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The two nonzero proper invariant subspaces areBoth are preserved by the rotation. To show there are no others, note thatis the linear projection onto $U$, and $I-P$ is the linear projection onto $W$. Every invariant subspace $L$ is preserved by $I,T^2$, and hence by both of these linear projections. ThereforeBecause $U$ has dimension one, $L\cap U$ is either $0$ or $U$. If $L\cap W$ contains a nonzero $v=(a,b,0)$, it also contains $Tv=(-b,a,0)$. These two vectors are linearly independent, because their two-dimensional determinant is $a^2+b^2>0$. They span $W$, so $L\cap W$ is either $0$ or $W$. The only possibilities for $L$ are therefore $0,U,W,V$.

Finally every $(x,y,z)$ has the unique decomposition $(0,0,z)+(x,y,0)$, with one summand in each of $U,W$. HenceThis also illustrates how invariant subspaces of a quarter-turn in three dimensions can be classified by linear projections.

Suppose a line $T$ through $\mathbf{y}$ meets each $L_i$. Its intersection with $L_i$ is different from $\mathbf{y}$, so $T$ lies in ${\Pi }_i$. A nonzero direction vector $\mathbf{d}$ for $T$ must therefore satisfyA square matrix admitting a nonzero vector $\mathbf{d}$ with $M\mathbf{d}=0$ has zero determinant. Expanding givesThus the necessary condition isThe condition is not sufficient. Take $\mathbf{y}=(2,1,0)$, which is on none of the three original lines. It satisfies the condition, since $2+1-2=1$. The three plane equations becomeTheir common intersection is the single line $T=\{(s,1,0):s\in \mathbb{R}\}$. Any proposed line through $\mathbf{y}$ meeting all three original lines would have to be this $T$. But $T$ never meets $L_1=\{(s,0,1):s\in \mathbb{R}\}$: their second and third coordinates differ. The zero determinant guarantees a common direction for the three planes; it does not guarantee the required intersections. This is a parallel obstruction to a transversal of three skew lines.

Write $c=\mathbf{k}\cdot \mathbf{e}$. Since $\mathbf{s}=\ell (\mathbf{k}-\mathbf{e})$, the total torque simplifies toTaking the dot product of the equation of motion with $\mathbf{e}$ gives $C\dot{n}=0$: both torque terms are perpendicular to $\mathbf{e}$, as are the other terms on the left. HenceTaking the dot product of ${\dot{\mathbf{L}}}_Q=\mathbf{\tau }$ with the fixed vector $\mathbf{k}$ givesThese are conservation of the axial spin and of the vertical component of angular momentum.

For energy, the kinetic energy and potential energy areDot the equation of motion, with $\dot{n}=0$, with $\mathbf{e}\wedge \dot{\mathbf{e}}$. The cross product identity $(\mathbf{e}\wedge \mathbf{u})\cdot (\mathbf{e}\wedge \mathbf{v})=\mathbf{u}\cdot \mathbf{v}-(\mathbf{e}\cdot \mathbf{u})(\mathbf{e}\cdot \mathbf{v})$ givesIt follows thatsoThe applied force is conservative even if $\kappa$ is negative: its potential energy is $\kappa |\mathbf{s}{|}^2/2$. This is a symmetric top with a quadratic alignment potential.

If $A$ has $m$ rows and $n$ columns, attach a nonnegative dual variable $y_i$ to each of the $m$ upper-bound constraints. The dual linear program isHere $A^T$ is the transpose of $A$, and all vector inequalities hold componentwise. Nonnegative dual variables preserve the direction of the original inequalities when taking weighted sums.

Apply the divergence theorem to $\varphi \mathbf{c}$, using ii. Because the vector $\mathbf{c}$ is constant, it can be taken outside both integrals:This holds for every $\mathbf{c}$, in particular each of the three Cartesian basis vectors. Equality of the three components proves the gradient volume-to-boundary identityIt suffices that $\varphi$ be a continuously differentiable function on a neighbourhood of the closure of $V$.

In Cartesian coordinates, write the constant vector as $\mathbf{c}=(c_1,c_2,c_3)$. The product rule and constancy of its components giveThe last expression is the dot product with the gradient, so

The Steinitz exchange lemma states that if $u_1,\dots ,u_r$ are linearly independent and $w_1,\dots ,w_s$ span a vector space $V$, then $r\le s$ and some $r$ of the $w_j$ can be replaced by $u_1,\dots ,u_r$ while preserving the spanning set property.

Here is the exchange argument. Suppose that after $k$ replacements, a spanning set iswhere the remaining $w_j$ have been relabelled if necessary. Express $u_{k+1}$ as a linear combination of this family. At least one remaining $w_j$ has a nonzero coefficient; otherwise $u_{k+1}$ belongs to the span of $u_1,\dots ,u_k$, contradicting linear independence. Relabel this entry $w_{k+1}$. Solving for $w_{k+1}$ shows that it lies in the span ofThe new family therefore still spans $V$. Starting at $k=0$ repeats this step until all $r$ independent vectors have been inserted. If $r>s$, after $s$ steps the next $u_{s+1}$ would lie in the span of $u_1,\dots ,u_s$, a contradiction. Thus $r\le s$, and the construction proves the replacement assertion as well.

The vectors are linearly independent ifThey span $V$ if every $v\in V$ is a linear combination $v={\sum }_{j=1}^n{a}_j{v}_j$ with real coefficients. They form a basis of $V$ if they are linearly independent and span $V$. Equivalently, every $v\in V$ has exactly one such representation: existence is the spanning set condition, and subtracting two representations shows that uniqueness is precisely linear independence.

Subtracting the first vector from each of the other two preserves the basis property by iii. The resulting family isIts last two vectors belong to the plane with first coordinate zero. By b, they are linearly independent there precisely whenIn that case their two-dimensional span is the whole plane, and adjoining $e_1$ gives a basis of ${\mathbb{R}}^3$. If the displayed expression vanishes, a nontrivial linear relation among those two vectors prevents a basis. Thus

If $\mathbf{b}$ is outside the image of a linear map $T_A$, there is no solution. Otherwise choose ${\mathbf{x}}_0$ with $A{\mathbf{x}}_0=\mathbf{b}$. By ii, a zero kernel gives a unique solution. If the kernel contains a nonzero vector $\mathbf{z}$, allare solutions. They are distinct, since $(t-s)\mathbf{z}=0$ forces $t=s$. Hence there are infinitely many. These alternatives exhaust all possibilities.

By i, a free choice of $x_3$ is possible precisely when $\lambda =5$ and $\mu =2$. ThusThe complete solution set of a linear system isThe direction vector spans the kernel, and the displayed straight line is the entire solution set of a linear system.

For $u\in U$, the vector $Tu$ already belongs to $\mathrm{im}T$, soTherefore $\ker (ST)=T^{-1}(\ker S^{\prime })$. Since $\ker S^{\prime }=\ker S\cap \mathrm{im}T$, its intersection with $\mathrm{im}T$ is itself. Applying a with $X=\ker S^{\prime }$ yieldsThe inequality follows from the inclusion $\ker S^{\prime }\subseteq \ker S$ and monotonicity of dimension for vector subspaces.

A real number $\lambda$ is an eigenvalue of $T$ if $Tv=\lambda v$ for some nonzero vector $v\in V$. Its eigenspace isIt is a vector subspace because it contains $0$ and $T(av+bw)=aTv+bTw=\lambda (av+bw)$ whenever $v,w\in E_{\lambda }$ and $a,b\in \mathbb{R}$.

If $v\in E_{\lambda }\cap E_{\mu }$, then $\lambda v=Tv=\mu v$, so $(\lambda -\mu )v=0$. Since $\lambda \ne \mu$, this forces $v=0$. Therefore

For this tripotent linear operator, the relation $T^3=T$ gives $T(T^2-I)=0$. Since $T^2\ne I$, there is $v$ with $(T^2-I)v\ne 0$, and that nonzero vector belongs to $\ker T$. Hence $n_T\ge 1$. Also $T\ne 0$, because $T=0$ would imply $T^2=T$. The rank-nullity theorem in dimension three therefore givesNow suppose $n_T=2$. Then $\mathrm{im}T$ is a line. On this line, $T^2$ is the identity operator: if $y=Tx$, then $T^{2}y=T^{3}x=Tx=y$. Consequently $T$ acts on this line by a scalar $c$ satisfying $c^2=1$. If $c=1$, then $T^{2}x=Tx$ for every $x$, contrary to $T^2\ne T$. Thus $c=-1$.

Choose a nonzero $w\in \mathrm{im}T$, so $Tw=-w$. It does not belong to $\ker T$. A basis $k_1,k_2$ of $\ker T$, followed by $w$, is therefore a basis of $V$, in which

For vectors $\mathbf{a},\mathbf{b},\mathbf{c}\in {\mathbb{R}}^3$, the scalar product, cross product and scalar triple product are respectivelyExpanding the dot product givesInterchanging $\mathbf{a}$ and $\mathbf{b}$ changes the sign of their cross product, so its dot product with $\mathbf{b}$ is also zero. Thus the cross product is orthogonal to both vectors, including when it is the zero vector.

The displayed scalar triple product is a determinant. A square matrix has nonzero determinant exactly when its rows are linearly independent. Hence

Use the point ${\mathbf{p}}_0=(-2,-2,4)$ and the orthonormal basis of directions in the cutting planeBoth vectors are orthogonal to the normal vector $(1,1,2)$, so every point of the plane has unique coordinatesSubstituting into the cone equation givesConsequently the intersection has equationBecause these are orthonormal coordinates, their lengths are the actual lengths in the plane. Thus this is an ellipse with semi-major axis $\sqrt{24}$ and semi-minor axis $\sqrt{8}$, not merely a projected ellipse. By a, its eccentricity is

The scalar product, also called the dot product, and the vector product, also called the cross product, areFor nonzero vectors, let $\vartheta \in [0,\pi ]$ be their angle. Their Euclidean norms satisfyThe cross product is orthogonal to both vectors, with its orientation determined by the right-hand rule. If either vector is zero, both products vanish, although the angle is undefined.

A real square matrix $A$ is an orthogonal matrix whenHere $A^T$ is the transpose and $I$ the identity matrix; equivalently, its columns form an orthonormal basis. If this holds, the squared Euclidean norm obeysso the linear transformation preserves length.

Conversely, suppose $|A\mathbf{v}|=|\mathbf{v}|$ for every vector $\mathbf{v}$. Put $B=A^{T}A-I$, a symmetric matrix. Then ${\mathbf{v}}^{T}B\mathbf{v}=0$ for every vector. Taking $\mathbf{v}={\mathbf{e}}_i$, the standard basis vectors, gives $B_{ii}=0$. Taking $\mathbf{v}={\mathbf{e}}_i+{\mathbf{e}}_j$ for $i\ne j$ gives $2B_{ij}=0$. Thus $B=0$, proving that the linear transformation preserves length if and only if $A$ is an orthogonal matrix.

Choose the first orthonormal basis vector to be ${\mathbf{e}}_1=\mathbf{a}$. If $\mathbf{b}$ is not parallel to $\mathbf{a}$, choose ${\mathbf{e}}_2$ along $\mathbf{b}-(\mathbf{a}\cdot \mathbf{b})\mathbf{a}$; otherwise choose any unit vector orthogonal to $\mathbf{a}$. Complete to an orthonormal basis with ${\mathbf{e}}_3$. The matrix with rows ${\mathbf{e}}_i^T$ is an orthogonal matrix, and expressing the three unit vectors in this basis givesThe last two components of ${\mathbf{c}}^{\prime }$ have polar coordinates of the displayed form; we may choose $w\ge 0$. If they vanish, $\theta$ is arbitrary.

An orthogonal transformation preserves dot products. Taking dot products with ${\mathbf{a}}^{\prime }$, and then using the unit Euclidean norms, givesThe remaining dot product givesAn orthogonal matrix has determinant $\pm 1$, so it preserves the absolute scalar triple product. In the new coordinates this scalar triple product is $yw\sin \theta$. ConsequentlyTaking the nonnegative square root provesNo division by $y$ or $w$ was used, so the proof includes parallel vectors. Equivalently, the expression under the square root is the Gram determinant of the three unit vectors.

Over the finite field $F={\mathbb{Z}}_p$, a basis of the hyperplane has $n-1$ vectors by ii. Each vector on the hyperplane has a unique expression as a linear combination of that basis, and each coefficient has $p$ possibilities. ThusEquivalently, choose freely all coordinates except $x_j$, where $a_j\ne 0$, and solve uniquely for $x_j$. For $n=1$ this counts just the zero vector, consistently with $p^0=1$.