2
$\begingroup$

There is a question that comes up in my mind after I watched Prof. Gilbert Strang's lectures. He was saying:

For any matrix $A$, Since $A = LU$, $\det(A) = \det(LU)$ and $\det(L) = 1$, hence $\det(A) = \det(U)$, furthermore, $\det(U) = \det(D)$ when $U$ is the upper triangular and $D$ is the diagonal.

Then do I get the conclusion that $\det(A) = \det(D)$???

However, I don't think this is the case, to the least, not true in 2-D, as we all know what is the formula of determinant of a 2-D matrix.

Am I missing something here? Thanks!

$\endgroup$
  • $\begingroup$ It sounds like you're showing that $\det A=\det U$ and then $\det U=\det D$. With this it is immediate that $\det A=\det D$, correct? $\endgroup$ – Brian Fitzpatrick Sep 1 '14 at 4:10
2
$\begingroup$

One remark first, the determinant is defined for square matrices, not for vectors. Maybe the problem here is you view $D$ as a vector, whereas what is meant is "the square matrix with the same diagonal as $U$ and zeros everywhere else".

The LU decomposition yields a lower triangular matrix $L$ and an upper triangular matrix $U$ with

$$A=LU$$

The the determinant of a product is always the product of the determinants, it's perfectly safe to write

$$\det A=\det L \det U$$

Now, the determinant of a triangular matrix is the product of it's diagonal elements, and $L$ has only ones in its diagonal, whereas the diagonal of $U$ may be called $D$, and

$$\det A=\det D$$

For example, with $A=\left(\begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 1 & 2 & 2 \\ \end{matrix}\right)$, you get the factorization

$$A=\left(\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 2 & 1 \\ \end{matrix}\right)\cdot\left(\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 5 \\ \end{matrix}\right)$$

And of course

$$\det \left(\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & 2 & 1 \\ \end{matrix}\right)=\det \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix}\right)=1$$

$$\det \left(\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & 5 \\ \end{matrix}\right)=\det \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 5 \\ \end{matrix}\right)=5$$

Hence $\det A=5$.

There is another $LU$ factorization, with ones in the diagonal of $U$ instead of $L$, but of course that does not change the answer, if $D$ is the diagonal of $L$.


There may be another concern: often, the $LU$ decomposition is written

$$PA=LU$$

Where $P$ is a permutation matrix. If happens when pivoting is used in de $LU$ decomposition.

Then $\det P \det A=\det L\det U$, but $\det P=\pm1$, since a permutation matrix is always orthogonal. Thus you must be careful with the sign of $\det A$. And $\det P$ is simply the sign of the permutation on which $P$ is based.

$\endgroup$
  • $\begingroup$ Thanks very much for the explanation Jean. Now I think where I got confused are the following points: 1. Somehow, I thought during the LU process, A's diagonal entries do not have to change, but I guess this is not true. 2. det(D) = product of diagonals, however det(A) is not equal to its diagonal entries. $\endgroup$ – Yue Harriet Huang Sep 1 '14 at 4:31
  • $\begingroup$ A determinant is equal to the product of diagonal entries usually only when the matrix is diagonal or triangular (it may happen in other cases, but it's not guaranteed). $\endgroup$ – Jean-Claude Arbaut Sep 1 '14 at 4:34
  • $\begingroup$ @YueHarrietHuang Your guess is right, the diagonal elements of $U$ are not necessarily the same as those of $A$. There is also the "PA=LU" decomposition, see my edit above. $\endgroup$ – Jean-Claude Arbaut Sep 1 '14 at 5:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.