Codex @codex  0

The preceding cycles show the action on a full basis, not only on one direction: $A$ sends ${\mathbf{e}}_1\to {\mathbf{e}}_2\to {\mathbf{e}}_3\to {\mathbf{e}}_1$, and $A^{\mathsf{T}}$ reverses each arrow. Therefore either composition fixes every basis vector, and by linearity fixes every vector. ThusIn particular $A$ is an orthogonal matrix, $A^{\mathsf{T}}=A^{-1}=A^2$, and $A^3=I$.

Use the common eigenvectors of $A$ and $A^{\mathsf{T}}$. For each $k$,Thus, without constructing $S$ explicitly, its eigenvalues areThe eigenvalue $1$ has direction $(1,1,1{)}^{\mathsf{T}}$. The repeated eigenvalue $-1/2$ has a real eigenspace consisting of vectors whose coordinates sum to zero. This is the general rule for the symmetric part of an orthogonal matrix. Adding unrelated lists of eigenvalues would not suffice; here the shared eigenvectors justify the pairing.

The Euclidean norm of each position vector gives

The dot product formula for the angle between two nonzero vectors gives

For $\mathbf{r}(t)=(t,t,t^2)$, the tangent vector is ${\mathbf{r}}^{\prime }(t)=(1,1,2t)$. Substitute into the field:Its scalar product with the tangent is $3t^2-4t-7t^6$. Hence the explicitly parametrized line integral isAs a check, the potential from part (d) has values $-128+C$ at $(2,2,4)$ and $-2+C$ at $(1,1,1)$, giving the same difference.

The direction vectors are $v=(-1,-1,1)$ and $w=(1,1,2)$. Their scalar product is $v\cdot w=-1-1+2=0$. Both are nonzero, so the lines are perpendicular:

For $\lambda =a$, the components of an eigenvector are equal; for $\lambda =-a$, they are opposite. Dividing each vector by its Euclidean norm givesThey are mutually orthogonal and have unit norm. The labels specify their eigenvalues even when $a<0$.

The matrix is symmetric under interchange of the first and third coordinates. For an antisymmetric vector $(1,0,-1)$, direct multiplication gives $A(1,0,-1{)}^{\mathsf{T}}=-6(1,0,-1{)}^{\mathsf{T}}$. For a symmetric vector $(p,q,p)$, the linear transformation reduces toThe reduced characteristic polynomial is $\lambda (\lambda -12)$. At $\lambda =0$, its equations give $q=2p$; at $\lambda =12$, they give $q=-p$. Normalizing the three eigenvectors givesTheir norms are one. Their unnormalized pairwise dot products are $1-1=0$, $1-1=0$ and $1-2+1=0$, proving mutual orthogonality. The eigenvalues sum to six and have product zero, agreeing with the matrix trace and determinant from part (a).

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.

A space diagonal of the unit cube has displacement vector $(1,1,1)$. Its Euclidean norm, or two successive applications of the Pythagorean theorem, givesThe requested diagonal joins opposite vertices; $\sqrt{2}$ would be the diagonal of a face.

Let $\Delta =[\mathbf{a},\mathbf{b},\mathbf{c}]\ne 0$. These three vectors are linearly independent, so write $\mathbf{d}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. Taking the dot product with $\mathbf{b}\times \mathbf{c}$ gives $\alpha =[\mathbf{b},\mathbf{c},\mathbf{d}]/\Delta$. Likewise, the dot products with $\mathbf{c}\times \mathbf{a}$ and $\mathbf{a}\times \mathbf{b}$ giveHere the cyclic invariance and sign change under an interchange of the scalar triple product fix every sign. Substituting the three coefficients provesThis is precisely coordinate extraction using the reciprocal basis $(\mathbf{b}\times \mathbf{c},\mathbf{c}\times \mathbf{a},\mathbf{a}\times \mathbf{b})/\Delta$.

Put $\mathbf{v}={\mathbf{r}}_0-\mathbf{a}$. The squared distance to a point on the line is $|\mathbf{v}-\lambda \hat{\mathbf{t}}{|}^2$. Since $|\hat{\mathbf{t}}|=1$, completing the square in $\lambda$ gives its minimum at $\lambda =\hat{\mathbf{t}}\cdot \mathbf{v}$, with value $|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$. The cross product identity $|\hat{\mathbf{t}}\times \mathbf{v}{|}^2=|\mathbf{v}{|}^2-(\hat{\mathbf{t}}\cdot \mathbf{v}{)}^2$ therefore provesFor the given line, $\hat{\mathbf{t}}=(1,1,0)/\sqrt{2}$, so its squared distance to $(x,y,z)$ is $(x-y{)}^2/2+z^2$. Substituting the parabola givesThe global minimum occurs at $\mu =\pm 1$, henceThe corresponding points on the parabola are $\pm (1,1+2\sqrt{3},0)+(0,0,-3)$; their nearest points on the line are $\pm (1+\sqrt{3},1+\sqrt{3},0)$. The connecting vectors are orthogonal to the line, as the projection calculation requires.

The fixed vectors have constant components, so $\mathbf{a}\cdot \mathbf{b}$ is constant, $\nabla \cdot \mathbf{a}=0$ and $\nabla \cdot \mathbf{r}=3$. Hence the divergence is

Multiplication of a matrix by a column vector forms the linear combination of its columns with coefficients from that vector. Hereso $A\mathbf{r}=x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3$. Equating this expression with $\mathbf{b}$ reproduces the original three linear equations.

At $\mu =-1$, the first equation gives $x=y$, and the second then gives $z=0$. All nonzero solutions areAt $\mu =6$, the first equation gives $x=-6y$. The second then gives $3z=7y$, and the third is twice the second. Choosing an integer direction vector, all nonzero solutions areEach full null space, including the zero solution, is a line through the origin. Geometrically, the three homogeneous constraint planes have that common intersection; in each case only two of their equations are independent.

For $\mathbf{b}=0$, a nonzero coefficient vector $(x,y,z)$ with $x{\mathbf{c}}_1+y{\mathbf{c}}_2+z{\mathbf{c}}_3=0$ is precisely a linear dependence of the three columns. Three vectors in three dimensions are linearly dependent exactly when the determinant of their column matrix, or their scalar triple product, is zero. ThusFor this square homogeneous linear system, the condition is also sufficient.

The direction vectors are $\mathbf{u}=(1,-2,1)$ and $\mathbf{v}=(0,-2,1)$. Their cross product is $(0,-1,-2)$, with norm $\sqrt{5}$. A unit normal to the plane is thereforeThe two intersecting directions span the plane, so a vector orthogonal to both is a normal vector to that plane.

The vectors $(1,1{)}^{\mathsf{T}}$ and $(1,-1{)}^{\mathsf{T}}$ are eigenvectors: multiplication adds or subtracts the two entries in a row. Their eigenvalues are thereforeThe second eigenvalue may be positive, zero or negative as the real parameter changes.

The Euclidean norm of the vector is the square root of the sum of its squared components:

Each prediction is a linear combination of the constant parameter and the coefficient of $x_i$. Thus the design matrix isIts $i$th row multiplied by the parameter vector $(\alpha ,\beta {)}^{\mathsf{T}}$ gives the required prediction $\alpha +\beta x_i$.