= Solution
The corresponding conditional <vectors> are $|0\rangle/\sqrt2$ and $|+\rangle/\sqrt2$. The second has a nonzero $|1\rangle$ component while the first does not, so they are <linearly independent>. Therefore \b[the third state is also <entangled>]. <Orthogonality> of these two conditional <vectors> is not required for this conclusion.
For an independent <matrix> check, expanding the second factor gives
$$
C=\begin{pmatrix}1/\sqrt2&0\\1/2&1/2\end{pmatrix},\qquad \det C=\frac1{2\sqrt2}\ne0.
$$
The original expression is normalized because its two first-wire factors are <orthogonal>: its squared <norm> is $(1+1)/2=1$, despite the overlap between $|0\rangle$ and $|+\rangle$ on the second wire.
Back to article page
