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  "body": "\n## Proof\n\nWe reduce both kinds of stability to the behaviour of the entries of $e^{tA}$, and compute these entries from the Jordan normal form of $A$.\n\nReduction to the matrix exponential.\n\n1. Let $\\norm{\\cdot}$ be the Euclidean norm on $\\C^n$. For a complex $n \\times n$ matrix $M = (m_{ij})$, the [[Cauchy–Schwarz inequality|0RFWW56Y22]] applied to each row gives $\\norm{M z} \\le \\left( \\sum_{i,j} \\abs{m_{ij}}^2 \\right)^{1/2} \\norm{z}$ for $z \\in \\C^n$.\n2. Let $e_1, \\dots, e_n$ be the standard basis. Then $\\abs{m_{ij}} \\le \\norm{M e_j}$, because $m_{ij}$ is the $i$-th coordinate of $M e_j$.\n3. By the theorem on the [[derivative of the matrix exponential|0YC8JT513J]], $t \\mapsto e^{tA} x_0$ is the unique solution of $x' = Ax$, $x(0) = x_0$ on $\\R$. So the [[flow|02903SSA48]] of the [[linear system|0VJF99CZ0Z]] is $\\varphi_t(x_0) = e^{tA} x_0$, defined for all $t \\in \\R$.\n4. Claim A: $0$ is [[Lyapunov stable|092S6VBECQ]] if and only if all entries of $e^{tA}$ are bounded on $[0, \\infty)$.\n5. Proof of \"if\". Let $c$ bound all entries. By step 1, $\\norm{e^{tA} x_0} \\le n c \\norm{x_0}$ for $t \\ge 0$. Given $\\varepsilon \u003e 0$, the number $\\delta = \\varepsilon / (nc + 1)$ works.\n6. Proof of \"only if\". Take $\\varepsilon = 1$ and its $\\delta \u003e 0$. The vector $y = (\\delta/2) e_j$ has $\\norm{y} \u003c \\delta$, so $\\norm{e^{tA} y} \u003c 1$ for $t \\ge 0$. By linearity $\\norm{e^{tA} e_j} \u003c 2/\\delta$, and step 2 bounds every entry by $2/\\delta$.\n7. Claim B: $0$ is [[asymptotically stable|05Y1JAM3Y5]] if and only if all entries of $e^{tA}$ tend to $0$ as $t \\to \\infty$.\n8. Proof of \"only if\". Let $\\eta \u003e 0$ be as in the definition. For $y = (\\eta/2) e_j$ we have $e^{tA} y \\to 0$. So the column $e^{tA} e_j$ tends to $0$, for each $j$.\n9. Proof of \"if\". By step 1, $e^{tA} x_0 \\to 0$ for every $x_0$. Each entry is continuous and tends to $0$. So it is below $1$ in modulus for $t \\ge T$, and bounded on $[0, T]$ by the [[extreme value theorem|0WS1A8E2P9]]. By Claim A, $0$ is also Lyapunov stable.\n\nEntries of $e^{tJ}$.\n\n10. By the [[Jordan normal form|085HQ5G5TE]] over $\\C$, there are an invertible complex matrix $P$ and a block diagonal matrix $J$ with $A = P J P^{-1}$. The diagonal blocks are [[Jordan blocks|0PBFS92AJP]] $J_{k_1}(\\lambda_1), \\dots, J_{k_r}(\\lambda_r)$.\n11. We have $(P J P^{-1})^m = P J^m P^{-1}$ for each $m \\ge 0$. The entries of $P X P^{-1}$ are fixed linear combinations of the entries of $X$. Taking limits of partial sums in the [[matrix exponential|0K665H5V1S]] gives $e^{tA} = P e^{tJ} P^{-1}$.\n12. Since also $e^{tJ} = P^{-1} e^{tA} P$, the entries of $e^{tA}$ are all bounded on $[0, \\infty)$ if and only if those of $e^{tJ}$ are. The same holds for convergence of all entries to $0$.\n13. Powers of a block diagonal matrix are block diagonal, with the powers of the blocks on the diagonal. So $e^{tJ}$ is block diagonal with blocks $e^{t J_{k_i}(\\lambda_i)}$.\n14. Fix a block $J_k(\\lambda) = \\lambda I + N$. Here $N$ has ones just above the diagonal and zeros elsewhere. By induction on $j$, $N^j$ has ones at the places $(p, p + j)$ and zeros elsewhere. So $N$ is [[nilpotent|0419GJC84P]] with $N^k = 0$.\n15. The matrices $t\\lambda I$ and $tN$ commute. By the [[exponential of a sum of commuting matrices|04BD23PSK5]], $e^{t J_k(\\lambda)} = e^{t \\lambda I} e^{tN} = e^{\\lambda t} \\sum_{j=0}^{k-1} \\frac{t^j}{j!} N^j$.\n16. So the entry at $(p, p + j)$ of $e^{t J_k(\\lambda)}$ is $e^{\\lambda t} t^j / j!$ for $0 \\le j \\le k - 1$, and all other entries are $0$.\n17. Let $\\mu = \\operatorname{Re} \\lambda$. By the [[product of exponentials|0WM7DJCM6G]] and [[Euler's formula|F6PH4JEGS5]], $\\abs{e^{\\lambda t}} = e^{\\mu t}$.\n18. If $\\mu \u003c 0$ and $j \\ge 0$, then $t^j e^{\\mu t} \\to 0$ as $t \\to \\infty$. Indeed, with $x = -\\mu t$ we get $t^j e^{\\mu t} = (-\\mu)^{-j} x^j e^{-x}$, and $e^x / x^j \\to \\infty$ because [[exponential growth beats polynomial growth|0CT6CBGTYY]].\n19. All entries of $e^{tJ}$ tend to $0$ if and only if every $\\lambda_i$ has $\\operatorname{Re} \\lambda_i \u003c 0$. \"If\" follows from steps 16 to 18. For \"only if\", a block with $\\mu \\ge 0$ has diagonal entries of modulus $e^{\\mu t} \\ge 1$.\n20. All entries of $e^{tJ}$ are bounded on $[0, \\infty)$ if and only if each block has $\\operatorname{Re} \\lambda_i \u003c 0$, or has $\\operatorname{Re} \\lambda_i = 0$ and size $k_i = 1$.\n21. Proof of step 20, \"if\". Entries of blocks with $\\mu \u003c 0$ tend to $0$, so they are bounded as in step 9. A block of size $1$ with $\\mu = 0$ is $e^{\\lambda t}$, of modulus $1$.\n22. Proof of step 20, \"only if\". If $\\mu \u003e 0$, a diagonal entry has modulus $e^{\\mu t}$, which is unbounded. If $\\mu = 0$ and $k \\ge 2$, the entry at $(1, 2)$ has modulus $t$, which is unbounded.\n\nMultiplicities.\n\n23. The eigenvalues and multiplicities are those of $A$ as a complex matrix. By [[invariance of the characteristic polynomial|0TYMASG33Y]], $A$ and $J$ have the same [[characteristic polynomial|039HF58KQ5]].\n24. The matrix $xI - J$ is upper triangular. By the [[determinant of a triangular matrix|0K1KKMFHYH]], this polynomial is $\\prod_{i=1}^r (x - \\lambda_i)^{k_i}$.\n25. So the [[eigenvalues|053QVHF6N7]] of $A$ are exactly $\\lambda_1, \\dots, \\lambda_r$. The algebraic multiplicity of an eigenvalue $\\lambda$, its [[multiplicity as a root|03XVP8Q4GY]], is the sum of the $k_i$ with $\\lambda_i = \\lambda$.\n26. The map $v \\mapsto P^{-1} v$ is a linear bijection from $\\ker(A - \\lambda I)$ onto $\\ker(J - \\lambda I)$. So these kernels have the same dimension, the [[geometric multiplicity|06YSJVR356]] of $\\lambda$.\n27. The kernel of the block diagonal matrix $J - \\lambda I$ is the direct sum of the kernels of its blocks $(\\lambda_i - \\lambda) I + N$.\n28. If $c = \\lambda_i - \\lambda \\ne 0$ and $(cI + N) v = 0$, then $c v_k = 0$ and $c v_p + v_{p+1} = 0$ for $p \u003c k$. Going upward from $p = k$ gives $v = 0$.\n29. If $\\lambda_i = \\lambda$, then $N v = (v_2, \\dots, v_k, 0)$, so $\\ker N$ is spanned by $e_1$ and has dimension $1$.\n30. So the geometric multiplicity of $\\lambda$ is the number of blocks with $\\lambda_i = \\lambda$. Each $k_i \\ge 1$. By step 25, the two multiplicities agree exactly when all these blocks have size $1$.\n31. For a real eigenvalue, the real kernel of $A - \\lambda I$ has the same dimension. Row reduction of the real matrix gives the same rank over $\\R$ and over $\\C$, and the [[rank–nullity theorem|TR48YQGCCA]] applies.\n\nConclusion.\n\n32. Part (1). By Claim B and steps 12, 19 and 25, $0$ is asymptotically stable exactly when every eigenvalue of $A$ has negative real part.\n33. Part (3). By Claim A and steps 12, 20 and 30, $0$ is Lyapunov stable exactly when two conditions hold. Every eigenvalue has $\\operatorname{Re} \\lambda \\le 0$. Each eigenvalue with $\\operatorname{Re} \\lambda = 0$ has equal algebraic and geometric multiplicities.\n34. Part (2): if some eigenvalue has $\\operatorname{Re} \\lambda \u003e 0$, the condition in step 33 fails. So $0$ is not Lyapunov stable, that is, it is unstable.\n\n## Idea\n\nThe solutions of $x' = Ax$ are $e^{tA} x_0$, so stability only asks whether the matrix $e^{tA}$ stays bounded or tends to zero. In Jordan form, each block contributes the terms $e^{\\lambda t} t^j$. Exponential decay beats every power of $t$ when $\\operatorname{Re} \\lambda \u003c 0$. On the imaginary axis the powers $t^j$ with $j \\ge 1$ grow, and they are absent exactly when every Jordan block has size one.\n",
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