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By the spectral theorem, the nonscalar unitary V has orthonormal eigenvectors $|v_0⟩,|v_1⟩$ with distinct eigenvalues ${\lambda }_0\ne {\lambda }_1$. Choose $|b⟩=(|v_0⟩+|v_1⟩)/\sqrt{2}$. Then $V|b⟩=({\lambda }_0|v_0⟩+{\lambda }_1|v_1⟩)/\sqrt{2}$ is not proportional to $|b⟩$.

On the product input $|+⟩\otimes |b⟩$, the controlled unitary gate producesIts two conditional target vectors are linearly independent, so the conditional-vector product-state criterion proves that the output is entangled. Equivalently, in the control computational basis and target eigenbasis its coefficient matrix is $\frac{1}{2}\left(\begin{array}{cc}1& 1\\ {\lambda }_0& {\lambda }_1\end{array}\right)$, whose determinant is $({\lambda }_1-{\lambda }_0)/4\ne 0$. Thus U is an entangling gate. The excluded scalar case would only apply a phase gate to the control and could not entangle the factors; this is the controlled-unitary entangling criterion.

For $\lambda =1$ and $\mu =1/2$, the third equation gives $x=z-1$. Combining it with $2x-z=-4$ gives $z=-2$ and $x=-3$, after which the second equation gives $y=12$. HenceThe coefficient matrix has determinant $\lambda -2=-1\ne 0$, proving uniqueness. Direct substitution gives the required right-hand sides $1$, $5$ and $-1$.

Gaussian elimination of the first two equations gives $x=2y-1$ and $z=3-4y$. The third then reduces to $(\alpha -2)y=\beta -3$. HenceIn the infinite family $y=t$, $x=2t-1$ and $z=3-4t$, for $t\in \mathbb{R}$. This makes the distinction between an inconsistent linear system and a free parameter explicit; its coefficient matrix has determinant $4-2\alpha$.

Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.

Row reduction, or the cofactor formula for a matrix inverse, givesMultiplying by the right-hand-side column $(5,13,35{)}^T$ gives, at $\lambda =3$,For general $\lambda$, the coefficient matrix has determinant $(\lambda -1)(\lambda -2)$. Thus other parameter values give a unique solution. If $\lambda =1$, subtracting the first equation from the second gives $y=8$, while subtracting it from the third gives $3y=30$, a contradiction. If $\lambda =2$, subtraction gives $y+z=8$ and $3y+3z=30$, again inconsistent. Hence

Collecting the coefficients gives the coefficient matrixA square homogeneous linear system has a nonzero solution precisely when its matrix is a singular matrix. Expanding the determinant down the first column,Hence the necessary and sufficient condition is $a=0$ or $b=c$ or $b=-c$. A zero determinant gives a nontrivial kernel of a linear map; otherwise multiplication by the matrix inverse would force the zero solution.

The square roots simplify to the absolute values $|a-b|$ and $|a+b|$. Put $s=\sin (\pi \sqrt{a^2+b^2})$. A nontrivial solution of the homogeneous linear system exists exactly when its coefficient matrix is singular:The second factor vanishes precisely when $(a-b{)}^2=(a+b{)}^2$, or $ab=0$, giving both coordinate axes. The first factor vanishes when $\sqrt{a^2+b^2}$ is a nonnegative integer. Hence the complete locus isThese are the two axes together with concentric circles of positive integer radii. The zero-radius case is the origin, already included on the axes; at that point both equations vanish and any nonzero pair $(x,y)$ works.

Let $\Psi \ne 0$ and take the reduced positive operator ${\rho }_1={\mathrm{Tr}}_2|\Psi ⟩⟨\Psi |$. Choose orthonormal eigenvectors ${\xi }_j$ for its positive eigenvalues ${\lambda }_j$. DefineThe defining identity of the partial trace gives $⟨v_i,v_j⟩={\lambda }_j{\delta }_{ij}$, so the ${\eta }_j$ are orthonormal. Expansion in a full eigenbasis of ${\rho }_1$ yields $\Psi ={\sum }_j{\xi }_j\otimes v_j$; zero-eigenvalue terms have zero squared norm and contribute nothing. ThusThe zero vector has the empty decomposition. Unit normalization would give ${\sum }_j{c}_j^2=1$, but no normalization is needed for the theorem itself.

For the given vector, the coefficient matrix is $M=\left(\begin{array}{cc}1& 0\\ 0& 1\\ 1& 1\end{array}\right)$. Its second reduced operator is $M^{\ast }M=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, whose eigenvalues are $3,1$ and eigenvectors ${\eta }_{+}=(E_1+E_2)/\sqrt{2}$, ${\eta }_{-}=(E_1-E_2)/\sqrt{2}$. The corresponding normalized first-factor vectors areTheir norms are one and their inner product is zero. HenceExpanding the two terms returns the original three contributions. The vector has squared norm $4$; if a normalized state is wanted, divide by two and the Schmidt coefficients become $\sqrt{3}/2$ and $1/2$.