\mathcal{T}(x)+\mathcal{T}(y)\text{(6)}& & \mathcal{T}(\alpha x)=\alpha \mathcal{T}(x),\end{array}$

where $\alpha \in \mathbb{R}$ (resp. $\alpha \in \mathbb{C}$) if $\mathcal{S}$ and $\mathcal{U}$ are vector spaces over $\mathbb{R}$ (resp. $\mathbb{C}$).

Properties of linear mappings

• The range ${\mathcal{R}}_{\mathcal{T}}$ of $\mathcal{T}$, which is the image of $\mathcal{S}$ under $\mathcal{T}$, is a linear subspace of $\mathcal{U}$.
• The nullspace or kernel of $\mathcal{T}$ is a linear subspace of $\mathcal{S}$. It is defined as the set ${\mathcal{N}}_{\mathcal{T}}$ of vectors of $\mathcal{S}$ whose image under $\mathcal{T}$ is zero,

${\mathcal{N}}_{\mathcal{T}}=\{x\in \mathcal{S}|\mathcal{T}(x)=0\}.$

• The dimensions of ${\mathcal{R}}_{\mathcal{T}}$ and ${\mathcal{N}}_{\mathcal{T}}$ are such that

$\text{dim}({\mathcal{R}}_{\mathcal{T}})+\text{dim}({\mathcal{N}}_{\mathcal{T}})=\text{dim}(\mathcal{S}).$

Matrices

Every linear mapping

$\begin{array}{llll}\mathcal{T}:& {\mathbb{R}}^n& \to & {\mathbb{R}}^m\\ & x& \mapsto & u\end{array}$

can be written as

$u_i=\sum _{j=1}^n{A}_{ij}{x}_{j}i=1,\dots ,m;u=\left(\begin{array}{c}{u}_1\\ u_2\\ ⋮\\ u_m\end{array}\right);x=\left(\begin{array}{c}{x}_1\\ x_2\\ ⋮\\ x_n\end{array}\right),$

where $A\equiv (A_{ij})$ is a $m\times n$ ($m$ rows, $n$ columns) matrix with real entries. By convention, $A_{ij}$ is the entry at the intersection of the $i^{\text{th}}$ row and $j^{\text{th}}$ column of $A$.

Note that once a basis has been chosen, every linear vector space of dimension $n$ is isomorphic to ${\mathbb{R}}^n$. We can then represent any linear mapping between two finite dimensional vector spaces by a matrix. In what follows, we will only consider matrices with real coefficients.

Definitions

• The transpose of the matrix $A$ is $A^T$ such that $(A_{ij}^T)=(A_{ji})$.
• The rank of the matrix $A$ associated with the linear transformation $\mathcal{T}$ is the dimension of ${\mathcal{R}}_{\mathcal{T}}$. It is also equal to the rank of $A^T$.
• The determinant of a $2\times 2$ matrix, $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is

$detA=\left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc.$

• The determinant of an $n\times n$ matrix $A=(A_{ij})$ can be calculated by means of the formula below, where $i$ is one row of $A$ and $M_{ij}$ is the matrix obtained from $A$ by deleting row $i$ and column $j$:

$detA=\sum _{j=1}^n(-1{)}^{i+j}{A}_{ij}det{M}_{ij}.$

• A similar formula exists for expanding $detA$ with respect to one column of $A$.
• The trace Tr($A$) of a square matrix $A$ is the sum of the diagonal entries of $A$.

Properties

• If $A$ is an $m\times n$ matrix, the system $Ax=b$ has at least one solution for every $b$ if and only if the columns of $A$ span ${\mathbb{R}}^m$. Then the rank of $A$, $r$, is such that $r=m$, which implies $m\le n$.
• The system $Ax=b$ has at most one solution for every $b$ if and only if the columns of $A$ are linearly independent, i.e. if and only if the nullspace of $A$ is trivial. Then, $r=n$, which implies $n\le m$.
• Let $A$ be an $n\times n$ matrix. Then, the following statements are equivalent.
  • The equation $AX=b$ has exactly one solution.
  • The range of $A$ is ${\mathbb{R}}^n$.
  • The nullspace of $A$ is trivial.
  • The matrix $A$ is invertible.
  • The determinant of $A$, $detA$, is non-zero.

Eigenvalues and eigenvectors

Definitions

Let $A$ be a real $n\times n$ matrix.

• The vector $h\in \mathcal{S}$ is an eigenvector of $A$ with eigenvalue $a\in \mathbb{C}$ if

$Ah=ah,h\ne 0.$

• The vector $f\in \mathcal{S}$ is a generalized eigenvector of $A$ with eigenvalue $a$ if, for some positive integer $m\ne 1$, we have

$(A-a{I}_n)f\ne 0,(A-a{I}_n{)}^{m}f=0,f\ne 0.$

In the above equation, $I_n$ is the $n\times n$ identity matrix. To find the eigenvalues and eigenvectors of a matrix, first note that if $u$ is an eigenvector of a matrix $A$ with eigenvalue $a$, then the equation

$(A-a{I}_n)u=0,(A1.1)$

has a non-trivial solution. This implies that

$det(A-aI)=0,(A1.2)$

and one can therefore find the eigenvalues of $A$ by solving this equation.

Properties

• The left-hand-side of (A1.2) is a polynomial of degree $n$ in $a$, called the characteristic polynomial of $A$.
• The characteristic polynomial of $A$ has $n$ complex roots, which are the eigenvalues of $A$.
• Since $A$ has real entries, if $a$ is an eigenvalue of $A$, so is its complex conjugate $a^{\ast }$. As a consequence, the eigenvalues of $A$ are either real, or complex conjugate pairs.
• The trace of $A$ is the sum of the eigenvalues of $A$.
• The determinant of $A$ is the product of the eigenvalues of $A$.

Once an eigenvalue is found, one needs to solve (A1.1) in order to obtain a corresponding eigenvector. There is not one such eigenvector, but a linear subspace thereof. Each of these eigenspaces is an invariant subspace of the linear transformation $\mathcal{T}$ associated with the matrix $A$. The vector space $\mathcal{S}$, or equivalently ${\mathbb{R}}^n$, can thus be viewed as the sum of the eigenspaces of $A$, and this decomposition gives a geometric picture of how $\mathcal{T}$ acts on $\mathcal{S}$.