If you use this algorithm in finite-precision environments (such as a computer) you should use the MAX pivot on the column, instead of an arbitrary non-zero one, in order to have a multiplier less than or equal 1, which is better because it will leave the algorithm stable and won't affect precision of results!
At 5:10, when multiplying the L matrix on both sides by a permuation matrix, it is important that the columns being swapped only have 1's on the diagonal. Imagine that you are switching two non-adjacent rows/columns, say 4 and 6. The column swap could moves the (5, 4) entry, below the diagonal, into the (5, 6) entry, above the diagonal. But, since we do this operation to columns that only have zeros except on the diagonal, the lower triangular structure is presereved.
@@hiuwakwan7377 Matrix A has to be decomposed into the form PLU. So in the equation A = (P_12)L(P_34)U, we have to make (L(P_34)) into a new L. And because the result of (L(P_34)) will not be a lower triangular matrix, we have to switch the row 3 and 4 again.
what if in the final step you did in order to put the permutation matrix in the right spot- the "L" matrix did not end up as a lower triangular matrix.
Hi, I have a question about the final result. Based on my textbook, the format is written as "PA=LU" Your final result becomes "A=PLU" So does that mean it mean that the P is also the same either in PA = LU or A = PLU, or in PA = LU, the P shall be the inverse from A = PLU? I also figured out that if P x P = I-Matrix (1, 0, 0; 0, 1, 0; 0, 0, 1) [This is calculated that both Ps are same and not from your examples but from random thoughts.] Which means P in inverse is exact the same as P also. I am looking forward for your help. Thank you! 🙏
Hi, I have a question. When we are doing the normal LU decomposition, we do the reverse order, which means we start at the end. However, when we are doing PLU decomposition, why do we start at the front?
Don't forget to download the pdf version for free!
tbsom.de/s/ov
PLU? More like "Please and thank you!" These videos are solid gold.
You’ve got kind heart... I suscribed.. from being a stubborn guy to a subscriber of bright side .. Thanks a lot
If you use this algorithm in finite-precision environments (such as a computer) you should use the MAX pivot on the column, instead of an arbitrary non-zero one, in order to have a multiplier less than or equal 1, which is better because it will leave the algorithm stable and won't affect precision of results!
Thanks for the comment. You are absolutely right. My algorithm is a pen-and-paper strategy and in general is not numerically stable.
@@brightsideofmaths but you made it easy, keep going ❤️
@@marianoaponte2518 can you please explain in detail your point ? I am not able to get it😢
@@readbhagwatgeeta3810 Computers can divide perfectly because rounding errors occur. So you want to avoid too many rounding errors in your algorithm.
Where would you pivot around for it to be the max? I don’t get the terminology here.
At 5:10, when multiplying the L matrix on both sides by a permuation matrix, it is important that the columns being swapped only have 1's on the diagonal. Imagine that you are switching two non-adjacent rows/columns, say 4 and 6. The column swap could moves the (5, 4) entry, below the diagonal, into the (5, 6) entry, above the diagonal. But, since we do this operation to columns that only have zeros except on the diagonal, the lower triangular structure is presereved.
Very good. Always remember that the PLU decomposition is just the Gaussian elimination in the correct order :)
I was so confused with how the row exchanges affect the L matrix, with just 10 seconds of the video i was finally capable to understand, thank you
Nice! Thank you :)
i am so confused about 5:06 (exchange the column of c3 and c4). would you explain deeply?
Just apply the matrix P_34 from the right. Write the calculation down and you see it :)
@@brightsideofmaths i donno why we need to change the column and row of lower triangular matrix.
@@hiuwakwan7377 Matrix A has to be decomposed into the form PLU. So in the equation A = (P_12)L(P_34)U, we have to make (L(P_34)) into a new L. And because the result of (L(P_34)) will not be a lower triangular matrix, we have to switch the row 3 and 4 again.
@@hiuwakwan7377you must do that since you wanna get rid of the p matrix stuck between L and U.
Thanks for the nice explanation!
Thanks saved me alot of time
Thanks for the support :)
Thank you so much! It is really helpful!
You're welcome! :)
what if in the final step you did in order to put the permutation matrix in the right spot- the "L" matrix did not end up as a lower triangular matrix.
What do you mean exactly?
that's awesome ! thanks!!!
is A=PLU different from PA=LU? At my Uni we are supposed to use PA=LU, how do I convert one form to the other?
Multiplying with P :) You can find that in my Linear Algebra course tbsom.de/s/la
Hi, I have a question about the final result.
Based on my textbook, the format is written as "PA=LU"
Your final result becomes "A=PLU"
So does that mean it mean that the P is also the same either in PA = LU or A = PLU, or in PA = LU, the P shall be the inverse from A = PLU?
I also figured out that if P x P = I-Matrix (1, 0, 0; 0, 1, 0; 0, 0, 1) [This is calculated that both Ps are same and not from your examples but from random thoughts.] Which means P in inverse is exact the same as P also.
I am looking forward for your help. Thank you! 🙏
Multiply P^{-1} to both side from the left to get the other result.
@@brightsideofmaths Ok, thank you very much!
More details in my book, see link in the description.
Nicely explained !!
Hi, I have a question. When we are doing the normal LU decomposition, we do the reverse order, which means we start at the end. However, when we are doing PLU decomposition, why do we start at the front?
What do you mean by starting at the end? You can check my video about the LU-decomposition to see the differences :)
This is such a helpful video, thank you so much
PLEASE I don t undestand why we have just 4 column in the L matrix and in the U matrix we have 5
U has the same form as A
I didn't really get why you were able to do those final row and column exchanges to the identity matrix without doing them to the matrix U
I have a linear algebra course where I go into the theoretical detail of row exchanges: tbsom.de/s/la
thank you so much
I did not understand aabout the squaring and column exchange.
I have a whole Linear Algebra course about it: tbsom.de/s/la
nice
Thanks