Is it possible to perform elementary row operations to every matrix in order to change its form into upper-triangle, without changing its determinant?

That is, by only adding multiples of other rows such that the determinant does not change. This would be of great help in for example proving that the determinant of a triangular block matrix is equal to the product of the determinants of its diagonal block matrices.

  • 1
    $\begingroup$ Any matrix is similar to an upper triangular matrix, so your first claim is true. $\endgroup$ – angryavian Mar 12 at 23:37
  • $\begingroup$ It's useful to notice that, although the elementary row operation of interchanging two rows is forbidden (as it changes the sign of the determinant), the interchange combined with reversing the sign of one of the two rows is OK. It can be achieved by adding multiples of a row to another: Subtract row 2 from row 1; then add row 1 to row 2; then again subtract row 2 from row 1. $\endgroup$ – Andreas Blass Mar 12 at 23:53
  • $\begingroup$ Thankyou! I really like the “trick” of interchanging rows but maintaining the same determinant. $\endgroup$ – Sam Mar 13 at 0:22
  • $\begingroup$ Just a note. The operation Andreas mentioned is a combination of 3 steps in my answer. $a_i' = a_i + a_j, \quad a_j' = a_j - a_i', \quad a_i'' = a_i' + a_j'$. $\endgroup$ – MoonKnight Mar 13 at 16:49

Short answer, yes.

According to the definition of matrix determinant : wiki $$ \det[a_1, ... , b\cdot a_j + c\cdot v, ... , a_n] = b\det[a_1, ... , a_j, ... , a_n] + c\det[a_1, ... , v, ... , a_n] $$

if $b=1$ and $v=a_k\{\text{with } j\neq k\}$ then,

  • obviously that is a basic column operation.
  • it does not change the determinant, since $\det[a_1, ...., a_k, ..., a_k,..., a_n] =0 $

The idea can easily also be applied to row operations. So a "valid" row operation that does not change the determinant of the matrix is $$ a_j' = a_j + \alpha \cdot a_k \quad (k\neq j) $$

The second part of the problem is how to use such row operations to get to a upper triangular matrix. It seems obvious to me, but I don't have a succinct way to write it down. Hope you can figure that out yourself.

  • $\begingroup$ Thank you very much for the help! $\endgroup$ – Sam Mar 13 at 0:19

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