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