Proving the Rank Equivalence of Adjoint Operators

In summary, an adjoint operator T* is defined as a function from W to V that satisfies <T(x),y>' = <x,T*(x)> for all x in V and y in W. To prove that rank(T*) = rank(T), one can choose orthonormal bases B and B' for V and W respectively, rewrite the equation in matrix form, and use Gram-Schmidt to conclude.
  • #1
typhoonss821
14
1
I have a question about the rank of adjoint operator...

Let T : V → W be a linear transformation where V and W are finite-dimensional inner product spaces with inner products <‧,‧> and <‧,‧>' respectively. A funtion T* : W → V is called an adjoint of T if <T(x),y>' = <x,T*(x)> for all x in V and y in W.

My question is how to prove that rank(T*) = rank(T)??

Can anyone give me some tips, thanks^^
 
Physics news on Phys.org
  • #2
Take a basis B of V and a basis B' of W in which the matrix of <,> and <,>' are both the identity. That is to say, pick B <,>-orthonormal and B' <,>'-orthonormal. This is always possible by Gram-Schmidt.

Then write the equation <T(x),y>' = <x,T*(x)> in matrix form with respect of these basis and conclude.
 

Related to Proving the Rank Equivalence of Adjoint Operators

1. What is the definition of a rank of an adjoint operator?

The rank of an adjoint operator is the dimension of its range, which is the set of all vectors that the operator can map to. In other words, it is the number of linearly independent vectors in the range of the operator.

2. How is the rank of an adjoint operator related to its nullity?

The rank and nullity of an adjoint operator are complementary, meaning that the sum of their dimensions is equal to the dimension of the vector space. In other words, if the rank of an adjoint operator is r, then its nullity will be n-r, where n is the dimension of the vector space.

3. Can the rank of an adjoint operator be greater than the dimension of the vector space?

No, the rank of an adjoint operator cannot be greater than the dimension of the vector space. This is because the range of an operator is a subspace of the vector space, and therefore, cannot have a dimension greater than the vector space itself.

4. How does the rank of an adjoint operator relate to its eigenvalues?

The rank of an adjoint operator is equal to the number of non-zero eigenvalues it has. This is because the eigenvalues of an operator are the values for which the operator maps a vector to a scalar multiple of itself, and the rank is the number of linearly independent vectors that the operator can map to.

5. Can the rank of an adjoint operator change depending on the basis of the vector space?

No, the rank of an adjoint operator is an intrinsic property and does not depend on the basis of the vector space. It will remain the same regardless of the basis chosen, as long as the vector space remains the same.

Similar threads

I Orthogonality of Eigenvectors of Linear Operator and its Adjoint
    • Linear and Abstract Algebra
Replies
3
Views
3K
I Does Axler's Spectral Theorem Imply Normal Matrices?
    • Linear and Abstract Algebra
Replies
3
Views
977
I If T is diagonalizable then is restriction operator diagonalizable?
    • Linear and Abstract Algebra
Replies
3
Views
1K
I Question Regarding Definition of Tensor Algebra
    • Linear and Abstract Algebra
Replies
10
Views
467
MHB Rank of composition of linear maps
    • Linear and Abstract Algebra
Replies
7
Views
1K
I A = norm-preserving linear map (+other conditions) => A = lin isometry
    • Linear and Abstract Algebra
Replies
4
Views
923
A Hilbert-adjoint operator vs self-adjoint operator
    • Linear and Abstract Algebra
Replies
2
Views
1K
I Shouldn't this definition of a metric include a square root?
    • Linear and Abstract Algebra
Replies
8
Views
1K
I Question from a proof in Axler 2nd Ed, 'Linear Algebra Done Right'
    • Linear and Abstract Algebra
Replies
3
Views
1K
MHB Proving Non-Degeneracy of Euclidean Inner Products
    • Linear and Abstract Algebra
Replies
8
Views
1K

Hot Threads

Recent Insights

Back
Top