In Nielsen and Chuang's book on quantum computing, the following proof for the spectral theorem (any normal matrix is diagonalizable with unitary matrices) is given.
I'm unable to follow the statement that QMQ is diagonalizable by an orthogonal basis for the space Q by induction. To me, QMQ is an operator going from V to V, so it being diagonalizable with unitary matrices relies on assuming that a normal matrix in V is diagonalizable with unitary matrices, but the induction hypothesis only suggests this for subspaces with a smaller dimension than V.
Any clarification would be appreciated
I agree this is deserving of some clarification. We can find an orthonormal basis $v_1, \dots, v_n$ for $V$ such that $v_1, \dots, v_k$ is a basis for $P$ and $v_{k+1}, \dots, v_n$ is a basis for $Q$. Now, lets look at the matrices $PMP$ and $QMQ$ in these bases. We have $v_i$ is an eigenvector $PMP$ with eigenvalue $\lambda$ for $i=1, \dots, k$ and in the kernel of $PMP$ for $i = k+1, \dots, n$. Thus, $P$ has the block diagonal form $$ PMP = \begin{pmatrix} \Lambda & 0 \\ 0 & 0\end{pmatrix}$$ where $\Lambda$ is the $k\times k$ matrix $\Lambda = \operatorname{diag}(\lambda, \dots, \lambda)$. This is what they meant by
Now, for $QMQ$, we know that $v_i$ is in the kernel for $i = k+1, \dots, n$, and the image of $QMQ$ is contained in the subspace $Q$, so we may write $$QMQ = \begin{pmatrix} 0 & 0 \\ 0 & A \end{pmatrix}$$ for some $(n - k) \times (n-k)$ matrix $A$. Now, $QMQ$ being normal implies $A$ is normal, so there is an orthonormal basis $\mathcal B = \{w_{k+1}, \dots, w_{n}\}$ for $Q \cong \mathbb C^{n-k}$ such that $A$ is diagonal. If $B : \mathbb C^{n-k} \to \mathbb C^{n-k}$ is the change of basis matrix from our old basis to $\mathcal B$ then $BAB^{-1}$ is diagonal, the basis $v_1, \dots, v_k, w_{k+1}, \dots, w_n$ is orthonormal, and the matrix $$C = \begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix} $$ is our desired change of basis matrix. In other words, $$CMC^{-1} = CPMPC^{-1} + CQMQC^{-1} = \begin{pmatrix} \Lambda & 0 \\ 0 & BAB^{-1}\end{pmatrix}.$$