Let $A$ be an $m \times n$ complex matrix (let's swap $m$ and $n$ to make the exercise consistent with Exercise 10.8).
a)
The matrix $n \times n$ matrix $A^{\dagger}A$ is Hermitian since $(A^{\dagger}A)^{\dagger} = A^{\dagger}A$, so it has real eigenvalues $\lambda_j$ and an orthonormal basis $\{\ket{u_j}\}$ ($j = 0,\ldots,n-1$).
Furthermore, the eigenvalues are non-negative since
b)
Think of $\ket{u_j}$ as a column $n$-vector in the standard basis and let $U$ be the $n \times n$ matrix formed by the columns $\ket{u_j}$. Since the vectors $\{\ket{u_j}\}$ are orthonormal, $U$ is unitary.
c)
Let $\ket{v_i} = \frac{A\ket{u_i}}{\sqrt{\lambda_i}}$ for $\lambda_i \neq 0$. Let $V$ be the $m \times m$ matrix formed by the column $m$-vectors $\{\ket{v_i}\}$. The columns $\ket{v_i}$ are orthonormal since
Therefore $V$ is unitary.
d)
Consider the $m \times n$ matrix $D = V^{\dagger}AU$. The $ij$'th entry is
The shows that $D$ is diagonal with the non-negative diagonal entries $\sqrt{\lambda_i}$.
e)
In particular, $A$ has a diagonal decomposition $A = VDU^{\dagger}$ with the singular values $\sqrt{\lambda_i}$ in the diagonal. If $m \leq n$, then some $\lambda_i$ will be 0, corresponding to the kernel of $A$. The non-zero $\lambda_i$ will be present in the diagonal of $D$: