[proofplan]\nWe use Hamilton's matrix Harnack inequality in its standard $M$-$P$-$R$ form, not as a direct inequality for $\partial_t\operatorname{Ric}$. At a fixed spacetime point, the theorem gives a nonnegative symmetric two-tensor built from the tensors $M_{ij}$, $P_{ijk}$, and the curvature term $R_{ikjl}V_kV_l$. Taking its trace and then using the contracted [second Bianchi identity](/theorems/1541) together with the scalar curvature evolution equation under Ricci flow produces exactly Hamilton's trace Harnack inequality.\n[/proofplan]\n\n[step:Apply Hamilton's matrix Harnack estimate in its standard $M$-$P$-$R$ form]\nFix $t\in(0,T]$, $x\in M$, and $V\in T_xM$. Let $\operatorname{Ric}_{g(t)}$ denote the [Ricci tensor](/page/Ricci%20Tensor) of $g(t)$, let $S(\cdot,t)$ denote its [scalar curvature](/page/Scalar%20Curvature), let $\nabla$ denote the Levi-Civita connection of $g(t)$, and let $R_{g(t)}$ denote the Riemann curvature tensor with the convention that tracing $R_{g(t)}(\cdot,V,\cdot,V)$ gives $\operatorname{Ric}_{g(t)}(V,V)$. Let $\Delta$ denote the scalar Laplacian of $g(t)$ acting componentwise on tensor components in [normal coordinates](/theorems/2713) at $x$, equivalently the trace $\operatorname{tr}_{g(t)}\nabla\nabla$ at $x$. Let $|\operatorname{Ric}_{g(t)}|_{g(t)}^2:=\sum_{i,j=1}^n \operatorname{Ric}_{ij}\operatorname{Ric}_{ij}$ denote the squared pointwise norm in any $g(t)$-[orthonormal basis](/page/Orthonormal%20Basis). Choose a $g(t)$-orthonormal basis $(e_1,\dots,e_n)$ of $T_xM$, and write $V_k:=\langle V,e_k\rangle_{g(t)}$ for $1\le k\le n$. We use the same sign and index normalization as [Hamilton's Matrix Harnack Inequality](/theorems/1172): the matrix tensor is formed as $M_{ij}-2P_{ikj}V_k+R_{ikjl}V_kV_l$, with the curvature convention above.\n\nDefine the components $M_{ij}$ and $P_{ijk}$ at $(x,t)$ by\n\begin{align*}\nM_{ij}:=\Delta \operatorname{Ric}_{ij}-\frac{1}{2}\nabla_i\nabla_j S+2\sum_{k,l=1}^n R_{ikjl}\operatorname{Ric}_{kl}-\sum_{k=1}^n \operatorname{Ric}_{ik}\operatorname{Ric}_{jk}+\frac{1}{2t}\operatorname{Ric}_{ij}.\n\end{align*}\n\begin{align*}\nP_{ijk}:=\nabla_i\operatorname{Ric}_{jk}-\nabla_j\operatorname{Ric}_{ik}.\n\end{align*}\nHere all components are taken in the chosen orthonormal basis and all contractions use $g(t)$ at the point $x$.\n\nThe hypotheses of [Hamilton's Matrix Harnack Inequality](/theorems/1172) are satisfied: the [Ricci flow](/page/Ricci%20Flow) is complete on $(0,T]$, has bounded curvature, and has nonnegative [curvature operator](/page/Curvature%20Operator). If the cited matrix Harnack theorem is stated on a time interval beginning at its initial time, fix $\varepsilon\in(0,t)$ and define the shifted Ricci flow $\tilde g_\varepsilon:[0,T-\varepsilon]\to \Gamma(\operatorname{Sym}^2T^*M)$ by $\tilde g_\varepsilon(s):=g(s+\varepsilon)$ for $s\in[0,T-\varepsilon]$. The shifted flow is complete, has bounded curvature, and has nonnegative curvature operator because these properties are inherited from $g(t)$ on $[\varepsilon,T]$. Applying the matrix Harnack theorem to $\tilde g_\varepsilon$ at shifted time $s=t-\varepsilon$ produces the same tensor expression except that the coefficient of $\operatorname{Ric}_{ij}$ in $M_{ij}$ is \begin{align*} \frac{1}{2(t-\varepsilon)}. \end{align*} Letting $\varepsilon\downarrow0$ and using smooth dependence of the curvature tensors on time gives the normalization \begin{align*} \frac{1}{2t}. \end{align*} Therefore the symmetric two-tensor with components\n\begin{align*}\nZ_{ij}:=M_{ij}-2\sum_{k=1}^n P_{ikj}V_k+\sum_{k,l=1}^n R_{ikjl}V_kV_l\n\end{align*}\nis nonnegative definite at $(x,t)$. In particular, for every $W\in T_xM$, if $W_i:=\langle W,e_i\rangle_{g(t)}$, then\n\begin{align*}\n\sum_{i,j=1}^n Z_{ij}W_iW_j\ge 0.\n\end{align*}\n\n[guided]\nFix one spacetime point $(x,t)\in M\times(0,T]$ and one tangent vector $V\in T_xM$. The desired inequality is pointwise, so proving it for this fixed data proves the theorem after we observe that the data were arbitrary. Let $\operatorname{Ric}_{g(t)}$ be the [Ricci tensor](/page/Ricci%20Tensor), let $S(\cdot,t)$ be the [scalar curvature](/page/Scalar%20Curvature), let $\nabla$ be the Levi-Civita connection of $g(t)$, and let $R_{g(t)}$ be the Riemann curvature tensor with the convention that tracing $R_{g(t)}(\cdot,V,\cdot,V)$ gives $\operatorname{Ric}_{g(t)}(V,V)$. Let $\Delta:=\operatorname{tr}_{g(t)}\nabla\nabla$ denote the scalar Laplacian of $g(t)$ acting componentwise on tensor components in normal coordinates at $x$. Let $|\operatorname{Ric}_{g(t)}|_{g(t)}^2:=\sum_{i,j=1}^n \operatorname{Ric}_{ij}\operatorname{Ric}_{ij}$ denote the squared pointwise norm in a $g(t)$-orthonormal basis. Choose a $g(t)$-orthonormal basis $(e_1,\dots,e_n)$ of $T_xM$ and define $V_k:=\langle V,e_k\rangle_{g(t)}$.\n\nHamilton's matrix Harnack inequality is not the assertion that $\partial_t\operatorname{Ric}+2\nabla_V\operatorname{Ric}+2R(\cdot,V,\cdot,V)+t^{-1}\operatorname{Ric}$ is nonnegative. Its standard form uses two auxiliary tensors. In the chosen orthonormal basis, define\n\begin{align*}\nM_{ij}:=\Delta \operatorname{Ric}_{ij}-\frac{1}{2}\nabla_i\nabla_j S+2\sum_{k,l=1}^n R_{ikjl}\operatorname{Ric}_{kl}-\sum_{k=1}^n \operatorname{Ric}_{ik}\operatorname{Ric}_{jk}+\frac{1}{2t}\operatorname{Ric}_{ij}.\n\end{align*}\n\begin{align*}\nP_{ijk}:=\nabla_i\operatorname{Ric}_{jk}-\nabla_j\operatorname{Ric}_{ik}.\n\end{align*}\nThese are exactly the tensors whose trace identities produce the scalar Harnack inequality. The $M$ term carries the time derivative of scalar curvature after using the scalar curvature evolution equation, the $P$ term carries the gradient of scalar curvature after using the contracted second Bianchi identity, and the curvature term carries $\operatorname{Ric}(V,V)$ after tracing.\n\nWe now verify the external theorem's hypotheses. [Hamilton's Matrix Harnack Inequality](/theorems/1172) applies to complete [Ricci flows](/page/Ricci%20Flow) with bounded curvature and nonnegative [curvature operator](/page/Curvature%20Operator). The theorem statement assumes precisely those three conditions on $(M^n,g(t))_{t\in(0,T]}$. If the cited theorem is phrased for an interval starting at its initial time, we make the time shift explicitly. Fix $\varepsilon\in(0,t)$ and define the shifted flow $\tilde g_\varepsilon:[0,T-\varepsilon]\to \Gamma(\operatorname{Sym}^2T^*M)$ by $\tilde g_\varepsilon(s):=g(s+\varepsilon)$ for $s\in[0,T-\varepsilon]$. This is again a complete Ricci flow with bounded curvature and nonnegative curvature operator, because those hypotheses hold for the original flow at all positive times and hence on the subinterval $[\varepsilon,T]$. We apply Hamilton's matrix Harnack inequality to $\tilde g_\varepsilon$ at shifted time $s=t-\varepsilon$. The time-normalized term in the matrix tensor is then \begin{align*} \frac{1}{2s}=\frac{1}{2(t-\varepsilon)}, \end{align*} not a new geometric term; all curvature, Ricci, scalar curvature, and covariant derivative tensors are exactly those of $g(t)$ after translating back to the original time variable. Since the curvature tensors are smooth in time along the flow and \begin{align*} \frac{1}{2(t-\varepsilon)}\to\frac{1}{2t} \end{align*} as $\varepsilon\downarrow0$, the limiting tensor has precisely the displayed \begin{align*} \frac{1}{2t} \end{align*} coefficient. Hence Hamilton's matrix Harnack inequality gives that the symmetric two-tensor\n\begin{align*}\nZ_{ij}:=M_{ij}-2\sum_{k=1}^n P_{ikj}V_k+\sum_{k,l=1}^n R_{ikjl}V_kV_l\n\end{align*}\nis nonnegative definite at $(x,t)$. Equivalently, for every vector $W\in T_xM$, defining $W_i:=\langle W,e_i\rangle_{g(t)}$ gives\n\begin{align*}\n\sum_{i,j=1}^n Z_{ij}W_iW_j\ge 0.\n\end{align*}\nThis is the correct matrix input. The trace Harnack inequality is obtained by tracing $Z_{ij}$, not by replacing the matrix Harnack tensor with a time derivative of the Ricci tensor.\n[/guided]\n[/step]\n\n[step:Trace the standard matrix Harnack tensor and simplify each contraction]\nSince $Z_{ij}$ is nonnegative definite and $(e_1,\dots,e_n)$ is $g(t)$-orthonormal, its trace is nonnegative:\n\begin{align*}\n0\le \sum_{i=1}^n Z_{ii}.\n\end{align*}\nWe compute the three trace contributions separately. First, tracing the definition of $M_{ij}$ gives\n\begin{align*}\n\sum_{i=1}^n M_{ii}=\frac{1}{2}\Delta S+|\operatorname{Ric}_{g(t)}|_{g(t)}^2+\frac{S}{2t}.\n\end{align*}\nThe scalar curvature evolution equation for Ricci flow is $\partial_t S=\Delta S+2|\operatorname{Ric}_{g(t)}|_{g(t)}^2$, so\n\begin{align*}\n\sum_{i=1}^n M_{ii}=\frac{1}{2}\partial_t S+\frac{S}{2t}.\n\end{align*}\nSecond, the contracted second Bianchi identity gives $\sum_{i=1}^n \nabla_i\operatorname{Ric}_{ki}=\frac{1}{2}\nabla_kS$. Therefore\n\begin{align*}\n-2\sum_{i,k=1}^n P_{iki}V_k=\sum_{k=1}^n (\nabla_kS)V_k=\langle\nabla S,V\rangle_{g(t)}.\n\end{align*}\nThird, by the curvature convention fixed above,\n\begin{align*}\n\sum_{i,k,l=1}^n R_{ikil}V_kV_l=\operatorname{Ric}_{g(t)}(V,V).\n\end{align*}\nCombining these three identities with $0\le\sum_i Z_{ii}$ gives\n\begin{align*}\n0\le \frac{1}{2}\partial_t S(x,t)+\langle\nabla S(x,t),V\rangle_{g(t)}+\operatorname{Ric}_{g(t)}(V,V)+\frac{S(x,t)}{2t}.\n\end{align*}\nMultiplying by $2$ yields\n\begin{align*}\n\partial_tS(x,t)+2\langle\nabla S(x,t),V\rangle_{g(t)}+2\operatorname{Ric}_{g(t)}(V,V)+\frac{S(x,t)}{t}\ge 0.\n\end{align*}\nSince $x$, $t$, and $V\in T_xM$ were arbitrary, the asserted pointwise trace Harnack inequality follows.\n[/step]