# Show that A and its transpose have the same eigenvalues

• Mr Davis 97
In summary, to show that ##A## and ##A^T## have the same eigenvalues, we need to show that ##|A - \lambda I| = 0## iff ##|A^T - \lambda I| = 0##. This is because if they have the same eigenvalues, then ##Ax = \lambda x## iff ##A^T x = \lambda x##. From the properties of transpose, we know that ##(A - \lambda I)^T = A^T - \lambda I##. Also, it is a property of transposes that ##A^T## is invertible iff ##A## is invertible. Therefore, we can conclude that ##A - \
Mr Davis 97

## Homework Statement

Show that ##A## and ##A^T## have the same eigenvalues.

## The Attempt at a Solution

If they have the same eigenvalues, then ##Ax = \lambda x## iff ##A^T x = \lambda x##. In other words, we have to show that ##|A - \lambda I| = 0## iff ##|A^T - \lambda I| = 0##. From the properties of transpose, we see that ##(A - \lambda I)^T = A^T - \lambda I##. It's a property of transposes that ##A^T## is invertible iff ##A## is also invertible. So we have shown that ##A - \lambda I## is invertible iff ##A^T - \lambda I## is also invertible. So this shows that they have the same eigenvalues.

I think that this is the correct solution, but I am a little confused about the beginning part of the proof. Does this imply that A and its transpose also have the same eigenvectors? Or do they have the same eigenvlues but different eigenvectors?

Mr Davis 97 said:

## Homework Statement

Show that ##A## and ##A^T## have the same eigenvalues.

## The Attempt at a Solution

If they have the same eigenvalues, then ##Ax = \lambda x## iff ##A^T x = \lambda x##. In other words, we have to show that ##|A - \lambda I| = 0## iff ##|A^T - \lambda I| = 0##. From the properties of transpose, we see that ##(A - \lambda I)^T = A^T - \lambda I##. It's a property of transposes that ##A^T## is invertible iff ##A## is also invertible. So we have shown that ##A - \lambda I## is invertible iff ##A^T - \lambda I## is also invertible. So this shows that they have the same eigenvalues.

I think that this is the correct solution, but I am a little confused about the beginning part of the proof. Does this imply that A and its transpose also have the same eigenvectors? Or do they have the same eigenvlues but different eigenvectors?
You don't need to consider invertibility here. And you have already said everything you need. Just clean it up a little.
What are the eigenvalues of ##A##?
What are the eigenvalues of ##A^T##?
And how are ##\det B## and ##\det B^T## related for any matrix ##B##? (Esp. for ##B=A-\lambda I##.)

fresh_42 said:
You don't need to consider invertibility here. And you have already said everything you need. Just clean it up a little.
What are the eigenvalues of ##A##?
What are the eigenvalues of ##A^T##?
And how are ##\det B## and ##\det B^T## related for any matrix ##B##? (Esp. for ##B=A-\lambda I##.)
##det(A) = det(A^T)## for all square ##A##. So we have that ##det(A - \lambda I) = det((A - \lambda I)^T) = det(A^T - \lambda I)##

fresh_42
fresh_42 said:
You don't need to consider invertibility here. And you have already said everything you need. Just clean it up a little.
What are the eigenvalues of ##A##?
What are the eigenvalues of ##A^T##?
And how are ##\det B## and ##\det B^T## related for any matrix ##B##? (Esp. for ##B=A-\lambda I##.)
So as a last note do ##A## and ##A^T## also have the same eigenvectors?

## What does it mean to have the same eigenvalues?

Having the same eigenvalues means that two matrices have the same set of numbers that, when multiplied with the matrix, result in the same vector. In other words, the matrices have the same characteristic polynomial.

## Why is it important to show that A and its transpose have the same eigenvalues?

This is important because it provides us with useful information about the properties of the matrix. It also helps us understand the relationship between a matrix and its transpose, which can be useful in various mathematical applications.

## How can we show that A and its transpose have the same eigenvalues?

To show that A and its transpose have the same eigenvalues, we can use the fact that the characteristic polynomial of a matrix is independent of its orientation. This means that the characteristic polynomial of A and its transpose will be the same, and therefore, they will have the same eigenvalues.

## What are some real-world applications of matrices with the same eigenvalues?

Matrices with the same eigenvalues have various real-world applications, such as in physics, engineering, and computer science. For example, they can be used to model physical systems, analyze data, and solve optimization problems.

## Are there any exceptions where A and its transpose may not have the same eigenvalues?

Yes, there are some exceptions. For example, if the matrix A is not square, then it will not have the same eigenvalues as its transpose. Additionally, if A is not a symmetric matrix, then it may not have the same eigenvalues as its transpose.

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