A linear operator U on a finite dimensional inner product space V is a partial isometry is there exists a subspace W of V that ||U(x)||=||x|| for all $x \in W$ and U(x)=0 for all $x \in W^{\perp}$. W need not be U-invariant. Suppose U is such an operator and ${v_1,...v_k}$ is an orthonormal basis for W. Prove there exists an orthonormal basis $\gamma$ for V such that the first k columns of $[U]_{\gamma}$ form an orthonormal set and the remaining columns are zero.
Try:Let $\gamma={v_1,...,v_n}$ be the orthonormal basis of U and let $A=[U]_{\gamma}$, then we know that $U(v_l)=0$ for $l>k$ based on text.
So let $U(v_j)=\sum_{i=1}^n U_{ij} v_i$, then we have for any g, h less than k-th column, we have $$<U(v_g),U(v_h)>=<\sum_{i=1} U_{ig} v_i,\sum_{i=1} U_{ih}v_i>=0$$ and $$<U(v_g),U(v_g)>=<\sum_{i=1} U_{ig} v_i,\sum_{i=1} U_{ig}v_i>=1$$.
I am not sure if that is the right argument.
As I understood, you extend the $(v_1,\dots, v_k)$ to an orthonormal basis $\gamma=(v_1,\dots, v_n)$ of $V$. Let $U=\|U_{ij}\|$ be the matrix of the operator $U$ in this basis. Then $U_{ij}=\langle Uv_j, v_i\rangle$ for each $1\le i,j\le n$. So if $j>k$ then $Uv_j=0$ and thus $U_{ij}=0$ for each $i$, implying that the $j$-th column $U_j$ of the matrix $U$ is zero. For each $1\le j,j’\le k$ we have $$\langle U_j, U_{j’}\rangle=\sum_{i=1}^n U_{ij}U_{ij’}=\sum_{i=1}^n \langle Uv_j, v_i\rangle\langle Uv_{j’}, v_i\rangle.$$
The last expression $A$ is an inner products of the vectors $Uv_j$ and $Uv_j$, calculated with respect to the basis $\gamma$. Since $v_j,v_{j’}\in W$ and $\|U(x)\|=\|x\|$ for each $x\in W$, it follows that $\langle Ux, Uy\rangle=\langle x, y\rangle$ for each $x, y\in W$, so $A$ equals $1$, if $j=j’$ and equals $0$, otherwise.