# Uniqueness of LU decomp for non-singular matrices

Is the LU decomp for invertible matrices unique? I am reading a book that asks to prove it, but I can't because I don't think it makes sense that it could be unique. I looked online and at other posts it doesn't say it is anywhere. If it is, can anyone give me some intuition behind this?

Thank you.

The entries of the matrix cannot be unique, as we can always take $\underbrace{aL}_{\hat{L}}\underbrace{\frac{1}{a}U}_{\hat{U}} = A$ for $a \neq 0$. But suppose we impose a restriction that the diagonal entries of $L$ are some fixed value, say $1$.
Now suppose there were two unique such decompositions, $A = L_1 U_1$ and $A = L_2 U_2$.
Then we have $Ax = b \implies L_1U_1x = b = L_2U_2x$ being solved by a unique non-zero $x$.
Take $(L_1U_1 - L_2U_2)x = 0$. Now, this implies that $U_1 - L_1^{-1}L_2U_2 = 0$, meaning that the product $L_1^{-1}L_2$ must be a strictly-diagonal matrix.
Since the diagonal entries of $L_1$ and $L_2$ are restricted to be $1$, and since the diagonal elements in the product of two lower-triangular matrices are simply the product of the corresponding diagonal elements of the factors, then we have that $L_1^{-1}L_2 = I$. By uniqueness of the inverse, $L_1^{-1} = L_2^{-1}$ so $L_1 = L_2$.
• Ah, so you have to put some restriction on L. And if A is singular the $x$ is not unique and then how does that cause this to fail? – dylan7 Jan 8 '15 at 23:02
• @dylan7 If $A$ is singular, then $(L_1U_1-L_2U_2)x =0$ can hold for some nonzero $x$, but $L_1U_1 - L_2U_2$ need not be the zero matrix. – Emily Jan 9 '15 at 5:26