Let $M$ be an $n\times n$ invertible matrix. Show that $M=U\Delta$ where $\Delta$ is an upper triangular matrix and $U$ is a unitary matrix.
So I started by using Gram-Shmidt since the columns of $M$ form a basis for our vector space. Then if $v_k$ is the $k $-th column vector of $M$ then by Gram-Schmidt we have $$v_k=u_k+\sum_{j=1}^{k-1}\frac{\langle v_k,u_j\rangle}{\langle u_j,u_j\rangle}u_j$$ for $1\leq k\leq n$ and where the $u_k$'s form an orthogonal basis generated by Gram-Shmidt.
From this I think we will have our unitary matrix $U$ but how do we get $\Delta$?
Best Answer
You are in the right direction. From the formulation you have, observe that we can rewrite it as
\begin{align}v_1 &= \Delta_{11}u_1 \\ v_2 &= \Delta_{12}u_1+\Delta_{22}u_2 \\ v_2 &= \Delta_{13}u_1+\Delta_{23}u_2 + \Delta_{33}u_3 \\ &\dots~\dots~\dots\end{align} where I define $$\Delta_{ij} = {<v_i,u_j> \over <u_j,u_j>}~,~\forall i\leq j~,~\{\mbox{0 elsewhere}\}$$ Now try to see if $\Delta$ is the upper triangular matrix you are looking for.