MHB Proving A is Zero Matrix if B is Invertible & Same Size as A

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If A and B are square matrices of the same size and B is invertible, then the equation AB = 0 implies that A must be the zero matrix. This is because multiplying by an invertible matrix (B) allows us to isolate A, leading to A = B^(-1) * 0, which simplifies to A = 0. The relationship between A and B is crucial, as the invertibility of B ensures that no non-zero matrix A can satisfy the equation. Thus, the conclusion is that A is indeed the zero matrix when B is invertible and AB equals zero. The proof hinges on the properties of matrix multiplication and the definition of invertibility.
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Show that if A and B are square matrices of the same size such that B is an
invertible matrix, then A must be a zero matrix.
 
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blueman11 said:
Show that if A and B are square matrices of the same size such that B is an
invertible matrix, then A must be a zero matrix.
We need information about how A and B are related. ie. AB = 0 or something.

-Dan
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...
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