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The two-vector Gram matrix isThe upper bound follows because the subtracted square is nonnegative. If either vector is zero, the determinant is zero. Otherwise apply the non-strict version of the calculation in i: $\Vert v_1-t{v}_2{\Vert }^2\ge 0$ with $t=⟨v_1,v_2⟩/\Vert v_2{\Vert }^2$ gives $⟨v_1,v_2{⟩}^2\le \Vert v_1{\Vert }^2\Vert v_2{\Vert }^2$. ThereforeEquality on the left holds exactly when the two vectors are linearly dependent; equality on the right holds exactly when they are orthogonal. The Gram determinant measures the squared area of their parallelogram: in the nonzero case, $\det A=\Vert v_2{\Vert }^2\Vert v_1-t{v}_2{\Vert }^2$, the square of base times perpendicular height.

By the spectral theorem for real symmetric matrices, there is an orthogonal matrix $Q$ such thatFor clarity, the diagonalization can also be justified recursively. The characteristic polynomial has a complex root of a polynomial by the fundamental theorem of algebra. By ii its eigenvalue is real; taking a nonzero real or imaginary part of an associated eigenvector in ${\mathbb{C}}^n$ gives a real eigenvector, which can be normalized. Its orthogonal complement is invariant, because $⟨Ax,z⟩=⟨x,Az⟩=\lambda ⟨x,z⟩=0$. Restriction to this linear subspace is again symmetric, and induction on dimension gives an orthonormal basis of eigenvectors.

All ${\lambda }_i\ge 0$ by ii. The permitted invariance of determinant and matrix trace under change of basis, applied to the diagonal matrix, yields

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

Let $\mathbf{r}(u,v)$ parametrize $S$. Then $A\mathbf{r}$ parametrizes its image, with tangent vectors $A{\mathbf{r}}_u$ and $A{\mathbf{r}}_v$. An orthogonal matrix preserves dot products, so the Lagrange identity for the cross product gives, for any $\mathbf{p},\mathbf{q}$,The area element of a surface is therefore the same pointwise before and after applying $A$. Integrating provesThis includes orthogonal transformations with determinant $-1$: reversal of orientation does not affect the Euclidean norm in the surface area formula.

First find the fixed axis of rotation by solving $(M-I)\mathbf{n}=0$. After removing a nonzero common factor, the linear equations areThey give $n_2=n_3=-n_1$. Choose the following unit vectors:Their dot products show that they form an orthonormal basis; hence $P=({\mathbf{p}}_1{\mathbf{p}}_2{\mathbf{p}}_3)$ is an orthogonal matrix. Since $M$ is an orthogonal matrix fixing ${\mathbf{p}}_1$, it preserves the orthogonal complement of ${\mathbf{p}}_1$. Its restriction there has determinant $1$, so it is a planar rotation.

The trace gives $1+2\cos \theta =\mathrm{tr}M=9/7$, so $\cos \theta =1/7$. To determine the sign of the sine, rather than just its magnitude, calculateThus a suitable angle and change of basis areIn fact direct matrix multiplication gives

A possible center $\mathbf{c}$ must be a fixed point, so $(I-R_{\theta })\mathbf{c}=\mathbf{b}$. Let $c=\cos \theta$ and $s=\sin \theta$. The determinant iswhen $\theta$ is not an integer multiple of $2\pi$. The unique center is thereforeThe fixed point equation implies $\mathbf{b}=\mathbf{c}-R_{\theta }\mathbf{c}$, whenceRelative to $\mathbf{c}$, this is precisely a planar rotation through $\theta$. This also explains the excluded case: when $\theta \in 2\pi \mathbb{Z}$, $T$ is a translation, with no fixed point unless $\mathbf{b}=0$.

We use the monic polynomial convention for the characteristic polynomial of an $n\times n$ matrix:It has degree $n$, since the only contribution to the coefficient of $t^n$ in the determinant is the product of the $n$ diagonal terms $t$. The alternative convention $\det (A-t{I}_n)=(-1{)}^n{\chi }_A(t)$ has the same polynomial roots.

For the linear map $T_A:x\mapsto Ax$ on the finite-dimensional vector space ${\mathbb{R}}^n$, the rank-nullity theorem implies that injectivity and surjectivity are equivalent. Consequently $T_A$ is invertible if and only if $\ker T_A=\{0\}$.

Using the permitted equivalence between invertibility and a nonzero determinant, we obtain

Write $B=[b_1,\dots ,b_n]$. Since $AB=[A{b}_1,\dots ,A{b}_n]$ and $b\mapsto Ab$ is a linear map, the multilinearity of the determinant gives, for any column position $j$,If $b_i=b_j$ for $i\ne j$, the corresponding columns $A{b}_i,A{b}_j$ also agree, so $D(B)=0$. Thus $D$ is an alternating multilinear map of the columns of $B$, even when $A$ is not invertible.

We have $D(I_n)=\det A$. Define a function of the columns of $B$ byIt is an alternating multilinear map, because both summands are, and $F(I_n)=\det A+1-\det A=1$. Hence $F$ is a normalized determinantal function. By the asserted uniqueness of the determinant, $F(B)=\det B$ for every $B$. Rearranging yieldsThis argument also covers $\det A=0$ without dividing by it.

Yes. The sums of the columns of $A$ are the sums of the rows of its transpose $A^T$. By the previous argument, $A^{T}1=\lambda 1$, so $\det (\lambda I_n-A^T)=0$. The determinant is unchanged by transpose, givingTherefore $\boxed{\lambda \text{ is an eigenvalue of }A}$. The vector $1$ need not itself be an eigenvector of $A$ in the column-sum case.

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.

Row reduction, or the cofactor formula for a matrix inverse, givesMultiplying by the right-hand-side column $(5,13,35{)}^T$ gives, at $\lambda =3$,For general $\lambda$, the coefficient matrix has determinant $(\lambda -1)(\lambda -2)$. Thus other parameter values give a unique solution. If $\lambda =1$, subtracting the first equation from the second gives $y=8$, while subtracting it from the third gives $3y=30$, a contradiction. If $\lambda =2$, subtraction gives $y+z=8$ and $3y+3z=30$, again inconsistent. Hence

Expand the determinant along the first row:The minus sign on the middle cofactor is essential.

The two-by-two determinant is the difference of the diagonal products:

The homogeneous linear system has coefficient matrixNonzero solutions exist exactly when its determinant vanishes. Expanding along the first row givesThus $\lambda =1$ or $-6$. The first equation always gives $x=3z$. For $\lambda =1$, the second then gives $y=-2z$; the third imposes the same relation. For $\lambda =-6$, the second gives $y=3z/2$, again consistent with the third. Therefore the nonzero solutions areFor every other $\lambda$, the matrix is invertible and only the zero solution exists.

Here the vertical bars denote determinants. Since $B=B^{-1}$, taking determinants gives $(\det B{)}^2=1$. The condition $\det B>0$ selects $\det B=1$. By the multiplicative property of the determinant,The hypotheses determine this determinant, while allowing many possible matrices $A$ and $B$.

Let $N=\mathbf{a}{\mathbf{b}}^T$. The orthogonality condition gives $N^2=\mathbf{a}({\mathbf{b}}^T\mathbf{a}){\mathbf{b}}^T=0$: this is the nilpotent part of a transvection when both vectors are nonzero. In two dimensions,Expanding its determinant, the fourth-degree terms cancel:These formulas also cover either vector being zero.

The required matrix form of the linear system isSubtract the first row from the third and expand the determinant along the resulting row:Thus the linear system has a unique solution, for every real $\mu$, precisely when $\boxed{\lambda \ne 1,2}$.

For $\lambda =0$, the second equation gives $x=\mu$. Subtracting the first equation from the third gives $-y=1$, and the first equation then determines $z$:For $\lambda =1$, the first and third left-hand sides are identical but their right-hand sides are $1$ and $2$. HenceFor $\lambda =2$, subtracting the first equation from the third gives $y=1$. The first equation now gives $x+2z=0$, while the second requires $x+2z=\mu$. ConsequentlyThe free variable in a linear system in the consistent singular case describes a line of solutions. These direct reductions distinguish an inconsistent linear system from one with infinitely many solutions, which a zero determinant alone cannot do.

Direct matrix multiplication gives $MN=\left(\begin{array}{cc}5& 18\\ 7& 25\end{array}\right)$. The three determinants are $\det M=8-9=-1$, $\det N=4-3=1$, and $\det (MN)=125-126=-1$. Hence $\boxed{\det (MN)=(-1)(1)=\det M\det N}$, as required.