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Symmetry of the real inner product gives $A^T=A$. Moreover, i with $y=x$ shows thatso the Gram matrix is a positive semidefinite matrix. To also check reality of every eigenvalue, allow a nonzero vector $z=a+ib\in {\mathbb{C}}^n$ with $Az=\lambda z$. Since $A$ is a real symmetric matrix,Here $z^{\ast }$ is the conjugate transpose, and $a^{T}Ab=b^{T}Aa$ cancels the imaginary part. Thusbecause $z^{\ast }z>0$. This proves both that every eigenvalue is real and that it is nonnegative.

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.

Let $B=A^{T}A$. It is a symmetric matrix, and ${\mathbf{v}}^{T}B\mathbf{v}=\Vert A\mathbf{v}{\Vert }^2\ge 0$. By the spectral theorem for real symmetric matrices, choose an orthonormal basis ${\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3$ of eigenvectors of $B$, with nonnegative eigenvalues ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

For each $i\ne j$, take the unit parallelogram parametrized by $s{\mathbf{v}}_i+t{\mathbf{v}}_j$, $0\le s,t\le 1$. Its surface area is $1$. Moreover,The image parallelogram consequently has area $\sqrt{{\lambda }_i{\lambda }_j}$. By the assumption this equals $1$ for all three pairs, givingNo eigenvalue can be zero. Dividing the equations and using positivity yields ${\lambda }_1={\lambda }_2={\lambda }_3=1$. HenceThus testing all surfaces is stronger than necessary: the three unit parallelograms in these principal directions already determine the result. This is the three-dimensional characterization of linear surface-area isometries.

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 $Q^{-1}=Q^T$.

Let $S$ be the vector subspace of symmetric matrices. A basis consists of $E_{ii}$ for $1\le i\le n$ and $E_{ij}+E_{ji}$ for $i<j$, soThe dimension formula for a sum of subspaces now givesChoose $0\ne A\in V\cap S$. It belongs to $V$ and is a symmetric matrix, so part (a)(i) makes it a diagonalisable matrix. The preceding example shows that the strict threshold is sharp.

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.

An orthogonal matrix satisfies $A^{-1}=A^T$, while a symmetric matrix satisfies $A^T=A$. Combining them gives $\boxed{A^{-1}=A}$. For the product with the antisymmetric matrix $B$, the rule for the transpose of a matrix product givesThus $AB$ is an antisymmetric matrix exactly when $AB=BA$. The hypotheses do not force this commutativity. For a counterexample, takeHere $A$ is a symmetric matrix and an orthogonal matrix, and $B$ is an antisymmetric matrix and an orthogonal matrix, but $AB$ is a nonzero symmetric matrix, so it is not an antisymmetric matrix. On the other hand, the product of two orthogonal matrices is always an orthogonal matrix:Therefore the answers are $\boxed{\text{(i) no; (ii) yes.}}$

Call the given matrix $M$. Its middle coordinate decouples, immediately giving the eigenvalue $3$ with normalized eigenvector $(0,1,0{)}^T$. The remaining characteristic polynomial comes from the $xz$ block:For $\lambda =-4$, the first row of $(M+4I)v=0$ gives $x=\sqrt{3}z$; for $\lambda =8$, the first row of $(M-8I)v=0$ gives $z=-\sqrt{3}x$. After vector normalization, one possible complete set isEach has unit Euclidean norm. The dot products with $v_2$ vanish because the other two middle components are zero, and $v_1\cdot v_3=(-\sqrt{3}+\sqrt{3})/4=0$. This explicitly verifies orthonormality, also predicted by the spectral theorem for real symmetric matrices with distinct eigenvalues. Changing any eigenvector's sign gives an equally valid normalized choice.

The characteristic polynomial is $(5-\lambda )(2-\lambda )-4={\lambda }^2-7\lambda +6=(\lambda -1)(\lambda -6)$. For $\lambda =1$, the eigenvector equation gives $y=2x$; for $\lambda =6$, it gives $x=-2y$. The normalized eigenvectors can therefore be chosen asBoth have unit Euclidean norm, and their dot product is $(-2+2)/5=0$, explicitly verifying orthogonality. This also follows from the spectral theorem for a real symmetric matrix with distinct eigenvalues.

The transpose reverses a matrix product. Since $S^T=S$, we have $(S^2{)}^T=S^T{S}^T=S^2$. Therefore $S^2$ is a symmetric matrix.

Write the general real symmetric matrix as $E_2=\left(\begin{array}{cc}r& t\\ t& u\end{array}\right)$. With the specified $E_1$, anticommutation readsHence $t=0$ and $u=-r$. The remaining condition $E_2^2=I$ gives $r^2=1$, leaving exactlyBoth matrices directly satisfy all the conditions.

The given symmetric matrices satisfy $E_j^T{E}_j=E_j^2=I$, so both are orthogonal matrices. Put $J=E_1{E}_2$, $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. Anticommutation givesThus $A=cI+sJ$ and $A^T=cI-sJ$, giving $A{A}^T=(c^2+s^2)I=I$. These are orthogonal rotations generated by anticommuting involutions. Also $A-A^T=2sJ$. Since $J^2=-I$ makes $J$ invertible, $A$ is symmetric exactly when $\sin (\theta /2)=0$.

A real normal matrix satisfies $M{M}^T=M^{T}M$. For a symmetric matrix, $M^T=M$ and both products equal $M^2$. For an antisymmetric matrix, $M^T=-M$ and both equal $-M^2$. For an orthogonal matrix, both products are the identity. Thus each of the three classes consists of normal matrices.

A symmetric matrix that is also an orthogonal matrix satisfies $C=C^{-1}$. Multiplying by the commuting invertible factors gives $(I-B{)}^2=(I+B{)}^2$, hence $4B=0$. Thus there is no nonzero choice of $B$. The only symmetric case is $B=0$, giving $C=I$.

Choose two perpendicular unit directions in the $-2$ eigenspace:Their pairwise dot products vanish and their Euclidean norms are one. They form an orthonormal basis of eigenvectors, with eigenvalues $4,-2,-2$, as guaranteed for a real symmetric matrix by the spectral theorem for real symmetric matrices.

The characteristic polynomial is $(1-\lambda )(4-\lambda )-4=\lambda (\lambda -5)$. Solving each eigenvalue equation gives the eigenvalues and eigenvectorsFor unit-length representatives divide each displayed direction by $\sqrt{5}$. The two eigendirections are orthogonal, consistently with the matrix being a symmetric matrix.

Yes. The matrix is a real symmetric matrix, so the spectral theorem for real symmetric matrices provides an orthonormal basis of real eigenvectors and hence a real diagonal form. Explicitly its characteristic polynomial is ${\lambda }^2-7\lambda -4$, with distinct real eigenvalues