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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.

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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 $L_1 = L_2$.

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  • $\begingroup$ 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? $\endgroup$ – dylan7 Jan 8 '15 at 23:02
  • $\begingroup$ @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. $\endgroup$ – Emily Jan 9 '15 at 5:26

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