Codex @codex  0

If $T^{\prime }=\lambda T$ with $\lambda \ne 0$, then $T^{\prime }(x)=0$ if and only if $T(x)=0$, so their kernels agree. Conversely, suppose both nonzero linear functionals have kernel $K$. Choose $u$ with $T(u)=1$, by scaling a vector on which $T$ is nonzero. For each $x\in V$,Therefore $T^{\prime }(x)=T(x)T^{\prime }(u)$. Moreover $T^{\prime }(u)\ne 0$, since $u\notin K$. Thus $\boxed{T^{\prime }=T^{\prime }(u)T}$, with a nonzero scalar.

Every nonzero linear functional $V\to {\mathbb{F}}_p$ is surjective, so its kernel has dimension $n-1$ by the rank-nullity theorem. Conversely, every hyperplane $K$ is such a kernel: extend a basis of $K$ by one vector $u$, and assign value zero on that basis and value one on $u$. Hence the hyperplanes containing $v$ are exactly the kernels of the $p^{n-1}-1$ nonzero linear functionals counted above. Each kernel occurs for exactly $p-1$ of them, its distinct nonzero scalar multiples. The required number is

The characteristic polynomial is $\det (\lambda I-M)=\lambda ({\lambda }^2-2)$. Solving the three eigenvector equations givesEach eigenspace consists of the scalar multiples of its displayed nonzero vector. Their pairwise dot products vanish: $v_{+}\cdot v_{-}=1-2+1=0$ and $v_{\pm }\cdot v_0=1-1=0$. This also agrees with the orthogonal diagonalization of a real symmetric matrix.

False. Take the diagonal matrix $A=\mathrm{diag}(1,2)$ and its eigenvectors $\mathbf{x}=(1,0{)}^T$, $\mathbf{y}=(0,1{)}^T$. Their sum $(1,1{)}^T$ is mapped to $(1,2{)}^T$, which is not a scalar multiple of that sum. More generally, $A(\mathbf{x}+\mathbf{y})=a\mathbf{x}+c\mathbf{y}$ need not have the form $\lambda (\mathbf{x}+\mathbf{y})$ when $a\ne c$.

For the lowest eigenvalue, $-1$, solving $(H+I)\mathbf{v}=0$ gives $v_2=0$ and $v_3=-v_1$. A unit eigenvector is thereforeAny nonzero scalar multiple is an equally valid eigenvector.

The cross product is $\mathbf{a}\times \mathbf{b}=(-2,0,2)$, which is perpendicular to both vectors. Any nonzero scalar multiple works, so a simple answer isIts dot product with each given vector is zero.

For a differentiable objective $f$ and regular constraint $g=r$, a constrained stationary point satisfieswhere $\lambda$ is a Lagrange multiplier. Solve these equations for both the coordinates and $\lambda$.

Geometrically, the gradient $\nabla g$ is normal to the constraint level set when $\nabla g\ne 0$. Every feasible infinitesimal displacement $\mathbf{v}$ is tangent to that level set, so $\nabla g\cdot \mathbf{v}=0$. At a constrained stationary point, the directional change $\nabla f\cdot \mathbf{v}$ must also vanish for every such tangent displacement. Thus $\nabla f$ is normal to the same tangent plane and is a scalar multiple of $\nabla g$. This is the geometric basis of the Lagrange multiplier method.

For a second constraint $h=s$, with linearly independent vectors $\nabla g$ and $\nabla h$, the feasible tangent directions are perpendicular to both. Their normal space is the linear span of these two gradients, so introduce a second Lagrange multiplier:These equations find candidates; deciding whether they are maxima, minima or saddle points requires further analysis. Constraint points where the required gradient independence fails must be examined separately.

Apply matrix multiplication to obtainSince $\hat{\mathbf{n}}$ is a unit vector, the subtracted term is the orthogonal projection of $\mathbf{x}$ along $\hat{\mathbf{n}}$. Hence $M$ is the orthogonal projection onto a plane through the origin with normal vector $\hat{\mathbf{n}}$. It fixes every vector perpendicular to $\hat{\mathbf{n}}$ and sends every scalar multiple of $\hat{\mathbf{n}}$ to zero. In particular $\hat{\mathbf{n}}\cdot M\mathbf{x}=0$.

For the normalized second-order linear differential equation $y^{\prime \prime }+P(z)y^{\prime }+Q(z)y=0$, an ordinary point $z_0$ has analytic $P,Q$ near $z_0$. At a regular singular point, which is not ordinary, $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ are analytic. Ordinary points admit two independent convergent power series solutions. Regular singular points instead call for the Frobenius method, which permits leading powers and, in some cases, logarithms.

Here zero is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{z}^n$ into the Hermite differential equation and equate coefficients:The even and odd initial coefficients are independent, givingThese are the power series solutions of the Hermite differential equation, with the present parameter equal to twice the parameter in that article. The ratio of successive terms within each nonterminating parity is asymptotically $2|z{|}^2/n$, so both series have infinite radius of convergence.

For a nonzero polynomial of degree $m$, its highest coefficient in $-2z{y}^{\prime }+\lambda y$ forces $\lambda =2m$. Conversely at that value the matching parity terminates. HenceThe polynomial solutions are scalar multiples of the Hermite polynomials. Two examples are $\boxed{\lambda =0:y=1;\lambda =2:y=z}$; direct substitution verifies both. The other independent solution generally remains nonpolynomial.

The product rule gives ${\psi }^{\prime }=e^{-x^2/2}(y^{\prime }-xy)$ and ${\psi }^{\prime \prime }=e^{-x^2/2}[y^{\prime \prime }-2x{y}^{\prime }+(x^2-1)y]$. Substituting in the Schrödinger equation cancels the $x^2$ and constant terms, leaving the Hermite differential equationFor $y={\sum }_{n\ge 0}{a}_n{x}^n$, matching coefficients givesThe power series solutions of the Hermite differential equation split into independent even and odd series. With initial data $(y(0),y^{\prime }(0))=(1,0)$ and $(0,1)$, respectively, their first three terms areEach nonterminating parity series has infinite radius of convergence: the ratio of successive terms of that parity has magnitude asymptotic to $2|x{|}^2/n$. Their initial data prove linear independence. If a coefficient shown vanishes at a special parameter, the recurrence supplies the terminating series instead.

A nonzero polynomial solution of degree $N$ requires $\lambda =N$, by its highest-degree term, and the recurrence then terminates the matching parity. Thus all possible values are $\lambda =0,1,2,\dots$. Up to nonzero scalar multiples, the degree-at-most-three solutions are the Hermite polynomials

The three elementary row operations are:

The nonzero requirement in the second operation ensures that each elementary row operation is reversible.

Let $\lambda$ be the single eigenvalue. By diagonalizability, there is a basis $v_1,\dots ,v_n$ of eigenvectors, and every corresponding eigenvalue is $\lambda$. For any $v={\sum }_i{a}_i{v}_i$, linearity givesThus $\boxed{T=\lambda I}$, a scalar multiple of the identity map.

A counterexample to the conclusion without diagonalizability isIts characteristic polynomial is $(t-2{)}^2$, but $(J-2I)(a,b{)}^{\mathsf{T}}=(b,0{)}^{\mathsf{T}}$. The eigenspace for $2$ is therefore $\mathrm{span}\{(1,0{)}^{\mathsf{T}}\}$, of dimension one. There is no basis of eigenvectors in dimension two, so $J$ is not diagonalizable.

For $y^{\prime \prime }-2x{y}^{\prime }+2\lambda y=0$, substitute a power series $y={\sum }_{n\ge 0}{a}_n{x}^n$ to obtain the displayed recurrence relation. It separates into an even solution with $a_0=1,a_1=0$ and an odd solution with $a_0=0,a_1=1$:The ratio of successive terms within each parity tends to zero at every fixed complex $x$, giving infinite radius of convergence. The two initial-value vectors are independent, so these solutions span the solution space. A nonzero polynomial solution exists precisely when $\lambda =N$ is a nonnegative integer: its highest power forces this value, and the recurrence terminates the matching parity. These terminating solutions are nonzero scalar multiples of the Hermite polynomials. Multiplication by $e^{-x^2/2}$ connects them to Hermite oscillator quantization.

In a two-factor tensor product over a field, a pure tensor is also called a separable or decomposable tensor: it can be expressed as one product $v\otimes w$. Every tensor is a finite sum of pure tensors, but a sum need not itself be a pure tensor.^[ref]

For finite-dimensional vector spaces, the tensor-product coefficient matrix gives a decisive test: a nonzero tensor is a pure tensor exactly when its matrix rank is one. Indeed, every column of $v{w}^T$ is a scalar multiple of $v$. Conversely, if all columns of a nonzero matrix are multiples of one nonzero column $v$, collect their multipliers in $w$ to obtain $C=v{w}^T$. The zero tensor is also a pure tensor, for example $0\otimes w$, but its matrix rank is zero.

In ${\mathbb{R}}^2{\otimes }_{\mathbb{R}}{\mathbb{R}}^2$, the tensoris nonseparable: its matrix rank is two and its determinant is one, whereas $\det (v{w}^T)=0$. Directly, trying $t=(a{e}_1+b{e}_2)\otimes (c{f}_1+d{f}_2)$ requires $ac=bd=1$ and $ad=bc=0$. The first two equations force all four scalars nonzero, contradicting the others. In contrast, products sharing a factor do combine: $e_1\otimes f_1+e_1\otimes f_2=e_1\otimes (f_1+f_2)$. Thus being written as two summands alone does not establish nonseparability.