# LU factorization of a 2X2 singular matrix [duplicate]

My book is asking me to show that every matrix of the form $A=\begin{pmatrix}0&a\\0&b\end{pmatrix}$ has a $LU$-factorization and to show that even if $L$ is a unit lower triangular matrix that the $LU$ factorization is not unique.

Doesn't this contradict the theorem for $LU$ factorization which says for a matrix $A$: If the leading principal minors of $A$ are all nonsingular, then $A$ has an $LU$ decomposition. Clearly, the leading principal minors of $A$ are singular.

What does this mean? What is the $LU$ factorization?

## marked as duplicate by M.H, rschwieb, Amzoti, user67258, Cameron BuieJul 19 '13 at 3:06

• Just yesterday: math.stackexchange.com/questions/445457/… – Amzoti Jul 19 '13 at 1:09
• That makes sense for the second part of my question, but I do not understand how that gives the first part of my question. That assumes we have an $LU$ factorization. – CodeKingPlusPlus Jul 19 '13 at 1:17

Notice that $A$ is upper triangular, so you can just set $L = {\rm I}_2$, $U = A$.

Without that, we can do some computation. I denote elements of $L$ as $l_{ij}$ and elements of $U$ as $u_{ij}$ (both for $i=1,2$ and $j=1,2$).

If the first row of $L$ is zero, then the first row of $A$ is zero. So, $l_{11} \ne 0$, i.e., $l_{11} = 1$. By definition, $l_{12} = 0$ and $u_{21} = 0$. We now have: \begin{align*} 0 &= l_{11}u_{11} + l_{12}u_{21} = u_{11} &\Rightarrow \quad u_{11} = 0, \\ a &= l_{11}u_{12} + l_{12}u_{22} = u_{12} &\Rightarrow \quad u_{12} = a, \\ 0 &= l_{21}u_{11} + l_{22}u_{21} = 0, \\ b &= l_{21}u_{12} + l_{22}u_{22} = l_{21}a + l_{22}u_{22}. \end{align*} Since there are no more constraints, we can choose $l_{21} = 0$, $l_{22} = 1$, $u_{22} = b$ (because such choice is convenient). Finally, we get the solution from the begining of this answer: $$\begin{bmatrix} 0 & a \\ 0 & b \end{bmatrix} = A = LU = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & a \\ 0 & b \end{bmatrix}.$$ However, a different choice of $l_{21}, l_{22}, u_{22}$ would have produced a different LU factorization of $A$.

Doesn't this contradict the theorem for $LU$ factorization which says for a matrix $A$: If the leading principal minors of $A$ are all nonsingular, then $A$ has an $LU$ decomposition. Clearly, the leading principal minors of $A$ are singular.

It says "if $A$ has something, then it has LU". It doesn't say that $A$ doesn't have LU if some of the leading minors are singular.