Demonstration of the uniquenes of a QL matrix factorisation

Question

One of my teacher say to us that the QL decomposition is unique. I am not convinced. How can I demonstrate the decomposition is unique?

Answer 1

You can use the Gram-Schmidt process starting from the last vector of the matrix. Consider a matrix $$A=\left [ \begin{array}{c|c|c|c} v_1 & v_2 & \ldots &v_n \end{array} \right ]$$ Applying Gram-Schmidt,

$$ \begin{array} u u_n=v_n & e_{n}=\frac{u_n}{||u_n||} \\ u_{n-1}=v_{n-1}-(v_{n-1}\cdot e_n)e_n &e_{n-1}=\frac{u_{n-1}}{||u_{n-1}||} \\ \vdots & \vdots \\ u_{1}=v_1-(v_1\cdot e_2)e_2-\ldots -(v_{1}\cdot e_n)e_n & e_{1}=\frac{u_1}{||u_1||} \end{array}$$

Since $\langle v_n \rangle = \langle e_n\rangle, \ldots , \langle v_n,\ldots, v_1\rangle =\langle e_n,\ldots,e_1\rangle$ by construction, there exist real numbers $a_{(1j},\ldots a_{jj}$ for $1\le j \le n$ such that

$$v_{n-j}=\sum_{i=0}^{j}a_{((n-i),(n-j))}e_{n-i} \qquad 1\le j < n$$

We can suppose that these numbers are positive: if $a_{ij}$ is negative, it suffices to exchange $u_j$ with $-u_j$ and consider $-a_{ij}$. Then,

$$\left [ \begin{array}{c|c|c|c} v_1 & v_2 & \ldots &v_n \end{array} \right ]=\left [ \begin{array}{c|c|c|c} e_1 & e_2 & \ldots &e_n \end{array} \right ]\begin{pmatrix} a_{1,1} & 0 & \ldots & 0 \\ a_{2,1} & a_{2,2} & \ldots & 0\\ \vdots & \vdots & \ddots & 0 \\ a_{n,1} & a_{n,2} & \ldots & a_{n,n} \end{pmatrix}=QL$$

If the matrix has full rank all the entries of the matrix $L$ are strictly positive.

Now, suppose that $$A=Q_1L_1= Q_2L_2$$ with $Q_1,Q_2$ orthogonal, $L_1,L_2$ lower triangular. Then $$L_1^TL_1=L_1^TQ_1^TQ_1L_2=A^TA=L_2^TQ_2^TQ_2L_2=L_2^TL_2$$ Hence $$(L_2^{-1})^TL_1^T=L_2L_1^{-1}$$ Since $L_1^{-1},L_2$ are lower triangular, $L_2^{-1},L_1^T$ are upper triangular. This implies that the matrix $L_2L_1^{-1}$ is diagonal (it's upper and lower triangular). The above identity shows that if $a_i$,$b_i$ are the diagonal values of $L_1,L_2$, then $$\frac{a_i}{b_i}=\frac{b_i}{a_i}\implies a_i=b_i$$ because $a_i$, $b_i$ are positive. But then $$L_2L_1^{-1}=\mathbb{I} \implies L_2=L_1$$ Since $Q_1L_1=Q_2L_2$ and $L$ is invertible, $Q_1=Q_2$.