Show Invertibility of Similar Matrices: A & B

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In summary: You've shown something you think is B^(-1) times B is I. That proves that the thing that you think is B^(-1) is, in fact, B^(-1). If CB=I then C is B^(-1). I think this is how the exercise wants you to show...That B^(-1) exists and is unique when A and B are similar matrices.
  • #1
pyroknife
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If A and B are similar matrices, show that A has an inverse IFF B is invertible.

A=P^-1 * B * P
Where P is an invertible matrix.

(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P

Does this show what the question wanted?
 
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  • #2
pyroknife said:
If A and B are similar matrices, show that A has an inverse IFF B is invertible.

A=P^-1 * B * P
Where P is an invertible matrix.

(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P

Does this show what the question wanted?

Not until you explain in words how it does.
 
  • #3
This shows that B^-1 is similar to A^-1?I forgot to put this in the statement, the problem specifically asked to "use the "definition" of similar matrices to show this."

My original proof was that if A and b are similar then they must have the same determinant. If B is not invertible then it's determinant is 0 and A will have a determinant of 0 and is therefore not invertible. If B is intervible then it's determinant is not 0 and...A is invertible. But the teacher said I needed to use the definition.
 
  • #4
pyroknife said:
This shows that B^-1 is similar to A^-1?I forgot to put this in the statement, the problem specifically asked to "use the "definition" of similar matrices to show this."

My original proof was that if A and b are similar then they must have the same determinant. If B is not invertible then it's determinant is 0 and A will have a determinant of 0 and is therefore not invertible. If B is intervible then it's determinant is not 0 and...A is invertible. But the teacher said I needed to use the definition.

Ok, then use the definition. If A is invertible then A^(-1) exists. Use what you've got to give a formula for B^(-1). The show it works. And vice versa.
 
  • #5
Dick said:
Ok, then use the definition. If A is invertible then A^(-1) exists. Use what you've got to give a formula for B^(-1). The show it works. And vice versa.

Alright so, after this:
(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P
P^A^-1*P^-1=B^-1

Ugh, I guess I don't know what I need to do or maybe that I already have all the stuff, but not have worded it.
 
  • #6
pyroknife said:
Alright so, after this:
(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P
P^A^-1*P^-1=B^-1

Ugh, I guess I don't know what I need to do or maybe that I already have all the stuff, but not have worded it.

Ok, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?
 
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  • #7
Dick said:
Ok, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?

Wait from the definition, doesn't B=P^-1 * A *P?
 
  • #8
pyroknife said:
Wait from the definition, doesn't B=P^-1 * A *P?

No. P is a fixed matrix once you've used the definition. You said A=P^-1 * B * P. So B=PAP^(-1). You are misunderstanding what the definition of similarity means.
 
  • #9
Dick said:
No. P is a fixed matrix once you've used the definition. You said A=P^-1 * B * P. So B=PAP^(-1). You are misunderstanding what the definition of similarity means.

(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P
P^A^-1*P^-1=B^-1


B*P^A^-1*P^-1=B*B^-1=I
PAP^(-1)*P^A^-1*P^-1=I
PA*I*A^-1*P^-1=I
PP^-1=I
I=I


Ok, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?
 
  • #10
pyroknife said:
(A)^-1 = (P^-1 * B * P)^-1
A^-1 = P^-1 * B^-1 * P
P^A^-1*P^-1=B^-1B*P^A^-1*P^-1=B*B^-1=I
PAP^(-1)*P^A^-1*P^-1=I
PA*I*A^-1*P^-1=I
PP^-1=I
I=IOk, so you are claiming B^(-1)=PA^(-1)P^(-1). Check it. B=PAP^(-1) right, from your definition? Multiply your expression for B^(-1) by the expression for B using those forms. What do you get? How does that prove part of what you want?

Any words to got along with all of those typos to indicate whether you get it or not? This sequence is rightish. PAP^(-1)*P^A^-1*P^-1=PA*I*A^-1*P^-1=PP^-1=I. So?
 
  • #11
Dick said:
Any words to got along with all of those typos to indicate whether you get it or not? This sequence is rightish. PAP^(-1)*P^A^-1*P^-1=PA*I*A^-1*P^-1=PP^-1=I. So?

If P*P^-1=I Then that means P is invertible. I think I'm just stating the obvious. This showed that multipling B^-1 by B gives the identity matrix which means B is invertible.
 
  • #12
pyroknife said:
If P*P^-1=I Then that means P is invertible. I think I'm just stating the obvious. This showed that multipling B^-1 by B gives the identity matrix which means B is invertible.

You've shown something you think is B^(-1) times B is I. That proves that the thing that you think is B^(-1) is, in fact, B^(-1). If CB=I then C is B^(-1). I think this is how the exercise wants you to show it.
 
  • #13
By the way, in a more abstract sense, a matrix always represents a linear transformation in some basis. A "similar" matrix represents the same linear transformation in a different basis. Of course, whether or not a linear transformation is invertible is independent of the specific basis in which it is represented.
 

FAQ: Show Invertibility of Similar Matrices: A & B

1. What is the definition of similar matrices?

Similar matrices are two matrices A and B that have the same size and their corresponding entries follow a specific relationship. This relationship is that matrix B can be obtained from matrix A by performing a series of row and column operations, such as scaling, swapping, or adding rows/columns.

2. How do you show the invertibility of similar matrices A and B?

To show the invertibility of similar matrices A and B, we need to prove that both matrices have the same determinant. This means that if matrix A is invertible, then matrix B will also be invertible. We can also show this by finding the inverse of matrix A and using it to obtain the inverse of matrix B.

3. What is the importance of proving the invertibility of similar matrices?

Proving the invertibility of similar matrices is important because it allows us to perform operations on the matrices without affecting their properties. For example, if we have a system of linear equations represented by a matrix A, and we find that A is similar to another matrix B, we can use this relationship to solve the system of equations by performing operations on B instead of A.

4. Can two matrices be similar if they have different dimensions?

No, two matrices cannot be similar if they have different dimensions. Similarity is a property that is only defined for matrices of the same size. If two matrices have different dimensions, they cannot have the same entries and thus cannot satisfy the relationship required for similarity.

5. What are some real-world applications of similar matrices?

Similar matrices have many real-world applications, especially in fields that involve linear algebra, such as engineering, physics, and economics. For example, in physics, similar matrices are used to represent physical systems and perform calculations on them. In economics, similar matrices can be used to analyze the relationships between different economic variables. They can also be used to simplify calculations in computer graphics and image processing applications.

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