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For a linear map $\alpha :{\mathbb{R}}^m\to {\mathbb{R}}^n$, its kernel and image of a linear map areThe kernel is a subspace of the domain and the image a subspace of the codomain. Given the stated bases, define the matrix of a linear map byIts $j$th column is the coordinate vector of $\alpha (e_j)$; if $v={\sum }_j{v}_j{e}_j$, the output coordinates are $(\alpha v{)}_i={\sum }_j{A}_{ij}{v}_j$.

For the change of basis, write $e_j^{\prime }={\sum }_k{P}_{kj}{e}_k$ and $f_i^{\prime }={\sum }_{\ell }{Q}_{\ell i}{f}_{\ell }$. The matrices $P,Q$ are invertible. Input coordinates change by $v_{\mathrm{old}}=P{v}_{\mathrm{new}}$ and output coordinates by $w_{\mathrm{old}}=Q{w}_{\mathrm{new}}$, soThe domain and codomain basis changes need not be the same matrix.

For $\beta$, the two input vectors form a basis of ${\mathbb{R}}^2$, so its image is the span of $v=(1,2,3{)}^T$ and $w=(6,4,2{)}^T$. The vector $n=(1,-2,1{)}^T$ is orthogonal to both, since $n\cdot v=n\cdot w=0$. Every possible output therefore satisfies $n\cdot \beta x=0$, whereas the target itself is $n$ and $n\cdot n=6$. Hence there is no $x$ with the required output. This uses an image obstruction by an annihilating functional, rather than an inconsistent guessed input.

For $\gamma$, use the input basis $v_1=(1,2,0{)}^T$, $v_2=(0,1,1{)}^T$, $v_3=(0,1,0{)}^T$. Its determinant is $-1$, so it is genuinely a basis. Write $x=s{v}_1+t{v}_2+w{v}_3$. ThenSubstitution back into physical coordinates givesFor a direct check, its standard matrix is $\left(\begin{array}{ccc}1& 0& -2\\ 1& 1& 0\end{array}\right)$, which annihilates exactly that line.

For a Cartesian change of basis represented by an orthogonal matrix $Q$, vectors transform as $E_i^{\prime }=Q_{ia}{E}_a$ and $B_i^{\prime }=Q_{ia}{B}_a$. Their squared norms are invariant and $Q_{ia}{Q}_{jb}{\delta }_{ab}={\delta }_{ij}$. Substitution into the given expression therefore givesthe transformation law of a Cartesian second-rank tensor. The magnetic field's extra axial sign under an improper physical reflection, if included, appears twice and cancels in its quadratic contribution.

Write the Maxwell stress tensor as $T=\mathbf{EE}+\mathbf{BB}-\frac{1}{2}(E^2+B^2)I$. Its divergence isThe identity $(\mathbf{A}\cdot \nabla )\mathbf{A}-\frac{1}{2}\nabla A^2=-\mathbf{A}\times (\nabla \times \mathbf{A})$ follows by contracting two Levi-Civita symbols, or directly by differentiating components. Using Maxwell's equations consequently yieldsHence the local conservation of electromagnetic momentum isThe two subtracted terms are the Lorentz force density, while $\mathbf{E}\times \mathbf{B}$ is the electromagnetic momentum density in the units used here.

Let $P_B,P_{B^{\prime }},P_C,P_{C^{\prime }}$ have the corresponding basis vectors as columns in standard coordinates. From the PDF,The matrix representation of a linear map in standard coordinates is $P_{C}A{P}_B^{-1}$. The change of basis formula therefore givesMultiplication yields

By the real spectral theorem, the real symmetric matrix $A$ has an orthonormal basis of eigenvectors. In that basis the associated quadratic form isRescaling the coordinates belonging to nonzero eigenvalues changes every positive coefficient to $+1$ and every negative coefficient to $-1$. Hence the diagonal normal form has one positive square for each positive eigenvalue and one negative square for each negative eigenvalue. By Sylvester's law of inertia, these counts do not depend on the diagonalizing basis, so

The numerical eigenvalues are not invariant under a general change of basis, because the matrix changes by congruence rather than similarity. For example, on a one-dimensional space let $q(e)=1$. Its matrix in the basis $(e)$ is $(1)$, whereas in the basis $(2e)$ it is $(4)$. The eigenvalue changes from $1$ to $4$, although its sign, and therefore the signature, is unchanged.

Let $f_1,f_2\in V^{\ast }$ satisfy $f_1(v)f_2(v)=0$ for every vector $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne 0$. For every $v\in \ker f_1$ and every real number $t$,so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Sincewe obtain $f_2=0$. Thus one of the two linear functionals is zero.

Sylvester's law of inertia says that, in a suitable basis, every real quadratic form isThe rank is $r=p+m$, and the signature is $\sigma =p-m$; both are independent of the chosen basis.

If $q=f_1{f}_2$, its polar form has image contained in $\mathrm{span}\{f_1,f_2\}$, so $r\le 2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma =0$. If $r=1$, then $|\sigma |=1$. In every case,Conversely, this inequality leaves onlyup to a linear change of coordinates. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectorswith $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the radical. The $p+m+k$ vectorsare linearly independent and all satisfy $q(v)=0$. They form the required basis of isotropic vectors.