The base is $2e^{i\pi /4}$. Its complex logarithms are $\ln 2+i(\pi /4+2\pi k)$, $k\in \mathbb{Z}$. Exponentiating $i$ times each logarithm gives all values of the multivalued power:The values form a discrete geometric sequence along the Argand ray of angle $\ln 2$. Consecutive radii differ by $e^{-2\pi }$. The origin is an accumulation point but is not itself a value. If the power is instead defined using the principal logarithm, only $k=0$ is retained. The phrase “all values” requires this branch distinction.

The right-hand side has modulus two and argument $\pi /3$. Its fifth roots therefore haveThey are the vertices of a regular pentagon on the circle of radius $2^{1/5}$, beginning at angle $\pi /15$. These five roots are distinct, and a degree-five polynomial has no further complex roots.

Write $\mathbf{w}=\mathbf{a}-\mathbf{c}$, $B=\mathbf{b}\cdot \mathbf{b}$, $D=\mathbf{d}\cdot \mathbf{d}$ and $E=\mathbf{b}\cdot \mathbf{d}$. Differentiating $|\mathbf{w}+\lambda \mathbf{b}-\mu \mathbf{d}{|}^2$ with respect to the two parameters givesFor nonparallel directions, $BD-E^2=|\mathbf{b}\times \mathbf{d}{|}^2>0$. Solving this linear system yieldsThe quadratic being minimized is strictly convex in the two parameters under this nonparallel assumption, so the critical point gives the unique pair of closest points. Parallel lines do not select a unique value of $\lambda$.

For nonparallel direction vectors, let $\mathbf{n}=(\mathbf{b}\times \mathbf{d})/|\mathbf{b}\times \mathbf{d}|$. Every joining vector has the same normal component, $\mathbf{R}\cdot \mathbf{n}=(\mathbf{a}-\mathbf{c})\cdot \mathbf{n}$. Its norm is therefore at least the absolute value of that component. Minimizing $|\mathbf{R}{|}^2$ makes it perpendicular to both directions, and hence parallel to $\mathbf{n}$, attaining the bound:The expression printed without absolute values is an oriented separation, not a nonnegative distance. Reversing either line's parametrization can flip its sign. For parallel nonzero directions, this normal is undefined and the distance is instead $|(\mathbf{a}-\mathbf{c})\times \mathbf{b}|/|\mathbf{b}|$.

A vector from the point on $L_2$ to the point on $L_1$ isChoosing the opposite direction negates this vector and does not change its norm or the shortest distance.

Assume $\mathbf{a}\ne 0$ and allow all $\mathbf{R}\ne 0$. Put $\hat{\mathbf{a}}=\mathbf{a}/|\mathbf{a}|$ and $\mathbf{u}=\mathbf{R}/|\mathbf{R}|$. Every resulting vector is perpendicular to $\mathbf{a}$ and has $|\mathbf{r}|=|c|\sin \theta \le |c|$. Conversely, for $c\ne 0$ and any $\mathbf{q}=\mathbf{r}/c$ perpendicular to $\hat{\mathbf{a}}$ with $|\mathbf{q}|\le 1$, chooseThe vector triple product gives $\mathbf{u}\times \hat{\mathbf{a}}=\mathbf{q}$, and $|\mathbf{u}|=1$. Thus every point of the closed disk centered at the origin, perpendicular to $\mathbf{a}$, of radius $|c|$ is attained. The boundary circle alone would omit allowed directions of $\mathbf{R}$. For $c=0$ the locus is the origin; for $\mathbf{a}=0$ the original expression is undefined.

For $c>0$, this is the sphere centered at $\mathbf{a}$ with radius $c$. If $c=0$, its locus is the singleton $\{\mathbf{a}\}$; if $c<0$, it is empty, because a Euclidean norm cannot be negative. The equation describes the surface, not the enclosed ball.

For ${\beta }^2=1$, the reduced equation is $u^{\prime \prime }+u=1+\cos x$. The constant forcing has particular solution one, while the resonant cosine requires $u_p=(x/2)\sin x$. ThusThe secular factor $x$ is the signature of resonance. It must replace the divergent nonresonant coefficient $1/({\beta }^2-1)$; substituting ${\beta }^2=1$ directly into that coefficient is invalid.

The exponential substitution removing a constant first derivative, $y=e^{x}u$, reduces the equation to $u^{\prime \prime }+{\beta }^{2}u=1+\cos x$. For $\beta \ne 0$ and ${\beta }^2\ne 1$, solve the homogeneous harmonic equation and use a constant plus a cosine for the particular solution:There is a further exceptional case inside the printed condition ${\beta }^2\ne 1$: $\beta =0$ must be treated separately. It gives $u^{\prime \prime }=1+\cos x$, henceThis retains the constant forcing's resonance with the zero-frequency homogeneous equation. Substitution verifies both branches.

Contract the Levi-Civita symbol identity ${\epsilon }_{ijk}{\epsilon }_{i\ell m}={\delta }_{j\ell }{\delta }_{km}-{\delta }_{jm}{\delta }_{k\ell }$. Using Einstein summation convention,This Lagrange identity for the cross product includes zero and parallel vectors without a separate division by their norms.

The linear span is $\{\lambda \mathbf{a}+\mu \mathbf{b}:\lambda ,\mu \in \mathbb{R}\}$, a vector subspace of ${\mathbb{R}}^3$. Its dimension is zero if both vectors vanish, one if at least one is nonzero and they are linearly dependent, and two if they are linearly independent. The corresponding geometric sets are the origin, a line through the origin, and a plane through the origin.

Write $\Delta =⟨\mathbf{u},\mathbf{v},\mathbf{w}⟩=\mathbf{u}\cdot (\mathbf{v}\times \mathbf{w})$. The displayed definitions require $\Delta \ne 0$, equivalently linear independence of the original vectors. The vector triple product identity givesHere the first scalar triple product is $\Delta$ by cyclic permutation, and the second is zero because a cross product is orthogonal to both its factors. ThusThis is the reciprocal basis involution identity. When $\Delta =0$, the proposed reciprocal basis is undefined; for example, three vectors in the $xy$ plane cannot be inserted into the printed quotients.

Expand the dot product and cross product in Cartesian components:The two expressions are the same scalar triple product, or determinant with columns $\mathbf{u},\mathbf{v},\mathbf{w}$:The equality also holds when this determinant is zero.

Yes. The old matrix is triangular, with determinant zero and trace $0+0+1+1=2$. The upper-left two-by-two block of the new matrix has determinant zero, so the whole determinant remains zero. Its trace isThusThis verifies both change of basis invariants from part (b).

Choose $\boxed{v_4=e^t(1-t)}$, and denote the new ordered basis by $\mathcal{C}=(b_1,b_2,b_3,b_4)$. The change of basis matrix, with columns expressed in $\mathcal{B}$, isThis nonzero determinant proves that the proposed fourth function completes a basis. Express the derivatives in that basis:Reading off columns therefore gives

Write the ordered basis as $\mathcal{B}=(1,t,e^t,t{e}^t)$. Differentiation givesThe corresponding coordinate vectors form the columns of the representing matrix:This is an example of differentiation on an exponential-polynomial space.

The coefficients are real, so the supplied polynomial root $2+2i$ has its complex conjugate $2-2i$ as another polynomial root. Their monic polynomial factor of degree two isPolynomial division then givesThus all five polynomial roots, with multiplicity of a root, areTheir sum is $5$ and their product is $8\cdot 1\cdot (-1)=-8$, agreeing with solution and solution.

For a monic polynomial of degree five, Vieta's formulas give the product of its polynomial roots as $(-1{)}^5$ times its constant coefficient. Therefore

By Vieta's formulas, the sum of all five polynomial roots, counted with multiplicity of a root, is minus the coefficient of $z^4$ divided by the leading coefficient. Hence

Induction from $M^2=M$ gives $M^j=M$ for every integer $j\ge 1$. Separate the zeroth term of the matrix exponential series:Therefore the exponential of an idempotent linear map isThe last form shows its action directly: it multiplies the image of a linear map of $M$ by $e^{\alpha }$ and fixes its kernel of a linear map. No Hermiticity assumption on the idempotent matrix is needed.