I see. So then AB has a non-trivial kernel, which means that AB is not invertible.fresh_42 said:If B is not invertible, it has a non-trivial kernel. Take a vector from it and apply AB.
The same. Since A is invertible, we can find a y to an element x of B's kernel, such that x=Ay. Now Bx=0=BAy and y is in the kernel of BA.Mr Davis 97 said:I see. So then AB has a non-trivial kernel, which means that AB is not invertible.
What about if we wanted to show that BA is not invertible, given that B is not invertible?
Ah! Makes perfect sense. Thanksfresh_42 said:The same. Since A is invertible, we can find a y to an element x of B's kernel, such that x=Ay. Now Bx=0=BAy and y is in the kernel of BA.
For a matrix to be invertible, it means that it has an inverse matrix that, when multiplied together, results in the identity matrix.
No, a matrix cannot be invertible if its determinant is zero. The determinant is a key factor in determining if a matrix has an inverse.
If matrix B is not invertible, then matrix AB cannot be invertible. This is because the non-invertibility of B means that there is no inverse matrix that can cancel out its effects when multiplied with A.
A matrix is invertible if and only if its row operations result in the identity matrix. This means that the row operations must be able to transform the matrix into a simpler form that can be easily inverted.
No, a matrix cannot be invertible if it has non-zero entries in its null space. The null space represents the vectors that get mapped to the zero vector when multiplied by the matrix, and an invertible matrix cannot map any non-zero vector to the zero vector.