I am reading a book on quantum computing. The author is constructing an arbitrary 2-Qbit state from unitary transformations. I need help understanding on step in his logic. For those unfamiliar with with quantum computing,
$$ \begin{align} |0\rangle &= \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad |1\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \end{align} $$
And
$$ \begin{align} |00\rangle &= \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}, \quad |01\rangle = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, \quad |10\rangle = \begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \quad |11\rangle = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \end{align} $$
He starts by noting that the general 2-Qbit state has the form
$$ |\Psi\rangle = a_{00}|00\rangle + a_{01}|01\rangle + a_{10}|10\rangle + a_{11}|11\rangle $$
This can also be written as:
$$ |\Psi\rangle = |0\rangle \otimes |\psi\rangle + |1\rangle \otimes |\phi\rangle $$ $$ |\psi\rangle = a_{00}|0\rangle + a_{01}|1\rangle $$ $$ |\phi\rangle = a_{10}|0\rangle + a_{11}|1\rangle $$
So far so good. Next the author says apply $ \textbf{u} \otimes \textbf{1} $ to $ |\Psi\rangle $, where $ \textbf{u} $ is a linear transformation, whose action on the computational basis is of the form:
$$ \textbf{u}|0\rangle = a|0\rangle + b|1\rangle, \quad \textbf{u}|1\rangle = -b^*|0\rangle + a^*|1\rangle; \quad |a|^2 + |b|^2 = 1 $$
The author doesn't state this, but I assume $ a,b \in \mathbb{C} $ and $ a^* $ and $ b^* $ are the complex conjugates of $ a $ and $ b $ respectively. Also unstated but assumed by me is that $ \textbf{u} $ is a $ 2x2 $ matrix and $ \textbf{1} $ is the $ 2x2 $ identity.
Now comes the part I don't understand. The author states:
$$ (\textbf{u} \otimes \textbf{1})|\Psi\rangle = (a|0\rangle + b|1\rangle) \otimes |\psi\rangle + (-b^*|0\rangle + a^*|1\rangle) \otimes |\phi\rangle $$
In order for the above to be true, it would seem that
$$ (\textbf{u} \otimes \textbf{1})|\Psi\rangle = \textbf{u} |0\rangle \otimes |\psi\rangle + \textbf{u}|1\rangle \otimes |\phi\rangle $$
But I don't understand why.