Real symmetric matrices have real eigenvalues

A real matrix is symmetric if A^t=A. I will show in this post that a real symmetric matrix have real eigenvalues.

I will need a dot product for the prof and I’ll use the basic dot product for two vectors X and Y: <X,Y>=X^t\overline{Y}, where \overline{Y} is the complex conjugate of the vector Y.

The useful property of this dot product is that <AX,Y>=<X,{A^t}Y>, for any matrix A

And considering that A is real, a simple proof is:

<AX,Y>=(AX)^t\overline{Y}=X^tA^t\overline{Y}=X^t\overline{A^tY}=<X,{A^t}Y>

An eigenvalue have a correspondent eigenvector: AX=\lambda X.

We have <AX,X>=<\lambda X, X>=\lambda <X,X> and considering that A is symmetric <AX,X>=<X,{A^t}X>=<X,AX>=<X,\lambda X>=\overline{\lambda}<X,X>.

From \lambda<X,X>=\overline{\lambda}<X,X> and because X is not a zero vector it results that the imaginary part of \lambda is zero, so the eigenvalue is a real number.

Advertisements
This entry was posted in Linear Algebra, Math. Bookmark the permalink.

3 Responses to Real symmetric matrices have real eigenvalues

  1. Pingback: Real symmetric matrices are diagonalisable | AlfaNotes

  2. Pingback: Real symmetric matrices are diagonalizable | AlfaNotes

  3. Pingback: SVD – a simple proof | AlfaNotes

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s