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Unmodified Hebbian learning reinforces correlated input and output but has no mechanism limiting synaptic strength. The second rule adds activity-dependent negative feedback and stabilizes the norm at an input-scale-independent value, provided the input keeps exciting the output. The third has a cubic weight penalty independent of the instantaneous output; its equilibrium norm instead depends on the strength of the input second moment.

For the third averaged rule, an equilibrium along eigenvector $i$ has radial linearized eigenvalue $-2{\lambda }_i/\tau$ and transverse eigenvalues $({\lambda }_j-{\lambda }_i)/\tau$. Thus a positive simple largest eigenvalue gives stable equilibria in its two opposite directions, whereas lower eigen-directions are unstable to larger components. With generic initial weights, it learns the dominant correlation direction, linking the rule to principal component analysis. These are idealized normalization and feature-extraction mechanisms, not claims that biological synapses implement these exact equations.

The fixed points are $(0,0)$ and $(\pm 1,1)$. The Jacobian matrix isAt the origin its eigenvalues are $1,-1$, so it is a saddle with stable direction the y-axis. At either other fixed point, the characteristic polynomial is ${\lambda }^2+\lambda +2$, giving eigenvalues $(-1\pm i\sqrt{7})/2$. Thus both are stable foci. The nullclines are $x=0$, $y=1$ and $y=x^2$. In the right half-plane $x$ grows below $y=1$ and decreases above it; in the left half-plane these horizontal arrows reverse. Vertically, $y$ increases below $y=x^2$ and decreases above it.

A global argument determines the requested omega-limit sets, rather than guessing from the local phase portrait. For $x\ne 0$, use the quadratic-logarithmic Lyapunov function for a feedback systemIts sublevel sets are compact and bounded away from $x=0$, so trajectories starting in either half-plane exist for all positive time and stay in such a set. Since $L$ decreases, it tends to a limit, and is constant on the omega-limit set. Every point of that set therefore has $y=1$. Invariance then requires $\dot{y}=x^2-1=0$, so only the corresponding focus can occur. Also $x(t)=x(0)\exp ({\int }_0^t(1-y(s))ds)$ preserves its sign.

ConsequentlyOn $x=0$, the exact solution is $y(t)=y(0)e^{-t}$. The phase portrait below shows this stable axis, the nullclines and the spiral attraction in each half-plane.

If $v=\pm u$, then $Au=u$. On the perpendicular plane, $u{u}^T$ vanishes and $[v{]}_{\times }$ is a signed right-angle rotation, whose square is minus the identity. Hence its complex eigenvalues are $\pm i$, and

Let $[v{]}_{\times }$ denote the cross product matrix, with $[v{]}_{\times }x=v\times x$. ThenThe two eigenvalue calculations use an orthonormal basis adapted to $u,v$.

Substituting an exponential particular solution gives $\lambda u-Au=v$, or $(\lambda I-A)u=v$. Since $\lambda$ is not an eigenvalue of $A$, the matrix is invertible. The exponential forcing of a constant-coefficient differential system therefore hasFor the specified matrix, $I-A=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$. Solving $u_2=1$, $u_1-u_2=1$ gives $u=(2,1{)}^T$, soDifferentiation verifies both component equations directly. If a general solution were desired, the homogeneous term $e^{tA}c$ would be added, with the matrix exponential here equal to $I+tA$ because $A^2=0$.

Decompose the electric field into its orthogonal parts $\mathbf{E}={\mathbf{E}}_{\perp }+{\mathbf{E}}_{\parallel }$, where ${\mathbf{E}}_{\parallel }=n(n\cdot \mathbf{E})$. The uniaxial conductivity tensor acts on them asIf $\mathbf{J}=0$, its perpendicular and parallel components each vanish. Since $\alpha \ne 0$ and $\alpha +\gamma \ne 0$, this forces both field components to be zero. The contrapositive givesEquivalently, the tensor has nonzero eigenvalues $\alpha ,\alpha ,\alpha +\gamma$ and determinant ${\alpha }^2(\alpha +\gamma )\ne 0$.

False; in fact no matrix with this spectrum is diagonalizable over $\mathbb{R}$. If a real invertible $P$ gave $P^{-1}AP=D$ with $D$ a real diagonal matrix, all eigenvalues of $A$ would be the real diagonal entries of $D$. But two of the required eigenvalues have nonzero imaginary parts. This contradiction provesThe real block form in the initial example is consistent with this: a two-dimensional rotation block is not a real diagonalization.

False. Consider the real matrixIt is block upper triangular, so its characteristic polynomial is the same as that of $A_0$ and it has the required eigenvalues. Its second column, however, has squared Euclidean norm $1+c^2+c^2=2$, whereas every column of an orthogonal matrix has norm one. ThusUnit-modulus eigenvalues alone do not imply orthogonality; they do not control the geometry of the eigenvectors.

True. All three eigenvalues are distinct in $\mathbb{C}$, so their nonzero eigenvectors are linearly independent. Here is a direct proof of that assertion. If $a_1{v}_1+a_2{v}_2+a_3{v}_3=0$, apply $(A-{\lambda }_{2}I)(A-{\lambda }_{3}I)$ to obtainDistinctness makes both scalar factors nonzero, hence $a_1=0$. Applying the analogous products gives $a_2=a_3=0$. The three eigenvectors therefore form a basis of ${\mathbb{C}}^3$. Taking them as the columns of $P$ givesThis proves that every such matrix is a diagonalizable matrix over $\mathbb{C}$, rather than merely showing it for the example.

Let $c=1/\sqrt{2}$. A concrete real matrix isIts first coordinate has eigenvalue $-1$. The lower block has characteristic polynomial $(\lambda -c{)}^2+c^2$, whose roots are $c\pm ic=(1\pm i)/\sqrt{2}$. Thus this example has all three required eigenvalues. It is also an orthogonal matrix, but orthogonality is not forced by its spectrum, as the second part shows.

Write $v=\dot{x}$. The simple pendulum becomes $\dot{x}=v$, $\dot{v}=-\sin x$. Its equilibria are $(k\pi ,0)$. The linearization is $\left(\begin{array}{cc}0& 1\\ -\cos (k\pi )& 0\end{array}\right)$. For even $k$ its eigenvalues are $\pm i$, and for odd $k$ they are $\pm 1$. The conserved energyhas a strict local minimum at each even multiple of $\pi$, proving nonlinear Lyapunov stability there. The odd multiples are saddles and are unstable. ThereforeThe stable centres are not asymptotically stable because energy is conserved.

The saddle energy is $H=2$, so the separatrix is determined byEnergy levels $0<H<2$ form closed libration curves around the centres; levels $H>2$ have velocity of constant sign and describe rotations. The separatrices join neighboring saddles in the unwrapped $(x,v)$ plane. On the angle cylinder, where $x$ is identified modulo $2\pi$, they are homoclinic to the single saddle.

At $x=0$ the positive separatrix has $\dot{x}=2$. Before reaching the neighboring saddle at $x=\pi$, the velocity is $2\cos (x/2)>0$, so the time to angle $a<\pi$ isAs $a\uparrow \pi$, $\cos (x/2)\sim (\pi -x)/2$. The integrand is therefore asymptotic to $1/(\pi -x)$, whose improper integral diverges logarithmically. Thus the pendulum separatrix approach takes infinite time:Starting with velocity $-2$ gives the reflected approach to $x=-\pi$ and the same infinite time. The velocity tends to zero but never reaches zero at a finite time.

In orthonormal Cartesian coordinates, a second-rank tensor has components transforming under a change of axes $Q$ as $M_{ij}^{\prime }=Q_{ik}{Q}_{j\ell }{M}_{k\ell }$, where $Q{Q}^T=I$. Its contraction is invariant:Thus $M_{ii}$ is a scalar. This is tensor contraction with the Euclidean metric. In general nonorthonormal coordinates the invariant expression is $g^{ij}{M}_{ij}$, not an unweighted sum of diagonal covariant components.

For the planar plate, every mass element has $z=0$. The inertia tensor is consequentlyIt follows directly that ${\mathbf{e}}_z$ is an eigenvector andThe real symmetric matrix has three real eigenvalues. Its trace is $2{\int }_D(x^2+y^2)\rho dS=2M_{\perp }$, so $M_1+M_2+M_{\perp }=2M_{\perp }$. Hence the perpendicular axis theorem gives $M_{\perp }=M_1+M_2$.

For a disc of constant areal density ${\rho }_0$, symmetry gives ${\int }_{D}xydS=0$, and polar integration gives ${\int }_D{x}^{2}dS={\int }_D{y}^{2}dS=\pi a^4/4$. Thereforewhere the scalar $M$ in the last expression denotes the total mass rather than the tensor.

Sylvester's law of inertia says that any real quadratic form has coordinates in which it is a sum of $p$ positive squares minus $m$ negative squares, with $z$ unused coordinates, and that the numbers $(p,m,z)$ are invariant under every invertible linear change of coordinates. Its rank is $p+m$; its signature of a quadratic form is commonly $p-m$, with $(p,m)$ also called its signature pair.

The symmetric matrix here is $J-I$, where $J=11^T$ and $1=(1,\dots ,1{)}^T$. On the line spanned by $1$, $J-I$ has eigenvalue $n-1$. If ${\sum }_i{x}_i=0$, then $Jx=0$, so that the orthogonal complement has eigenvalue $-1$ with multiplicity $n-1$. Thus the off-diagonal all-ones quadratic form hasFor $n=1$ the form vanishes, so rank and scalar signature are both zero, with inertia $(0,0,1)$. Including this case is necessary because the distinguished eigenvalue then becomes zero.