Eigensystems for Symmetric Matrices
Learning Goals: students see that the eigenvalues of symmetric matrices are all real, and that
they have a complete basis woth of eigenvectors, which can be chosen to be orthonormal.
Since they appear quite often in both application and theory, lets take a look at symmetric
matrices in light of eigenvalues and eigenvectors.
One special case is projection matrices.
We
know that a matrix is a projection matrix if and only if
P
=
P
2
=
P
T
.
So they are symmetric.
They also have only very special eigenvalues and eigenvectors.
The eigenvalues are
either 0 or 1.
Why?
If a vector is not in the column space, it gets sent to one that is, so the only
hope of it being an eigenvector is if it gets sent to zero (hence is in the nullspace, and is an
eigenvector with eigenvalue 0).
If a vector is in the column space, it gets sent to itself, so is an
eigenvector with eigenvalue 1.
Notice that the “eigenspaces” are the column space and the
nullspace.
And notice that they are orthogonal.
Usually, the nullspace is orthogonal to the row
space, but since
P
=
P
T
those two are the same!
It turns out this same thing happens for
any
symmetric matrix:
Example:
find the eigensystem of
2
1
0
1
1
1
0
1
2
!
"
#
#
#
$
%
&
&
&
.
We easily calculate the characteristic polynomial
to be 
λ
3
+ 5
λ
2
 6
λ
.
The eigenvalueeigenvector pairs are thus (0, (1, 2, 1)), (3, (1, 1, 1)), and
(2, (1, 0, 1)).
Note that the three eigenvectors are actually orthogonal.
If we assume that the matrix has a full set of eigenvectors, we could diagonalize it as
A
=
S
Λ
S
1
.
Note that
A
T
has this same factorization because
A
is symmetric.
But
A
T
is also
(
S
1
)
T
Λ
S
T
which suggests that maybe
S
1
=
S
T
—making
S
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 Spring '08
 Neely
 Linear Algebra, Eigenvectors, Matrices, Orthogonal matrix

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