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Theorem: Let A be a square matrix. there exists an invertible matrix B such that BA equals a triangular matrix.

My way:

case 1 :

if A = 0, then for every B : BA = 0, which is a triangular matrix.

case2 :

if A is regular then take B = A inverse, and then BA=I ,a triangular matrix

cas3 : (problem here )

if A does not equal to the zero matrix and also not invertible:

BA=C is the matrix whose columns are : Ba1 Ba2 ... Ban.

let B, be B inverse. so multiply by B' from the left you get : B'c1 = a1; B'c2 = a2; ...; B'cn = an;

B' is regular therefore row equivalent to In and therefore there's exists a uniqe solution for any system of equations of the form above. we built T 1-1 and onto such that T(B') = B'c = a'

a' is a column of a matrix A such that BA = C.

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  • $\begingroup$ didn't notice. changed. $\endgroup$ – idan di Jan 21 '16 at 20:01
  • $\begingroup$ What's the result of Gaussian elimination? Given that $B$ is a permutation matrix, i.e. the identity matrix with its $i$th and $jth$ row swapped - what's the result of $BA$? Additionally, consider the case where $B$ is the identitly matrix minus the matrix which is everywhere 0, except in the i-th row and j-th where it's $\lambda$. $\endgroup$ – Roland Jan 21 '16 at 20:08
  • $\begingroup$ If you know the Gram-Schmidt ortogonalization process this might help you... $\endgroup$ – Dac0 Jan 21 '16 at 21:21
  • $\begingroup$ @Dac0 , yes I know the process, but how that can help me ? $\endgroup$ – idan di Jan 21 '16 at 21:32
  • $\begingroup$ What you are searching for are elementary matrices. $\endgroup$ – Jendrik Stelzner Jan 22 '16 at 7:48
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Performing row operations is the same as multiplying on the left by an invertible matrix. Now a series of row operations can achieve a triangular matrix.

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  • $\begingroup$ so if I have a singular matrix A and I do row operations until got to a triangular matrix. It's the same as multiply A for the left by an invertible matrix B ? where can I find a proof for this. $\endgroup$ – idan di Jan 21 '16 at 21:34
  • $\begingroup$ @idandi Yes, its a pretty standard result in LA, its in Hoffman and Kunze for sure. $\endgroup$ – Rene Schipperus Jan 22 '16 at 18:41
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Start from the Jordan Canonical Form of the Matrix A. $$P^{-1} A P =J$$ $J$ is triangular so if we can find $AP=XA$ we are done since in this case your $B=P^{-1}X$. But since P is invertible the system of equations $$AP=XA$$ is well defined and has at least one solution (could be more then one)

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Thanks to all helpers.

All are correct.

Proof:

Let A be a square regular matrix. Any such matrix is row equivalent to a triangular matrix. Let perform elementary row operations until getting to a triangular matrix C. Arrange the rows to have a pivot entry at index (0,0). Make all below entries to zero. If the second column is all zeroes continue to the next columns and make all entries below the pivot to be zero. Continue. Now, from matrix A you got to a triangular matrix C.

Theorem: Let A and C be both a square matrix. If A and C are row equivalent then there exists a regular matrix B Such that BA = C;

By the theorem there's a regular matrix B such that BA = C.

C is a triangular matrix therefore BA Is a triangular matrix.

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