In section 2.3 of the Qiskit textbook (Phase Kickback), there's an example where a controlled-T gate is applied to $|1+\rangle$.
You're asked to attempt the same thing with $|0+\rangle$. I've done this by means of statevectors and successfully got the correct answer (that it has no effect): $$\text{Controlled-T}|0+\rangle$$ $$=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{\frac{i\pi}{4}} \end{pmatrix}\times\frac{1}{\sqrt{2}} \begin{pmatrix}1\\1\\0\\0\\\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\\0\\0\\\end{pmatrix}$$
The textbook's example (using $|1+\rangle$) performs the calculation in Dirac representation - i.e. by factoring out the individual qubits. I decided to try the same thing with the above. Here's what I got: $$|0+\rangle = |0\rangle \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$ $$=\frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)$$
$$\text{Controlled-T}|0+\rangle =\frac{1}{\sqrt{2}}(|00\rangle + e^{\frac{i\pi}{4}}|01\rangle)$$ $$=|0\rangle \otimes\frac{1}{\sqrt{2}}(|0\rangle + e^{\frac{i\pi}{4}}|1\rangle)$$ When I then multiply that out to get the statevector, I get: $$\frac{1}{\sqrt(2)}\begin{pmatrix}1\\e^{\frac{i\pi}{4}}\\0\\0\\\end{pmatrix}$$ which is clearly different. What am I doing wrong with the algebraic approach?
