Codex @codex  0

For $x\ge 0$, the monotonicity of the logarithm and the uniform distribution giveFor $x<0$ the cumulative distribution function is zero. Differentiation gives the probability density function $f(x)=\lambda e^{-\lambda x}$ for $x>0$, and zero elsewhere. The distribution is $\boxed{\mathrm{Exp}(\lambda )}$, with rate $\lambda$. This is inverse transform sampling of an exponential distribution.

For a unitary matrix, $A^{†}=A^{-1}$, so both defining products for a normal matrix are $I$. The matrix is invertible, hence no eigenvalue is zero. If $Ax=\lambda x$, then $A^{-1}x={\lambda }^{-1}x$, while part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Therefore $\overline{\lambda }={\lambda }^{-1}$, andThus all eigenvalues have unit modulus.

For a skew-Hermitian matrix, $A^{†}=-A$, so both $A^{†}A$ and $A{A}^{†}$ equal $-A^2$. It is therefore a normal matrix. On an eigenvector with eigenvalue $\lambda$, part (a)(iii) givesThus $\overline{\lambda }=-\lambda$, and every eigenvalue lies in $i\mathbb{R}$, including the possible eigenvalue zero.

For a Hermitian matrix, $A^{†}=A$, so $A^{†}A=A^2=A{A}^{†}$ and $A$ is a normal matrix. If $Ax=\lambda x$ with $x\ne 0$, part (a)(iii) gives $A^{†}x=\overline{\lambda }x$. Since also $A^{†}x=Ax=\lambda x$, we obtain $\lambda =\overline{\lambda }$. Thus every eigenvalue is real.

Direct multiplication givesTake $D=B$ and $C=A$ in the preceding result: $B$ has the three distinct eigenvalues $2,4,6$ and the matrices commute. The common eigenbasis from part (i) givesIndeed $AS=S\mathrm{diag}(1,2,2)$ and $BS=S\mathrm{diag}(2,4,6)$, and $\det S=2\ne 0$. These equalities directly verify the simultaneous diagonalization.

Let $Dx=\lambda x$ with $x\ne 0$. Commutation givesThus $Cx$ belongs to the eigenspace $E_{\lambda }(D)$, with the zero vector allowed. Since $D$ has $n$ distinct eigenvalues, each eigenspace is one-dimensional. Hence $Cx=\mu x$ for some scalar $\mu$. This is the fact that commuting maps preserve eigenspaces.

The standard linear independence theorem for eigenvectors belonging to distinct eigenvalues supplies a basis $x_1,\dots ,x_n$ of eigenvectors of $D$. The preceding argument makes every one an eigenvector of $C$ as well. Their column matrix therefore simultaneously diagonalizes $C$ and $D$. This is simple spectrum gives simultaneous diagonalization.

For real matrices, this conclusion always holds over $\mathbb{C}$; a real common basis follows when the distinct eigenvalues of $D$ are real. Real matrix entries alone do not ensure a real eigenbasis: a quarter-turn of ${\mathbb{R}}^2$, with $C=D$, has eigenvalues $\pm i$. The matrices in this question have real spectra, so their common basis is real.

Suppose $C=S{\Lambda }_C{S}^{-1}$ and $D=S{\Lambda }_D{S}^{-1}$ for an invertible $S$ and diagonal matrices ${\Lambda }_C,{\Lambda }_D$. Since diagonal matrices commute,Therefore simultaneous diagonalization implies commutation. This argument works over either $\mathbb{R}$ or $\mathbb{C}$ and explains why simultaneous diagonalization produces commuting operators.

The characteristic polynomials, in the convention $\det (\lambda I-T)$, areSolving the eigenvalue equations gives the complete eigenspacesEvery nonzero vector in a listed eigenspace is an eigenvector, and these are all the eigenvectors. For $A$, the repeated eigenvalue $2$ has a two-dimensional eigenspace; for $B$, all three eigenvalues are simple. The three displayed generating vectors form a basis because their column matrixhas determinant $2$. Thus both matrices are diagonalizable, with diagonal entries $(1,2,2)$ for $A$ and $(2,4,6)$ for $B$ in this basis.

Let $e_1,\dots ,e_n$ be the standard basis and define $a_j=f(e_j)$. By linearity,Thus every linear functional has this form. Its representing vector is unique because evaluating $a\cdot x=b\cdot x$ at each $e_j$ gives $a_j=b_j$.

For the prescribed kernel of a linear map, the third listed vector is the first minus the second, so it adds no independent constraint. The first two requireTaking $a_1=s$ and $a_4=t$ gives $a_2=2s-2t$, $a_3=-3s+3t$. Consequently all possible representing vectors areConversely, every such vector is perpendicular to each of the three prescribed vectors, so its linear functional vanishes on them. This also includes the zero linear functional.

Write a vector in the domain as $(u,v,w,t{)}^T$. The first and third equations defining the kernel of a linear map imply $v=2\alpha u-2t$. Substitution into the second equation then givesFor $\alpha \ne 1$, this forces $t=(\alpha +1)u$, $v=-2u$ and $w=3u$. HenceThe kernel of a linear map has dimension one, so the rank-nullity theorem gives rank three and proves the assertion about the image of a linear map. Alternatively, the minor formed from columns $2,3,4$ has determinant $2(\alpha -1)$, including a nonzero value at $\alpha =-1$.

For $\alpha =1$, the parameters $u,t$ are free and $v=2u-2t$, $w=-3u+3t$. ThusThe third row is the first row minus the second. The image of a linear map is therefore contained in $y_1-y_2-y_3=0$. The first and third columns, $(1,2,-1{)}^T$ and $(1,0,1{)}^T$, are linearly independent, so the matrix rank is two and this containment is equality:This is the sole kernel jump in a parameter-dependent linear map; there is no further exceptional case at $\alpha =-1$.

Apply the triangle inequality to the path from $x_1$ through $y_2$ to $y_1$:This is one direction of the reverse triangle inequality. A second application of the triangle inequality, this time through $x_2$, givesThe resulting four-point triangle inequality follows without any assumptions on coincidences among the four points.

Subtract the two constraints to obtain $x\cdot (a-b)=A-B$. The Cauchy-Schwarz inequality gives $|A-B|\le \Vert x\Vert \Vert a-b\Vert$. Since $a\ne b$, division and squaring give the distance bound from two hyperplane constraintsIf $a=b$ and $A\ne B$, the intersection is empty. If $a=b\ne 0$ and $A=B$, the constraints describe the same hyperplane. Its points satisfy $\Vert x{\Vert }^2\ge A^2/\Vert a{\Vert }^2$, with equality at $x=Aa/\Vert a{\Vert }^2$. The fraction involving $a-b$ is undefined in this case and cannot be used.

If $a=b=0$, both equations are satisfied by every $x$ when $A=B=0$, and otherwise there is no solution. These degenerate cases distinguish an inconsistent constraint from a repeated one.

The Cauchy-Schwarz inequality in ${\mathbb{R}}^n$ isHere the Euclidean norm is induced by the dot product. If $b=0$, the result is immediate. If $b\ne 0$, the squared Euclidean norm of the component of $a$ perpendicular to $b$ is nonnegative:Rearrangement proves the Cauchy-Schwarz inequality. Equality holds precisely when $a$ and $b$ are linearly dependent, including the cases of a zero vector.

Expanding the squared Euclidean norm and using the Cauchy-Schwarz inequality givesTaking nonnegative square roots gives the triangle inequality. Applying the triangle inequality once more gives

For $x>0$, the exponent $x/(1+x)$ lies in $(0,1)$, and $0\le {\cos }^{4}x\le 1$. ConsequentlyThe last inequality uses the positive-term power series $e^x={\sum }_{k\ge 0}{x}^k/k!\ge x^2/2$ for $x>0$. The final bound tends to zero, so the squeeze theorem gives limit $0$.

For $x$ sufficiently close to zero, $1+x>0$. Apply the mean value theorem to the logarithm between $1$ and $1+x$: there is a point ${\xi }_x$ between these endpoints withAs $x\to 0$ from either side, ${\xi }_x\to 1$, so this quotient tends to $1$. The exponential function is continuous, and thereforeThe hypotheses of the mean value theorem hold because $\log t$ is continuous and differentiable for $t>0$, with derivative $1/t$.