Symmetric matrix and diagonalization

In summary: The field of real numbers is not algebraically closed but it is certainly true that any symmetric matrix over the real numbers has real eigenvalues. In fact, any self-adjoint linear transformation on a vector space over the real numbers has all real eigenvalues.
  • #1
EvLer
458
0
This is a T/F question:
all symmetric matrices are diagonalizable.

I want to say no, but I do not know how exactly to show that... all I know is that to be diagonalizable, matrix should have enough eigenvectors, but does multiplicity of eigenvalues matter, i.e. can I say that if eignvalue has multiplicity > 1 then the matrix is not diagonalizable?
Thanks

EDIT: zero matrix is diagonalizable, right?
 
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  • #2
EvLer said:
This is a T/F question:
all symmetric matrices are diagonalizable.

I want to say no, but I do not know how exactly to show that... all I know is that to be diagonalizable, matrix should have enough eigenvectors, but does multiplicity of eigenvalues matter, i.e. can I say that if eignvalue has multiplicity > 1 then the matrix is not diagonalizable?
Thanks

EDIT: zero matrix is diagonalizable, right?

The zero matrix is already diagonal. So it's VERY diagonalizable. Think spectral theorem.
 
  • #3
Dick said:
...Think spectral theorem.

ummm... not there yet
 
  • #4
You must have something like it. Note that it also makes a big difference whether the entries are real or complex. If they are real then it's true, but a proof is just a proof of the finite dimensional version of the spectral theorem. If they are complex, then you just need a counter example. Because it's false.
 
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  • #5
oh, yeah, looked at further chapters, now I see the spectral theorem. Thanks a lot.
 
  • #6
Uhhh. Not true.

Remember eigenvalues have a lot to do with what field you're in.

Principal axis theorem is the theorem you want.
 
  • #7
A symmetric matrix is "self-adjoint"- that is, for a symmetric matrix A, the dot product (Ax).y= x.(Ay). It's fairly easy to prove, it may already be proven in your book, that any self-adjoint (and, so, any symmetric) n by n matrix has n independent eigenvectors. Using those eigenvectors as basis gives you a diagonal matrix with the eigenvalues on the diagonal.
 
  • #8
...over an algebraically closed field...
 
  • #9
HallsofIvy said:
A symmetric matrix is "self-adjoint"- that is, for a symmetric matrix A, the dot product (Ax).y= x.(Ay). It's fairly easy to prove, it may already be proven in your book, that any self-adjoint (and, so, any symmetric) n by n matrix has n independent eigenvectors. Using those eigenvectors as basis gives you a diagonal matrix with the eigenvalues on the diagonal.

matt grime said:
...over an algebraically closed field...

Was that in response to my post? The field of real numbers is not algebraically closed but it is certainly true that any symmetric matrix over the real numbers has real eigenvalues. In fact, any self-adjoint linear transformation on a vector space over the real numbers has all real eigenvalues.
 
  • #10
I should have said 'not necessarily over the field of definition of the matrix'. Or 'a field with sufficiently many elements'.
 

FAQ: Symmetric matrix and diagonalization

What is a symmetric matrix?

A symmetric matrix is a square matrix in which the elements above and below the main diagonal are mirror images of each other. In other words, if we were to draw a line from the top left corner to the bottom right corner, the elements on either side of the line would be equal. This means that AT = A, where AT represents the transpose of A.

How is a matrix diagonalized?

To diagonalize a matrix means to transform it into a diagonal matrix, which is a matrix with non-zero elements only on the main diagonal. This is done by finding the eigenvalues and eigenvectors of the matrix. The eigenvectors form the columns of a matrix P, and the eigenvalues are placed on the main diagonal of a diagonal matrix D. The diagonalized form of the original matrix is then P-1DP.

What is the significance of diagonalizing a matrix?

Diagonalization is useful because it simplifies many calculations involving matrices. It also allows us to easily find powers of a matrix, as raising a diagonal matrix to a power simply involves raising each element on the main diagonal to that power. Additionally, diagonalization can reveal important properties of a matrix, such as its determinant and trace.

Can all matrices be diagonalized?

No, not all matrices can be diagonalized. A matrix can only be diagonalized if it has n linearly independent eigenvectors, where n is the size of the matrix. This means that some matrices, such as non-square matrices, cannot be diagonalized. Also, some matrices may have repeated eigenvalues, which can make diagonalization impossible.

What is the relationship between symmetric matrices and diagonalization?

Symmetric matrices have a special property that makes them easier to diagonalize. They are always diagonalizable and have real eigenvalues. Additionally, the eigenvectors of a symmetric matrix are always orthogonal, which simplifies the diagonalization process. This is why symmetric matrices are commonly used in many applications, including in physics and engineering.

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