Proving Invertible Matrices: A and B are n × n Matrices

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In summary: In this case, you need to remember that A is invertible if and only if it can be reduced by elementary row operations to the identity matrix.In summary, if A is invertible and AB = 0, then B = 0. This can be shown by multiplying both sides of AB=0 by A^-1, reducing it to A^-1B=0, and then using the fact that A is invertible to reduce it to B=0. It is necessary to state the step of multiplying by A^-1 in the proof. When approaching proofs, it is important to have a strong understanding of definitions and theorems and use them to logically connect the hypothesis to the conclusion.
  • #1
charlies1902
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Let A and B be n × n matrices.
a. Show that if A is invertible and AB = 0, then
B = 0.



If A is invertible, it can be reduced to the I matrix.
Thus IB=0 (this is the part where I'm hesitant, can I say that IB=0?)
Thus B=0 since I≠0
 
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  • #2
charlies1902 said:
Let A and B be n × n matrices.
a. Show that if A is invertible and AB = 0, then
B = 0.



If A is invertible, it can be reduced to the I matrix.
Thus IB=0 (this is the part where I'm hesitant, can I say that IB=0?)
Thus B=0 since I≠0

How did you reduce A to the I matrix? What operation(s) did you perform?
 
  • #3
jbunniii said:
How did you reduce A to the I matrix? What operation(s) did you perform?

Multiplying A^-1 to both sides. Is this necessary to state in the proof?

When I got the BI=0, I think I got it for the wrong reasons. I'm not very good at proofs. It took me awhile to think of multiplying A^-1 to both sides that that's how it justified BI=0. Typically when you go about doing proofs, how do you think in the "right" direction? Proofs seem to be so open ended compared to computation problems, that if you're not thinking in the "right" direction, it might take you a very long time to get it. Do you have any tips?
 
  • #4
charlies1902 said:
Multiplying A^-1 to both sides. Is this necessary to state in the proof?
Yes, it is necessary! That is, in fact, exactly what you are doing.

When I got the BI=0, I think I got it for the wrong reasons. I'm not very good at proofs. It took me awhile to think of multiplying A^-1 to both sides that that's how it justified BI=0.
No, you do NOT get "BI= 0". You had AB= 0 and you multiply each side by A on the left (remember that matrix multiplication is NOT commutative) so you get "A^{-1}(AB)= A^-1A)B (matrix multiplication IS associative)= IB= A^{1}0.

Typically when you go about doing proofs, how do you think in the "right" direction? Proofs seem to be so open ended compared to computation problems, that if you're not thinking in the "right" direction, it might take you a very long time to get it. Do you have any tips?
Yes, proofs require thinking rather than just following what you were told to do. To do that you going to have to know the precise statement of definitions and other theorems and use the to "build a bridge" from the hypothesis to the conclusion.
 

What is an invertible matrix?

An invertible matrix, also known as a non-singular matrix, is a square matrix that has a unique solution for every system of linear equations with non-zero determinant. This means that it can be inverted or multiplied by another matrix to produce the identity matrix.

How do you know if a matrix is invertible?

A matrix is invertible if its determinant is non-zero. This means that the matrix has a unique solution for every system of linear equations and can be inverted to produce the identity matrix. In other words, if the determinant is equal to zero, the matrix is not invertible.

What is the inverse of an invertible matrix?

The inverse of an invertible matrix is another matrix that, when multiplied together, produces the identity matrix. It is denoted by A-1, where A is the original invertible matrix.

How do you find the inverse of a matrix?

The inverse of a matrix can be found by using the Gauss-Jordan elimination method, where the matrix is manipulated into reduced row echelon form. Alternatively, the inverse can also be found by using the adjugate matrix method, where the inverse is found by taking the transpose of the cofactor matrix divided by the determinant of the original matrix.

Why are invertible matrices important?

Invertible matrices are important in various areas of mathematics, including linear algebra, differential equations, and geometry. They allow for the representation of linear transformations and the solution of systems of equations, and are also used in applications such as cryptography and computer graphics.

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