row reducing matrix,

[shelly89]

When looking for a possible new pivot, why do we only search from row k to n in column k? In other words, what unwanted thing could happen if we used a row above row k as the new pivot row?





am not really sure why, we choose the next pivot to be below the kth row, can someone explain mathematically why it works?

for example


\(\displaystyle \begin{bmatrix} 2 & 4 & -5 & 1 \\ 0 & 0 & -3 & 2 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & -4 & 1 \end{bmatrix} \) so the current pivot entry is at 3,3 what would happen if I choose a pivot entry above it?

 

[daon2]

When looking for a possible new pivot, why do we only search from row k to n in column k? In other words, what unwanted thing could happen if we used a row above row k as the new pivot row?





am not really sure why, we choose the next pivot to be below the kth row, can someone explain mathematically why it works?

for example


\(\displaystyle \begin{bmatrix} 2 & 4 & -5 & 1 \\ 0 & 0 & -3 & 2 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & -4 & 1 \end{bmatrix} \) so the current pivot entry is at 3,3 what would happen if I choose a pivot entry above it?

I originally misread what you had asked. But I agree with you. For example

\(\displaystyle \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \)

Why would you look for a pivot position in row 3 at all here? Is it possible you are misunderstanding the directions?

 

[shelly89]

I originally misread what you had asked. But I agree with you. For example

\(\displaystyle \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \)

Why would you look for a pivot position in row 3 at all here? Is it possible you are misunderstanding the directions?

I think that one is already row reduced? I dont think I am misunderstanding the directions?

 

[daon2]

I think that one is already row reduced? I dont think I am misunderstanding the directions?

The pivot position in column k is always between 1 and k, not k and n. If you meant to search entries >= k for a nonzero entry, then there are still counterexamples for this.

e.g. in the below matrix if you begin searching column four starting at entry 4 and down, you will not find any possible "swap" points.

\(\displaystyle \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)

So I don't understand what your teacher is asking you to do.

 

[shelly89]

The pivot position in column k is always between 1 and k, not k and n. If you meant to search entries >= k for a nonzero entry, then there are still counterexamples for this.

e.g. in the below matrix if you begin searching column four starting at entry 4 and down, you will not find any possible "swap" points.

\(\displaystyle \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}\)

So I don't understand what your teacher is asking you to do.

You see in my example, the next pivot is at -2, i.e along the diagonal. However the qustion says if we were to choose the next pivot entry above it, i.e -3 what would happen?

 

[daon2]

You see in my example, the next pivot is at -2, i.e along the diagonal. However the qustion says if we were to choose the next pivot entry above it, i.e -3 what would happen?

The pivot positions of a matrix are fixed, even before you being row-reducing. It doesn't matter which non-zero number you decide to move up.

 

[Ishuda]

When looking for a possible new pivot, why do we only search from row k to n in column k? In other words, what unwanted thing could happen if we used a row above row k as the new pivot row?





am not really sure why, we choose the next pivot to be below the kth row, can someone explain mathematically why it works?

for example


\(\displaystyle \begin{bmatrix} 2 & 4 & -5 & 1 \\ 0 & 0 & -3 & 2 \\ 0 & 0 & -2 & 0 \\ 0 & 0 & -4 & 1 \end{bmatrix} \) so the current pivot entry is at 3,3 what would happen if I choose a pivot entry above it?

Pivoting not only has to do with solutions to equations but with numerical accuracy and stability, see
http://en.wikipedia.org/wiki/Pivot_element
for example. So, my take on the answer to the 'don't look back' choice is that, supposibly, you have already optimized the previous rows so don't mess them up.

Note also that row swapping with the row above you would generally introduce a non zero element into a column less than k. The only reason it works in this particular example is that element 2,2 is zero. Normally, if there were to be a unique solution to the system, this would not occur and element 2,2, to use your example, would not be zero.