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Represent the quadratic form by the symmetric matrixThe spectral theorem for real symmetric matrices gives an orthogonal matrix $Q$ and real eigenvalues ${\lambda }_1,{\lambda }_2$ such that $Q^{T}MQ=\mathrm{diag}({\lambda }_1,{\lambda }_2)$. In the orthogonal change of coordinates $\mathbf{x}=Q\mathbf{u}$, the equation becomes ${\lambda }_1{u}_1^2+{\lambda }_2{u}_2^2=1$.

If both eigenvalues are positive, this is an ellipse with semiaxes ${\lambda }_1^{-1/2}$ and ${\lambda }_2^{-1/2}$. Conversely, a zero eigenvalue makes the set either empty or unbounded along that coordinate, while two negative eigenvalues give the empty set. Opposite signs give an unbounded hyperbola. A negative eigenvalue together with a zero one also gives the empty set. None of those cases is an ellipse. Therefore the exact condition is that $M$ be a symmetric positive-definite matrix.

The sum and product of its eigenvalues are its matrix trace and determinant:Two real numbers are both positive precisely when their sum and product are positive. ThusAn orthogonal transformation preserves area. Using i gives the quadratic-form ellipse area

Eliminate $z$ using the plane equation. The orthogonal projection consists exactly of the points satisfyingsince each such $(x,y)$ lifts uniquely to $(x,y,1-x-y)$ on the intersection. Translating to the center of an ellipse with $X=x+1/3$ and $Y=y+1/3$ givesThe symmetric matrix of this quadratic form is $M=\left(\begin{array}{cc}1& 1/2\\ 1/2& 1\end{array}\right)$, whose eigenvalues are $3/2$ and $1/2$. Thus it is a symmetric positive-definite matrix, and the projected curve is an ellipse. The quadratic-form ellipse area, with the right-hand side scaled by $4/3$, isFor example, the orthogonal change of coordinates $u=(X+Y)/\sqrt{2}$, $v=(X-Y)/\sqrt{2}$ gives semiaxes $2\sqrt{2}/3$ and $2\sqrt{6}/3$, with the same product.

The quadratic form isThe symmetric matrix has eigenvalues $2,8$, with respective orthogonal eigenvectors $(1,-1{)}^T,(1,1{)}^T$. Normalize them to obtain the orthonormal basisThe given change of basis meansSubstitution, or orthogonal diagonalization of a real symmetric matrix, givesThus the semi-major and semi-minor axes have lengths $2\sqrt{2}$ and $\sqrt{2}$. The eccentricity and enclosed area areThe orthogonal change of basis preserves area, so this is also the area in the original Cartesian coordinates.

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

Choose the rotated Cartesian coordinatesTheir inverse is $x=(X-Y)/\sqrt{2}$, $y=(X+Y)/\sqrt{2}$. Consequently $x^2+y^2=X^2+Y^2$ and $xy=(X^2-Y^2)/2$, giving the canonical form of a quadricThe orthogonal diagonalization identifies the eigenvalues $1+a/2,1-a/2,1$ of the symmetric matrix of the quadratic form.

For $|a|<2$, all three coefficients are positive: the surface is an ellipsoid with semiaxes $(1+a/2{)}^{-1/2},(1-a/2{)}^{-1/2},1$. At $a=0$ it is a sphere.

For $a=2$, the equation is $2X^2+Z^2=1$, an elliptic cylinder parallel to the $Y$ axis. For $a=-2$, it is $2Y^2+Z^2=1$, an elliptic cylinder parallel to the $X$ axis. The absent coordinate is unrestricted; its disappearance does not turn the surface into an ellipse.

For $a>2$, the negative coefficient is that of $Y^2$: this is a one-sheet hyperboloid with axis $Y$. For $a<-2$, the negative coefficient is that of $X^2$: it is a one-sheet hyperboloid with axis $X$. This gives the signature transition from an ellipsoid to an elliptic cylinder and a one-sheet hyperboloid, including both transition values.

Write $N=(n_1,n_2,n_3)$. Restriction to the plane $z=0$ givesThis is an equation of degree at most two in two Cartesian coordinates, the general algebraic form of a conic, with possible degenerate conics and an empty set of real points. No determinant hypothesis is needed for this substitution.

For an arbitrary plane, choose a point ${\mathbf{x}}_0$ and an orthonormal basis $\mathbf{e},\mathbf{f}$ of its direction linear subspace. Put $B=(\mathbf{e}\mathbf{f})$ and $\mathbf{x}={\mathbf{x}}_0+B\mathbf{\xi }$, where $\mathbf{\xi }=(s,t{)}^T$. With $S=(M+M^T)/2$, the symmetric part of a matrix, the restricted equation isThis proves the plane-section result whenever “conic” allows the appropriate degeneracies. A rigid motion taking the chosen plane to $z=0$ gives the same argument.

The source's assumption $\det M\ne 0$ does not guarantee a nondegenerate conic. Even with $M=I$ and $f=x^2+y^2+z^2-1$, the section $z=1$ is a single point, and $z=2$ has no real points. With the nonsingular symmetric matrix giving $f=2xz+y^2$, the section $z=0$ is the repeated straight line $y=0$.

There is a further literal omission: the source does not assume $M$ is symmetric. Forone has $\det M=2$ but $f=2xz$. Thus the section $z=0$ is the entire plane, rather than a conic. Only $S$ controls the quadratic form; invertibility of an arbitrary $M$ cannot remove this exception. If the intended hypothesis is that $M$ is symmetric and nonsingular, the quadratic restriction to a plane cannot vanish identically: otherwise $B^{T}MB=0$, so $M$ would map a two-dimensional linear subspace injectively into its one-dimensional orthogonal complement, which is impossible. Degenerate and empty sections still remain possible.

Multiply the quadratic form out directly:Thus $b=5$, $c=-6$ and $d=5$. Both eigenvalues are positive, so this is a positive-definite quadratic form; its positive unit level set is an ellipse. The next part exhibits its principal axes explicitly.

The partial derivatives are $g_x=\sin x$ and $g_y=y$, so all stationary points are $(n\pi ,0)$ for integers $n$. Their Hessian matrix is $\mathrm{diag}(\cos x,1)$. At even multiples of $\pi$ it is a positive-definite matrix, whereas at odd multiples it has one positive and one negative eigenvalue. ConsequentlyThere are no local maxima. Since both $1-\cos x$ and $y^2/2$ are nonnegative, these are also global minima. In the requested open plotting window, the stationary points are $(0,0)$ and $(\pm \pi ,0)$; the local minima at $(\pm 2\pi ,0)$ lie on excluded edges.

For a level set $g=c$, solving for the vertical coordinate givesFor $0<c<2$ these level sets are closed ovals around each local minimum. At $c=0$ the level sets shrink to the local minima. The level set through the saddle points iswhere the two signed branches together describe the same set as $y=\pm 2|\cos (x/2)|$. They cross at $(\pm \pi ,0)$ and reach $(0,\pm 2)$. The crossing slopes at a saddle point are $\pm 1$, as also follows from the local quadratic form $(-\delta x^2+\delta y^2)/2$. For $c>2$, the upper and lower level sets remain separate and extend across the plotting window without meeting the horizontal axis.

Suppose $B\mathbf{x}=0$. The given quadratic form bound then givesso $\mathbf{x}=0$. Thus the kernel of $B$ is trivial. A square matrix with trivial kernel is an invertible matrix, proving the assertion. No symmetry assumption on $B$ is needed.

Expanding the quadratic form givesMatching coefficients of $xy$ and $y^2$ yields

Nilpotence and eigenvalues. If $Av=\lambda v$ with $v\ne 0$, then $A^{n}v={\lambda }^{n}v$. Thus $A^n=0$ forces $\lambda =0$ for every eigenvalue. Conversely, if all $n$ complex eigenvalues, counted with algebraic multiplicity, vanish, the monic characteristic polynomial is ${\chi }_A(t)=t^n$. The Cayley-Hamilton theorem then gives $A^n=0$. HenceThis is the criterion for a nilpotent matrix.

Symmetric case. Here a symmetric matrix is understood to be real. The spectral theorem gives $A=QD{Q}^T$ with $Q$ orthogonal and $D$ real diagonal. The previous result makes every diagonal entry zero, so $A=0$. Alternatively, choose $2^r\ge n$, so $A^{2^r}=0$. For every real vector $x$,It follows that $A^{2^{r-1}}=0$. Repeat until $A=0$. This uses positivity of the quadratic form $x^T{M}^{T}Mx$. The restriction to real symmetric, or complex Hermitian matrices, matters: a complex matrix satisfying only $A^T=A$ need not have this positivity.

The parameters and powers. The first two columns of $T$ coincide, so $\det T=0$ for all $\alpha ,\beta$. The sum of its principal two-by-two determinants, each a minor of a matrix, isTogether with $\mathrm{tr}T=2+\beta$, this givesFor every eigenvalue to vanish both remaining coefficients must be zero. ThereforeDirect multiplication at these values givesIndeed the first row of $T^2$ multiplied by $T$ is $3{\mathrm{row}}_1(T)+3{\mathrm{row}}_2(T)-6{\mathrm{row}}_3(T)=0$, and the second row is its negative.

Geometry of repeated application. For any parameters the image of a linear map lies in the span of $a=(1,1,1{)}^T$ and $b=(1,\alpha ,\beta {)}^T$, since $Tx=(x_1+x_2)a+x_{3}b$. Thus it lies in a fixed plane through the origin, independently of the input. When $a,b$ are independent its normal is $a\times b=(\beta -\alpha ,1-\beta ,\alpha -1)$; if they coincide, the image is a line and is still contained in a plane.

For $\alpha =-5,\beta =-2$, put $s=x_1+x_2$ and $t=x_3$. ThenThe columns span this entire plane, so $T$ has rank two. The next two steps areThus the image of a linear map contracts from a plane to a line and then to the origin under successive applications. The line from $T^2$ is precisely the kernel of $T$.

The spectral theorem gives a real orthonormal eigenbasis $v_1,\dots ,v_n$ for a real symmetric matrix $A$. Distinct eigenspaces are orthogonal; within a repeated eigenspace, choose an orthonormal basis, for example by Gram-Schmidt orthogonalization. Put these eigenvectors in the columns of $T$. Then $T^{-1}=T^T$ andFor a change of variables $y=Sx$, choose $\boxed{S=T^T}$. Since $x=Ty$, the quadratic form becomes $x^{T}Ax={\sum }_i{\lambda }_i{y}_i^2$; the ${\lambda }_i$ are precisely the eigenvalues of $A$.

In the numerical case, the off-diagonal entries are half the cross-term coefficients, soA direct eigenvector calculation gives the unit eigenvectors $(1,1,0)/\sqrt{2}$, $(1,-1,1)/\sqrt{3}$, $(1,-1,-2)/\sqrt{6}$, with eigenvalues two, three and minus six. They are pairwise orthogonal. Therefore the required orthogonal diagonalization isand the transformed quadratic form is $\boxed{2y_1^2+3y_2^2-6y_3^2}$.

Complete the square in the quadratic form:The real invertible linear transformationsets $x_1+2x_2=y_1$, $x_2=y_2$, $\sqrt{6}{x}_3=y_3$, giving $Q_1=y_1^2+y_2^2+y_3^2$. The same transformation givesIts signature is $(2,1)$. By Sylvester's law of inertia no real invertible change of variables can turn this second form into a sum of three positive squares.

For any Hermitian matrix $K$, the scalar $x^{†}Kx$ is real because its complex conjugate equals $x^{†}{K}^{†}x=x^{†}Kx$. Applying this to $H,S$ in part (a) givesThese are the real and imaginary quadratic forms associated with the numerical range of $A$.

Use the inner product $⟨u,v⟩=u^{†}v$ on ${\mathbb{C}}^n$. The distinct real eigenvalues permit a full orthonormal eigenvector expansion; the two child solutions establish that expansion and apply it to the proposed quadratic form.

Use the intended infinite square well interpretation: the kinetic-energy operator on $(-a,a)$ with zero Dirichlet boundary conditions. The variational energy is most safely evaluated through its quadratic form,The boundary term vanishes because the trial function vanishes at both walls. Its zero extension has derivative jumps there, so the form interpretation avoids assigning an unrestricted full-line second derivative to it.

Set $u=x/a$ and $b=\lambda a^2$. Apart from an irrelevant factor $a^2$, the trial function is $q_b(u)=(1-u^2)(1+b{u}^2)$. Polynomial integration givesHence the polynomial Rayleigh-Ritz trial space for a one-dimensional box has energyThe denominator is strictly positive, since $b^2+6b+21=(b+3{)}^2+12$. Differentiation givesThe two stationary parameters are $b_{\pm }=(-49\pm 4\sqrt{133})/13$. The smaller is a local maximum; the larger is a local minimum. Since the energy tends to $33{\hslash }^2/(4m{a}^2)$ as $|b|\to \infty$, the latter is the global minimum:The upper stationary value is $(7+\sqrt{133}/2){\hslash }^2/(m{a}^2)$. Exact roots give a slightly tighter bound than substituting the rounded hints.

There are two qualifications to the printed operator and comparison energy. For the box with Dirichlet boundary conditions and width $2a$, the actual ground-state energy iswhich lies just below the variational result. The coefficient $6/5$ printed as an exact eigenvalue is inconsistent with this box problem. Also, merely declaring the differential expression to be zero outside the interval does not impose hard-wall confinement or specify a self-adjoint domain. Read literally on a space admitting exterior-supported states, it has zero-energy states outside and does not have the claimed positive ground energy. The boxed variational result is the well-defined answer for the intended confined particle; a different full-line operator would need its domain and matching conditions supplied.

The $xy$ block has characteristic polynomial $(2-\lambda {)}^2-a^2$. A convenient orthonormal eigenbasis for the full real symmetric matrix iswith eigenvalues $2-a$, $2+a$, and $c$, respectively. Hence its orthogonal diagonalization isFor repeated eigenvalues, every nonzero combination within the corresponding eigenspace is also an eigenvector; the displayed basis remains valid. Introduce principal axes coordinatesThe quadratic form becomesThe principal-axis reduction of a quadric therefore has three regimes.

For $0<a<2$, all coefficients are positive: the surface is an ellipsoid centred at the origin, withFor $a=2$, the $X$ term disappears. The surface is an elliptic cylinder, unbounded along $e_{+}$, whose cross-section is $4Y^2+c{Z}^2=1$:For $a>2$, there is one negative coefficient, so the surface is a one-sheet hyperboloid:Its waist at $X=0$ is an ellipse with semiaxes $(a+2{)}^{-1/2}$ and $c^{-1/2}$. At other $X$, both waist radii are multiplied by $\sqrt{1+(a-2)X^2}$. There is no intercept on its longitudinal axis. Its asymptotic cone is $(a+2)Y^2+c{Z}^2=(a-2)X^2$. If positive eigenvalues coincide, the corresponding choice of principal axes is not unique, reflecting a rotational symmetry of the surface.

By the spectral theorem for real symmetric matrices, choose an orthonormal eigenbasis ${\mathbf{u}}_1,{\mathbf{u}}_2,{\mathbf{u}}_3$ corresponding to $6,\sqrt{3},-\sqrt{3}$, with ${\mathbf{u}}_1=(1,1,1{)}^T/\sqrt{3}$. Write $P=[{\mathbf{u}}_1{\mathbf{u}}_2{\mathbf{u}}_3]$ and $\mathbf{x}=P\mathbf{\xi }$. This orthogonal change of coordinates givesThe plane constraint is therefore ${\xi }_1=0$. On that plane the quadratic form vanishes exactly whenConsequently the entire set isThe two direction vectors are nonzero and not parallel, since ${\mathbf{u}}_2,{\mathbf{u}}_3$ are linearly independent. An invertible orthogonal transformation maps each full line to a full line. Thus the set consists of precisely two infinite lines through the origin.

A real symmetric matrix has a real orthonormal eigenbasis. Put those eigenvectors into an orthogonal matrix $Q$, so $Q^{T}AQ=\mathrm{diag}(a_1,\dots ,a_n)$. With the change of coordinates $x=Q{x}^{\prime }$, or $x^{\prime }=Q^{T}x$,Thus the coefficients are the eigenvalues and the new coordinate directions are the eigenvectors. This orthogonal diagonalization preserves Euclidean lengths and angles while removing cross terms from the quadratic form.

The four spring extensions are $x_1$, $x_2-x_1$, $x_3-x_2$ and $-x_3$. Hooke's law therefore gives potential energy $k[x_1^2+(x_2-x_1{)}^2+(x_3-x_2{)}^2+x_3^2]/2$, while the kinetic energy is $m{\sum }_i{\dot{x}}_i^2/2$. Thuswith the matrices in the requested quadratic formThe equilibrium offsets disappear because the expansion is in displacements from static equilibrium.