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Since $\mathbf{n}\cdot \mathbf{n}=1$, the vector $K\mathbf{r}$ is perpendicular to $\mathbf{n}$:It follows that $K^2\mathbf{r}=K\mathbf{r}$. If $\mathbf{r}=t\mathbf{n}$, then $K\mathbf{r}=0$; if $\mathbf{n}\cdot \mathbf{r}=0$, then $K\mathbf{r}=\mathbf{r}$. Thus $K=I-\mathbf{n}{\mathbf{n}}^T$ is the orthogonal projection matrix onto ${\mathbf{n}}^{\perp }$: it removes exactly the component normal to the plane.

For the equally inclined plane choose $\mathbf{n}=(1,1,1)/\sqrt{3}$. Reversing its sign gives the same plane; other sign choices give equivalent reflections of the coordinate axes. One suitable orthonormal basis of this plane isTheir dot products with $\mathbf{n}$ vanish, their lengths are one, and their mutual dot product is zero. The equal first two components of ${\mathbf{e}}_1$ give its equal inclinations to $\mathbf{i},\mathbf{j}$. Because ${\mathbf{e}}_j$ is already in the projection plane, ${\mathbf{e}}_j\cdot K\mathbf{r}={\mathbf{e}}_j\cdot \mathbf{r}$. The coordinates of the projected point are thereforeFor a circle of radius $R$ centered at the origin in $z=0$, its equation $x^2+y^2=R^2$ becomes $3u^2+v^2=R^2$. This is an ellipse with semiaxes $R/\sqrt{3}$ and $R$. Translating the original center only translates its orthogonal projection, so the result applies to every circle in that plane. ConsequentlyThe general reason is the orthogonal projection of a circle between planes: the direction common to the planes retains its length, while the perpendicular direction is shortened by the cosine of their angle.

Expanding the determinant gives $\det (\lambda I-M)=(\lambda -1)[(\lambda -1{)}^2-25]$. Solving each eigenvalue equation yieldsTheir pairwise inner products are $16-25+9=0$, $12-12=0$ and $12-12=0$. Their squared Euclidean norms are $50,50,25$, respectively, so division by $\sqrt{50},\sqrt{50},5$ gives an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices: symmetry makes eigenvectors for distinct eigenvalues orthogonal.

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.

Using the ordered orthonormal basis chosen in (c), the column matrix isOther signs or another orthonormal choice within the repeated eigenspace are equally valid, provided the same choice is retained in the remaining parts.

The eigenvector equations give $AB=B\mathrm{diag}(4,-2,-2)$. Multiplying on the left by $B^{-1}$ therefore givesThis is an orthogonal diagonalization: the linear map acts independently by multiplication by its eigenvalues in the new orthonormal basis. The repeated eigenvalue explains the freedom to rotate the last two basis directions.

A real square orthogonal matrix $Q$ satisfiesequivalently $Q^{-1}=Q^T$. Its columns, and also its rows, form an orthonormal basis. Consequently it preserves dot products and Euclidean norms.

The pairwise dot products of the normalized vectors areTheir squared norms are respectively $(1+2+1)/4$, $(1+1)/2$, and $(1+2+1)/4$, all one. Hence they form an orthonormal basis. This also follows from the spectral theorem for real symmetric matrices, because $D$ is symmetric and its three eigenvalues are distinct.

By the spectral theorem for real symmetric matrices, choose the normalized eigenvectors to form an orthonormal basis. With $P_n={\mathbf{e}}_n{\mathbf{e}}_n^{\mathsf{T}}$, orthonormality gives $P_n{P}_m={\delta }_{nm}{P}_n$ and ${\sum }_n{P}_n=I$. For every vector $\mathbf{x}$, expand $\mathbf{x}={\sum }_n({\mathbf{e}}_n^{\mathsf{T}}\mathbf{x}){\mathbf{e}}_n$ and apply $A$:Since this equality holds for every $\mathbf{x}$, the spectral decomposition isEach $P_n$ is the rank-one orthogonal projection onto its eigenvector direction. In a degenerate eigenspace, normalized eigenvectors must be chosen mutually orthogonal; normalization alone does not guarantee this resolution of the identity.

The third coordinate is decoupled, giving the unit eigenvector $(0,0,1{)}^{\mathsf{T}}$ with eigenvalue $1$. In the first two coordinates, the symmetric direction $(1,1)$ has eigenvalue $3-1=2$, and the antisymmetric direction $(1,-1)$ has eigenvalue $3+1=4$. Their normalized forms areThey are an orthonormal basis of ${\mathbb{R}}^3$; independent sign reversals are equally valid choices.

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

In $A^{\mathsf{T}}A$, every row sums to one, so $(1,1,1)$ is an eigenvector with eigenvalue $1$. The antisymmetric direction $(1,0,-1)$ gives eigenvalue $2$. A direction orthogonal to both is $(1,-2,1)$, and direct multiplication shows that it lies in the kernel of a linear map. Normalizing all three gives the complete systemThey form an orthonormal basis of ${\mathbb{R}}^3$. The extra zero eigenvalue reflects the one-dimensional kernel of a linear map for this rank-two rectangular matrix. The question's symbols ${\sigma }_m$ denote the eigenvalues of $A^{\mathsf{T}}A$; the nonzero singular values of $A$ are their square roots, namely $1$ and $\sqrt{2}$.

The three eigenvectors form an orthonormal basis, so an arbitrary vector has expansion $\mathbf{r}=a{\mathbf{v}}_{-}+b{\mathbf{v}}_0+c{\mathbf{v}}_{+}$. Multiplication by $A$ removes the zero-eigenvalue component:Its dot product with ${\mathbf{v}}_0$ is zero for every $a,b,c$. Thus one convenient answer is the nonzero vectorEquivalently, $A^{\mathsf{T}}\mathbf{u}=A\mathbf{u}=0$, so $(A\mathbf{r})\cdot \mathbf{u}=\mathbf{r}\cdot (A^{\mathsf{T}}\mathbf{u})=0$. Any nonzero scalar multiple of this vector in the null space also works.

The vector $(1,-1,0{)}^T$ is an eigenvector with eigenvalue $4$. The complementary subspace of vectors with first two components equal is preserved by $A$: a vector $(s,s,t{)}^T$ maps to $(4s+2t,4s+2t,4s+2t{)}^T$. There, the eigenvalue equations reduce to the $2\times 2$ matrix $\left(\begin{array}{cc}4& 2\\ 4& 2\end{array}\right)$, whose characteristic polynomial is $\lambda (\lambda -6)$. For $\lambda =6$, $t=s$; for $\lambda =0$, $t=-2s$.

Normalising by the Euclidean norm, a complete orthonormal basis of eigenvectors isSign changes of any of these eigenvectors are equally valid. The eigenvalues also check part (a): their sum is $10$ and the sum of their squares is $52$.

Expand the right-hand side in the orthonormal basis from part (b). Its coefficients areThus $\mathbf{b}=2\sqrt{2}{\mathbf{e}}_4+\sqrt{3}{\mathbf{e}}_6$. If $\mathbf{x}=c_0{\mathbf{e}}_0+c_4{\mathbf{e}}_4+c_6{\mathbf{e}}_6$, then matrix multiplication gives $A\mathbf{x}=4c_4{\mathbf{e}}_4+6c_6{\mathbf{e}}_6$. Matching coefficients fixes $c_4=\sqrt{2}/2$ and $c_6=\sqrt{3}/6$, while $c_0$ is arbitrary. Consequently all solutions areGeometrically, the solution set is an affine line in a vector space parallel to the kernel of $A$. The zero eigenvalue makes the component in that direction invisible to $A$; the other two components are uniquely determined. The displayed particular solution is perpendicular to the kernel, so it is also the point of this line closest to the origin.

The two prescribed directions have dot product zero. Normalize them toTheir cross product is the perpendicular unit vector ${\mathbf{q}}_3=(2,-1,-1{)}^T/\sqrt{6}$. Using these as columns givesThe columns form an orthonormal basis, so $Q^{T}Q=I$ and $Q$ is an orthogonal matrix.

The plane equation $x+z=0$ has normal vector $(1,0,1{)}^T=\sqrt{2}\hat{\mathbf{n}}$. Thus its normal vector is parallel to the cylinder's axis and the plane is perpendicular to that axis. It passes through the origin, so the intersection is a circle of radius $2$ centred at the origin in that plane.

Indeed, with the orthonormal basis $\mathbf{e}=(1,0,-1{)}^T/\sqrt{2}$ and $\mathbf{f}=(0,1,0{)}^T$ of the plane, write $\mathbf{r}=s\mathbf{e}+t\mathbf{f}$. The cylinder equation becomes $s^2+t^2=4$. Equivalently, the intersection has the parametric equation

The first three output components equal $x_1,x_2,x_3$, while the fourth is $x_4-{\sum }_{k=1}^4{x}_k=-x_1-x_2-x_3$. HenceIts outputs have component sum zero, and applying $P$ to any such output leaves it fixed. ThusThe first three columns give a basis of the image of a linear map:They are linearly independent because their first three coordinates are the standard basis. Their span is precisely the three-dimensional hyperplane $y_1+y_2+y_3+y_4=0$.

Apply the Gram-Schmidt process. First $q_1=v_1/\sqrt{2}$. Subtracting the $q_1$ projection from $v_2$ gives $(-1,2,0,-1{)}^T/2$, of length $\sqrt{3/2}$. Subtracting both preceding projections from $v_3$ gives $(-1,-1,3,-1{)}^T/3$, of length $2/\sqrt{3}$. Therefore an orthonormal basis isTheir pairwise dot products vanish and each has length one. Although this image admits an orthonormal basis, $P$ itself is an oblique projection: its kernel is the fourth coordinate axis, which is not perpendicular to the image, and $P\ne P^T$.

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 $\boxed{2y_1^2+3y_2^2-6y_3^2}$.

For $r>0$, the chain rule gives $\partial r/\partial x_i=x_i/r$. Differentiating the radial factor gives $d\varphi /dr=e^{ikr}(ikr-1)/(4\pi r^2)$, so in Cartesian coordinatesThe origin is excluded because the scalar field is singular there.

The Stokes theorem states ${\int }_S(\nabla \times \mathbf{F})\cdot \mathbf{n}dS={\oint }_{\partial S}\mathbf{F}\cdot d\mathbf{\ell }$ for a sufficiently smooth vector field and oriented surface with piecewise smooth boundary. The boundary orientation follows the right-hand rule relative to $\mathbf{n}$.

For the convective acceleration identity, use the Levi-Civita symbol contraction ${ϵ}_{ijk}{ϵ}_{k\ell m}={\delta }_{i\ell }{\delta }_{jm}-{\delta }_{im}{\delta }_{j\ell }$. It givesRearranging proves the required vector identity.

For orthogonal coordinates, define the scale factors of orthogonal coordinates $h_i=|\partial \mathbf{r}/\partial q_i|$ and tangent unit vectors ${\mathbf{e}}_i=h_i^{-1}\partial \mathbf{r}/\partial q_i$. They form an orthonormal basis, and $d\mathbf{r}={\sum }_i{h}_i{\mathbf{e}}_{i}d{q}_i$. Equating $d\psi ={\sum }_i{\partial }_{q_i}\psi d{q}_i$ with $\nabla \psi \cdot d\mathbf{r}$ identifies each component. Thus the gradient in orthogonal coordinates isThis holds on a regular coordinate chart where all scale factors are positive.

By the spectral theorem, the Hermitian matrix $A$ has a complete orthonormal eigenbasis. If $Av=\lambda v$, idempotence gives ${\lambda }^2=\lambda$, so every eigenvalue is zero or one. Since $\det A\ne 1$, at least one is zero. ThereforeThe trace of an idempotent equals its rank; here Hermiticity further identifies $A$ as an orthogonal projection matrix.

A complex inner product $⟨u,v⟩$ is conjugate-linear in its first argument, linear in its second, conjugate symmetric and positive definite: $⟨v,u⟩=\overline{⟨u,v⟩}$ and $⟨u,u⟩>0$ for nonzero $u$. An orthonormal basis is a spanning linearly independent family $e_i$ with $⟨e_i,e_j⟩={\delta }_{ij}$. The associated norm is $\Vert u\Vert =\sqrt{⟨u,u⟩}$.

For two distinct basis elements $x,y$, their orthogonality and unit norms give $\Vert x-y{\Vert }^2=⟨x,x⟩+⟨y,y⟩-⟨x,y⟩-⟨y,x⟩=2$. Hence their distance is $\sqrt{2}$; if the same element is selected twice, it is zero.

Finally, for $B^{†}=B$, the scalar quadratic form obeys $(x^{†}Bx{)}^{\ast }=x^{†}{B}^{†}x=x^{†}Bx$, proving $x^{†}Bx$ is real.