# How can I show that trace is Invariant under the change of basis?

• alphaneutrino
In summary, to show that trace is invariant under the change of basis, you can either explicitly express the change of basis as A' = B A B^{-1} and use the cyclic property of trace, or use the fact that the eigenvalues of a matrix are invariant under change of basis.
alphaneutrino
How can I show that trace is Invariant under the change of basis?

The easiest way is to explicitly express the change of basis as
$$A \to A' = B A B^{-1}$$
and use the cyclic property
$$\operatorname{Tr}(ABC) = \operatorname{Tr}(BCA) = \operatorname{Tr}(CAB)$$

Or you can use that
$$\operatorname{Tr}(A) = \sum \lambda$$
where $\lambda$ are the eigenvalues, and show that these are basis invariant
(for example, show that if v is a corresponding eigenvector, that $(B A B^{-1}) (B v B^{-1}) = \lambda v$ as well).

thanks CompuChip

$$(B A B^{-1}) (B v B^{-1}) = \lambda v$$
I should of course have said
$$(B A B^{-1}) (B v B^{-1}) = \lambda (B v B^{-1})$$
which simply says that if $A v = \lambda v[/tex] then [itex]A' v' = \lambda v'$ where the primes refer to matrices and vectors in the new basis (i.e. $A' = B A B^{-1}$ and $v' = B v B^{-1}$).

The trace of a matrix is defined as the sum of its diagonal elements. In other words, it is a measure of the magnitude of the matrix. To show that trace is invariant under the change of basis, we need to understand how the basis affects the matrix and its trace.

When we change the basis of a matrix, we are essentially changing the coordinate system in which the matrix is expressed. This means that the matrix itself will change, but its trace will remain the same. This is because the diagonal elements, which are used to calculate the trace, will remain the same regardless of the basis used.

To illustrate this, let's consider a matrix A expressed in the standard basis:

A = [a11 a12 a13
a21 a22 a23
a31 a32 a33]

The trace of this matrix is given by:

tr(A) = a11 + a22 + a33

Now, let's change the basis to a new basis B, where B = TA, where T is the transformation matrix. The new matrix B is given by:

B = [b11 b12 b13
b21 b22 b23
b31 b32 b33]

The trace of B is given by:

tr(B) = b11 + b22 + b33

But since B = TA, we can express the elements of B in terms of the elements of A as follows:

b11 = t11a11 + t12a21 + t13a31
b22 = t21a12 + t22a22 + t23a32
b33 = t31a13 + t32a23 + t33a33

Substituting these values in the expression for tr(B), we get:

tr(B) = (t11a11 + t12a21 + t13a31) + (t21a12 + t22a22 + t23a32) + (t31a13 + t32a23 + t33a33)
= (a11t11 + a12t21 + a13t31) + (a21t12 + a22t22 + a23t32) + (a31t13 + a32t23 + a33t33)
= a11(t11 + t22 + t33) + a22(t11 + t22 + t33) + a33(t11 + t22 + t33

## 1. How do I define trace?

Trace is defined as the sum of the diagonal elements of a square matrix. In other words, it is the sum of the elements on the main diagonal of a matrix.

## 2. What does it mean for trace to be invariant under change of basis?

If the trace of a matrix remains the same when the basis for the vector space is changed, then it is said to be invariant under change of basis. This means that the trace is independent of the choice of basis for the vector space.

## 3. How can I show that trace is invariant under change of basis?

To show that trace is invariant under change of basis, you can use the fact that the trace of a matrix is equal to the sum of its eigenvalues. Since eigenvalues are independent of the basis, the trace will remain the same regardless of the basis chosen.

## 4. Why is it important to show that trace is invariant under change of basis?

Proving that trace is invariant under change of basis is important because it is a fundamental property of matrices that allows for easier computation and manipulation. It also helps to simplify many calculations and proofs in linear algebra.

## 5. Can I use this property to simplify calculations involving trace?

Yes, knowing that trace is invariant under change of basis allows you to use different bases that may be more convenient for calculation purposes. This can help simplify calculations involving trace and make them easier to understand and work with.

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