- #1
Buffu
- 849
- 146
For a ##n\times n## matrix A, the following are equivalent.
1) A is invertible
2) The homogenous system ##A\bf X = 0## has only the trivial solution ##\mathbf X = 0##
3) The system of equations ##A\bf X = \bf Y## has a solution for each ##n\times 1 ## matrix ##\bf Y##.
I have problem in third part of the question so I skip proof of equivalence of first two parts as there are trivial equivalent.
If ##A## is invertible, the solutions of ##A \bf X = \bf Y## is ## A^{-1} \bf {Y} = \bf {X}##. Conversely suppose ##A\bf X = \bf Y## has a solution for each given ##\bf Y##. Let ##R## be a row reduced echelon matrix which is row equivalen to ##A##. We wish to show that ##R = I##. That amounts to showing that the last row is ##R## is not ##0##. Let ##E = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix}##
If the system ##R \bf X = \bf E## can be solved for ##X##, the last row of ##R## cannot be ##0##. We know that ##R = PA## where ##P## is invertible. Thus ##R\bf X = \bf E## if and only if ##A \mathbf X = P^{-1}\mathbf E##. According to (3) the latter system has a solution.
I get why last row should not be zero, if it is so then ##A## is not invertible, but I don't what is special in last row ? if any row(s) are zero ##A## will not be invertible, invalidating our claim.
1) A is invertible
2) The homogenous system ##A\bf X = 0## has only the trivial solution ##\mathbf X = 0##
3) The system of equations ##A\bf X = \bf Y## has a solution for each ##n\times 1 ## matrix ##\bf Y##.
I have problem in third part of the question so I skip proof of equivalence of first two parts as there are trivial equivalent.
If ##A## is invertible, the solutions of ##A \bf X = \bf Y## is ## A^{-1} \bf {Y} = \bf {X}##. Conversely suppose ##A\bf X = \bf Y## has a solution for each given ##\bf Y##. Let ##R## be a row reduced echelon matrix which is row equivalen to ##A##. We wish to show that ##R = I##. That amounts to showing that the last row is ##R## is not ##0##. Let ##E = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \\ 1 \end{bmatrix}##
If the system ##R \bf X = \bf E## can be solved for ##X##, the last row of ##R## cannot be ##0##. We know that ##R = PA## where ##P## is invertible. Thus ##R\bf X = \bf E## if and only if ##A \mathbf X = P^{-1}\mathbf E##. According to (3) the latter system has a solution.
I get why last row should not be zero, if it is so then ##A## is not invertible, but I don't what is special in last row ? if any row(s) are zero ##A## will not be invertible, invalidating our claim.