I need help with the following problem:
Let $A \in M_{n \times n}(\mathbb{R})$ be a symmetric matrix, and $\Lambda, \lambda$ the biggest and smallest eigenvalues respectively. Let $x$ be a solution for $x'=Ax$. Prove that for all $t>0$:
$||x(0)||^2e^{2\lambda t} \leq ||x(t)||^2 \leq ||x(0)||^2e^{2\Lambda t}$ with the euclidean norm.
We first develop a few useful and germane properties of the matrix $A$:
Since $A$ is symmetric, there exists an orthogonal matrix $O$ which diagonalizes $A$, and we have
$D = O^TAO, \tag 1$
where
$O^TO = OO^T = I, \tag 2$
and
$D = \text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n), \tag 3$
with the $\lambda_i \in \Bbb R$ being the eigenvalues of $A$. Writing $O$ in columnar form,
$O = [\vec e_1 \; \vec e_2 \; \ldots \; \vec e_n], \tag 4$
it follows that
$O^T = \begin{bmatrix} \vec e_1^T \\ \vec e_2^T \\ \vdots \\ \vec e_n^T \end{bmatrix}, \tag 5$
and it is easy to see that (2), (4), (5) imply
$\vec e_i \cdot \vec e_j = \delta_{ij}, \tag 6$
which shows that the columns $\vec e_i$ of $O$ (and hence the rows of $O^T$) form a set of orthonormal eigenvectors of $A$; from (1),
$OD = OO^TAO = IAO = AO, \tag 7$
and it follows via (3) that
$AO = A[\vec e_1 \; \vec e_2 \; \ldots \; \vec e_n] = [A\vec e_1 \; A\vec e_2 \; \ldots \; A\vec e_n] = OD$ $= [\vec e_1 \; \vec e_2 \; \ldots \; \vec e_n]\text{diag}(\lambda_1, \lambda_2, \ldots, \lambda_n)= [\lambda_1 \vec e_1 \; \lambda_2 \vec e_2 \; \ldots \; \lambda_n \vec e_n], \tag 8$
which is easily seen to be equivalent to
$A \vec e_i = \lambda_i \vec e_i, \; 1 \le i \le n. \tag 9$
We turn now to the equation linear, constant-coefficient ordinary differential equation
$\dot x(t) = Ax(t); \tag{10}$
with initial condition $x(0)$ at $t = 0$ the well-known solution is
$x(t) = e^{At}x(0); \tag{11}$
we expand $x(0)$ in terms of the $\vec e_i$
$x(0) = \displaystyle \sum_1^n x_i(0) \vec e_i, \tag{12}$
and substitute this into (11):
$x(t) = e^{At} \displaystyle \sum_1^n x_i(0) \vec e_i = \sum_1^n x_i(0)e^{At} \vec e_i; \tag{13}$
by virtue of (9),
$A^2 \vec e_i = A(A\vec e_i) = A(\lambda_i \vec e_i) = \lambda_i A \vec e_i = \lambda_i(\lambda_i \vec e_i) = \lambda_i^2 \vec e_i, \tag{14}$
and if we assume that
$A^k \vec e_i = \lambda_i^k \vec e_1 \tag{15}$
for some
$2 \le k \in \Bbb N, \tag{16}$
then
$A^{k + 1} \vec e_i = AA^k \vec e_i = A(\lambda_i^k \vec e_i) = \lambda_i^k A\vec e_i = \lambda_i^k \lambda_i \vec e_i = \lambda_i^{k + 1} \vec e_i; \tag{17}$
via (9),(14), (15) and (17) we conclude by induction that
$A^n \vec e_i = \lambda_i^n \vec e_i, \; \forall 0 \le n \in \Bbb N; \tag{18}$
then
$e^{At} \vec e_i = \left ( \displaystyle \sum_0^\infty \dfrac{A^n t^n}{n!} \right ) \vec e_i = \displaystyle \sum_0^\infty \dfrac{t^n A^n \vec e_i}{n!} = \sum_0^\infty \dfrac{t^n \lambda_i^n \vec e_i}{n!} = \left (\sum_0^\infty \dfrac{t^n \lambda_i^n}{n!} \right ) \vec e_i = e^{\lambda_i t} \vec e_i; \tag{19}$
in light of this, (13) yields
$x(t) = e^{At} \displaystyle \sum_1^n x_i(0) \vec e_i = \sum_1^n x_i(0)e^{\lambda_i t} \vec e_i, \tag{20}$
whence, using (6),
$\Vert x(t) \Vert^2 = \left( \displaystyle \sum_1^n x_i(0) e^{\lambda_i t} \vec e_i \right ) \cdot \left( \displaystyle \sum_1^n x_i(0) e^{\lambda_i t} \vec e_i \right )$ $= \displaystyle \sum_{i, j = 1}^n x_i(0)x_j(0) e^{(\lambda_i + \lambda_j)t} \vec e_i \cdot \vec e_j$ $= \displaystyle \sum_{i, j = 1}^n x_i(0)x_j(0) e^{(\lambda_i + \lambda_j)t} \delta_{ij} = \sum_1^n e^{2\lambda_i t} x_i^2(0); \tag{21}$
since the function $e^{\mu t}$ is monotonically increasing in $\mu$ for $t \ge 0$, (21) yields
$e^{2\lambda t} \Vert x(0) \Vert^2 = e^{2\lambda t}\displaystyle \sum_1^n x_i^2(0) = \sum_1^n e^{2\lambda t} x_i^2(0) \le \displaystyle \sum_1^n e^{2\lambda_i t} x_i^2(0)$ $= \Vert x(t) \Vert^2 \le \displaystyle \sum_1^n e^{2\Lambda t} x_i^2(0) = e^{2\Lambda t} \sum_1^n x_i^2(0) = e^{2\Lambda t} \Vert x(0) \Vert^2, \tag{22}$
from which we may extract the inequality
$e^{2 \lambda t} \Vert x(0) \Vert^2 \le \Vert x(t) \Vert^2 \le e^{2 \Lambda t} \Vert x(0) \Vert^2, \tag{23}$
which was to be proved.