- #1
Buffu
- 849
- 146
If ##A## is ##m \times n## matrix, ##B## is an ##n \times m## matrix and ##n < m##. Then show that ##AB## is not invertible.
Let ##R## be the reduced echelon form of ##AB## and let ##AB## be invertible.
##I = P(AB)## where ##P## is some invertible matrix.
Since ##n < m## and ##B## is ##n \times m## therefore there is a non trivial solution to ##B\mathbf X = 0##
Let it be ##\bf X_0##
##I\mathbf X = P(AB)\mathbf X_0 \implies \mathbf X_0 = PA (B \mathbf X_0) = PA \times 0 = 0##
Which means ##\mathbf X_0 = 0##, which is contradiction. Therefore ##AB## is not invertible.
Is this correct ?
Let ##R## be the reduced echelon form of ##AB## and let ##AB## be invertible.
##I = P(AB)## where ##P## is some invertible matrix.
Since ##n < m## and ##B## is ##n \times m## therefore there is a non trivial solution to ##B\mathbf X = 0##
Let it be ##\bf X_0##
##I\mathbf X = P(AB)\mathbf X_0 \implies \mathbf X_0 = PA (B \mathbf X_0) = PA \times 0 = 0##
Which means ##\mathbf X_0 = 0##, which is contradiction. Therefore ##AB## is not invertible.
Is this correct ?