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The parity-check matrix has rank $3$, so its kernel has dimension $7-3=4$ and size $16$; see linear code. Its columns are precisely the seven distinct nonzero vectors of ${\mathbb{F}}_2^3$. Thus no weight-one or weight-two word is in the kernel, while three dependent columns give a weight-three word. Hence $d=3$.

The normal linear model is $Y=X\beta +\epsilon$ with $\epsilon \sim N(0,{\sigma }^2{I}_n)$. Full column rank givesWith hat matrix $H=X(X^{T}X{)}^{-1}{X}^T$, the $i$th regression leverage is $h_{ii}$. Since $e=(I-H)Y=(I-H)\epsilon$ and $I-H$ is an orthogonal projection,High-leverage observations can move their own fitted values strongly and may exert disproportionate influence; residual size alone can therefore conceal them.

The displayed orthogonal matrix is a rotation through $2\alpha$, where ${\sin }^2\alpha =M/N$.

No. A pure product state has a pure reduced density matrix, whereas part (b) gives $I/2$.

With $p_{-1}=1,p_0=a_0$, $q_{-1}=0,q_0=1$ and $p_n=a_n{p}_{n-1}+p_{n-2}$, $q_n=a_n{q}_{n-1}+q_{n-2}$, induction gives $p_n{q}_{n-1}-p_{n-1}{q}_n=(-1{)}^{n+1}$. Matrix multiplication gives $[a_0,\dots ,a_n,\beta ]=(p_n\beta +p_{n-1})/(q_n\beta +q_{n-1})$; monotonicity in $\beta >0$ places it strictly between the adjacent convergents.

Choose a basis $u_1,\dots ,u_r$ of $U$, extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of $W$, and extend again to a basis of $V$ by adding $v_1,\dots ,v_t$. DefineEvery member of the full basis lies in $W$ or $W^{\prime }$, so the sum of vector subspaces is $W+W^{\prime }=V$. If $x\in W\cap W^{\prime }$, express it in both displayed bases. Uniqueness of the coordinates in the full basis forces its $w_j$ and $v_j$ coefficients to vanish. Thus $x\in U$. The reverse inclusion is built into the construction, proving $W\cap W^{\prime }=U$. This is a subspace supplement with a prescribed intersection.

For the matrix example, letThe skew-symmetric matrices are exactly $U=\mathrm{span}\{J\}$, and the traceless matrices form $W$. A suitable answer isThe matrix trace of $aI+bJ$ is $2a$, so $W\cap W^{\prime }=\mathrm{span}\{J\}=U$. Every $2\times 2$ real matrix has the decompositionwhose first term lies in $W$ and second in $W^{\prime }$. Therefore $W+W^{\prime }=M_2(\mathbb{R})$, as required.

An elementary row operation is one of three operations: interchange two rows, multiply one row by a nonzero scalar, or add a scalar multiple of one row to a different row. Each operation is reversed by another operation of the same kind.

Performing such an operation on an $m\times n$ matrix $A$ is equivalent to left multiplication by the $m\times m$ elementary matrix obtained by applying that operation to the identity matrix. Each elementary matrix is invertible, with inverse corresponding to the reverse operation. Thus a sequence of operations givesFor square matrices, if $A$ were invertible, then $B=PA$ would have inverse $A^{-1}{P}^{-1}$. Hence if $B$ is singular, $A$ must be singular as well. In fact the two are invertible together, because $A=P^{-1}B$ gives the converse.

Suppose $A^T=A$. For eigenvectors $v_i,v_j$ with distinct real eigenvalues, symmetry givesThus $({\lambda }_i-{\lambda }_j)v_i^T{v}_j=0$, and the eigenvectors are orthogonal. Divide each by its nonzero Euclidean norm to obtain an orthonormal basis. The matrix $P$ having these normalized columns satisfies $P^{T}P=I$, so it is an orthogonal matrix and $P^{-1}=P^T$. As in part (a),This proves the needed orthogonal diagonalization from symmetry and distinct eigenvalues, without invoking an unproved spectral theorem.

For the given symmetric matrix, $(1,1{)}^T$ is an eigenvector with eigenvalue $1$, and $(1,-1{)}^T$ is an eigenvector with eigenvalue $27$. Normalize them to obtainTake the real cube roots of the two diagonal entries and transform back:It is a real symmetric matrix, and $C^3=P\mathrm{diag}(1,27)P^T=A$ because $P^{T}P=I$. Direct multiplication also gives $C^2=\left(\begin{array}{cc}5& -4\\ -4& 5\end{array}\right)$ and then $C^3=\left(\begin{array}{cc}14& -13\\ -13& 14\end{array}\right)$.

For $\alpha \ne 2$, both reduced forms are $I_3$, so $A^{\prime \prime }=(A^{\prime }{)}^T$. At $\alpha =2$, the reduced column echelon form is the transpose of the reduced row echelon form of $A^T$, givingFor example, $C_2\mapsto -C_2$ followed by $C_3\mapsto C_3+C_1-C_2$, using the updated second column, produces the displayed $A^{\prime \prime }$ directly. Its third column is zero. The two matrices differ, so (C) is true exactly when $\alpha \ne 2$. Taking a transpose exchanges the kind of reduction, but also replaces the original $A$ by $A^T$; it does not generally identify the two reduced forms of $A$ itself.

A system of linear equations $T_{\alpha }x=b$ is solvable exactly when $b\in \mathrm{im}{T}_{\alpha }$. If $x_0$ is one solution, thenso its solution set is $x_0+\ker T_{\alpha }$, a translate of the kernel of a linear map.

For $\alpha \ne 2$, the matrix is an invertible matrix, andEvery $b$ is in the image of a linear map and has exactly one preimage. For $\alpha =2$, its first two columns span the image of a linear map, while the corresponding homogeneous linear system gives the kernel of a linear map:The prescribed $b=(-1,\beta ,1{)}^T$ satisfies the displayed condition precisely when $\beta =0$. If $\beta \ne 0$, it lies outside the image of a linear map and there is no solution. If $\beta =0$, it lies in the image of a linear map, and the nonzero kernel of a linear map supplies infinitely many solutions.

Put $p=(3x^2-2)/\sqrt{2}$ and use the orthonormal basis $\mathcal{B}=(1,\sqrt{3/2}x,p)$ from iii. The columns of the given matrix giveConsequentlyThe images of $\mathcal{B}$ are therefore $(p,\sqrt{3/2}x,-1)$, again an orthonormal basis. By i, $T$ is an orthogonal transformation. Its matrix in this orthonormal basis is

For the given matrix $A$, the characteristic polynomial isFor $\lambda =4$, the eigenvector equation gives $v_2=v_1$; for $\lambda =-1$, it gives $3v_1+2v_2=0$. Thus the eigenvalues and their corresponding eigenspaces areFor any eigenvector $v$ with eigenvalue $\lambda$, the vector-valued function $e^{\lambda x}v$ satisfies $Y^{\prime }=AY$, since its derivative is $\lambda e^{\lambda x}v=e^{\lambda x}Av$. The two displayed eigenvectors are linearly independent, so the general solution of the linear system of ordinary differential equations is

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.

If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.

Let $v\in S$. Choose a unit vector $u$ perpendicular to $v$, and put $w=v\times u$. The cross product identities show that $(v,u,w)$ is an orthonormal basis with positive orientation: $u\times (v\times u)=v$, soThus the matrix $A_v$ with these columns belongs to the special orthogonal group $G$ and satisfies $A_{v}N=v$. For any $v,z\in S$, the element $A_z{A}_v^{-1}$ sends $v$ to $z$. Therefore the group action on the unit sphere is transitive.

For $X\in {\mathrm{SO}}_3(\mathbb{R})$, the defining identity $X^{T}X=I$ givesThus an orthogonal transformation preserves the Euclidean norm and sends the unit sphere to itself. Restricting the given group action to this invariant set preserves both action axioms.

If $XN=N$, the first column of $X$ is $N=e_1$. The other columns are unit vectors perpendicular to it, so the first row is also $(1,0,0)$. Hence $X=\mathrm{diag}(1,B)$ with $B^{T}B=I_2$ and $\det B=1$. Conversely, every such matrix lies in $G$ and fixes $N$. Since every element of the special orthogonal group in two dimensions has the planar rotation matrix form, the stabilizer subgroup isGeometrically these are exactly the rotations in three dimensions about the first coordinate axis.

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

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.