Real Eigenvalues and Linear-System Solutions

Definition For an $m \times n$ matrix $A$ with entries $a_{ij}$ and a column $X = (x_1, \ldots, x_n)^{T}$, the product $AX$ is the column of length $m$ whose $i$th entry is

the sum of the entries of $X$ weighted by row $i$ of $A$. The product of an $m \times n$ matrix $A$ and an $n \times p$ matrix $B$ is the $m \times p$ matrix whose $j$th column is $A$ times the $j$th column of $B$. Multiplication by $A$ is linear, which means $A(aX + bY) = aAX + bAY$ for all columns $X$, $Y$ and scalars $a$, $b$. Matrix products are associative, but in general $AB \ne BA$. For a constant matrix $A$ and a differentiable vector function $X$ of $t$, the rule $(AX)' = AX'$ holds, because each entry of $AX$ is a fixed combination of the entries of $X$.

Definition The homogeneous linear system $X' = AX$ has a constant coefficient matrix when $A$ is a fixed real $n \times n$ matrix that does not depend on $t$. Such a system is linear and autonomous. Its coefficients are continuous on the whole line, so for every $t_0$ and every $X_0$ in $\mathbb{R}^n$ the problem $X' = AX$, $X(t_0) = X_0$ has exactly one solution, and it is defined for all real $t$. Sums and constant multiples of solutions are solutions, the zero function is always an equilibrium solution, and the solutions form a vector space of dimension $n$, since each solution is fixed by its value at $t_0$ and every value in $\mathbb{R}^n$ occurs. The exponential solutions built below from eigenvectors depend on $A$ being constant.

Definition Let $A$ be an $n \times n$ matrix. A number $\lambda$ is an eigenvalue of $A$ with eigenvector $v$ if

The condition says that $(A - \lambda I)v = 0$ has a nonzero solution, which happens exactly when $\det(A - \lambda I) = 0$. This determinant is a polynomial of degree $n$ in $\lambda$, the characteristic polynomial of $A$. For a $2 \times 2$ matrix it equals $\lambda^2 - (\operatorname{tr} A)\,\lambda + \det A$, where $\operatorname{tr} A$ is the sum of the diagonal entries. Every nonzero multiple of an eigenvector is again an eigenvector for the same eigenvalue, so an eigenvector really names a line through the origin. The zero vector is never an eigenvector, although the number $0$ can be an eigenvalue. A real eigenvalue of a real matrix always has real eigenvectors, and in this lesson every eigenvalue is real.

Definition If $A$ is a constant matrix and $Av = \lambda v$ with $v \ne 0$, then

solves $X' = AX$ for every real $t$, because $X' = \lambda e^{\lambda t}v = e^{\lambda t}Av = AX$. This is a modal solution, or mode, with rate $\lambda$ and direction $v$. It stays on the line through $v$ for all time, and its size is multiplied by $e^{\lambda t}$. Conversely, if $X(t) = a(t)\,v$ is a solution for an eigenvector $v$ and a differentiable scalar amplitude $a$, then $a'v = aAv = \lambda a\,v$, so $a' = \lambda a$ and $a(t) = a(0)\,e^{\lambda t}$.

Definition An eigenbasis for $A$ is a basis $v_1, \ldots, v_n$ of $\mathbb{R}^n$ made of eigenvectors of $A$, with $Av_k = \lambda_k v_k$. Eigenvectors for distinct eigenvalues are linearly independent, so $n$ distinct real eigenvalues always give a real eigenbasis. A repeated eigenvalue may or may not supply enough independent eigenvectors, and that has to be checked. When an eigenbasis exists, every solution of $X' = AX$ is

for exactly one list of constants. At $t = 0$ this reads $X(0) = Vc$, where $V$ is the matrix with columns $v_1, \ldots, v_n$ and $c = (c_1, \ldots, c_n)^{T}$. The columns are independent, so $\det V \ne 0$ and $c = V^{-1}X(0)$. The combination with these constants is a solution with the right starting state, so by uniqueness it is the solution.