10.3. Diagonanalisation of real symmetric matrices |
In the previous section we observed that a n×
n real symmetric matrix has n-
eigenvalues . The
eigenvectors corresponding to distinct eigenvalues of a real symmetric
matrix have a special property,
as given in the next theorem. |
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10.3.2 Corollary: |
Let A be n × n real symmetric
matrix such that all its eigenvales are distinct. Then, there
exists an orthogonal matrix P such that
,
where D is a diagonal matrix with diagonal entries being the
eigenvalues of A.
Proof
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We prove in the next theorem that the claim of
the above corollary always leads, even if not all the
eigenvalues of A are distinct. |
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Let
= ( 3 - λ )[( 3 - λ)( 5 - λ ) - 1] -1 [ 5 - λ
- 1 ] -1 [ -1 + 3 - λ ]
= [( 3 - λ )( 3 - λ)( 5 - λ ) - ( 3 - λ )] - ( 4 - λ ) - ( 2 - λ )
= [( 3 - λ )( 3 - λ)( 5 - λ ) - ( 3 - λ )] - 2 ( 3 - λ )
= ( 3 - λ )[( 3 - λ )( 5 - λ ) - 1 - 2 ]
= ( 3 - λ )[ 15 - 5λ -3λ +
-3 ]
= ( 3 - λ )(
-
8λ + 12 )
= ( 3 - λ )( λ - 6 )( λ - 2 )
Thus, λ = 2, 3, 6 are the eigenvalues of A .Let us find an
eigenvector corresponding to each
eigenvalue . For the eigenvalue λ = 2, since
A vector
will
be an eigenvector for eigenvalue λ = 2 if
i.e.,
=
0
+==
0.
If we choose
=1,
then
=
-1. Hence
is an eigenvector for λ = 2 For the eigenvalue λ = 3,
Thus
will be an eigenvector for the eigenvalue λ = 3, if
i.e.,
=
and
-=
0. Hence
is an eigenvector for λ = 3.
Finally, for λ = 6
Thus ,
will be an eigenvector for eigenvalue λ = 6 if
-4-2
= 0,
--+=
0.
Thus, if we take =
2 , then
=
-1 and
=-+=-1.
Hence
is an eigenvector for the eigenvalue λ = 6.
Note that for
the
columns of P are orthogonal.
To make P orthogonal, we normalize each
,
and ,
and define
It is easy to verify that
,
where
In fact,
.
Thus, we checks
.
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10.3.4
Theorem: |
Let A be any real symmetric matrix . Then there exists an
orthogonal matrix P such that
the following holds :
(i)
AP
= D , a diagonal matrix.
(ii) The diagonal entries of D are the eigenvalues of A .
(iii) The column vectors of P are the eigenvectors of the
eigenvalues of A.
Proof
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Let
Then A is a real symmetric matrix with eigenvalues given by
= - λ (
-
4 ) - 2 ( -2λ - 4 ) + 2 ( 4 + 2λ )
= - λ ( λ - 2 )( λ + 2 ) + 4 ( λ + 2 )
+ 4 ( λ + 2 )
= ( λ + 2 )( -
+ 2λ + 8 )
= - ( λ + 2 )( λ + 2 )( λ - 4 ).
Hence, λ = -2 , λ = 4 are two distinct eigenvalues of A. To find
eigenvectors for λ = -2, since
If we choose
= 0
and
= 1 then =
-1. And for =
1 and
= 0 we get =
-1.
Thus
are two mutually orthogonal eigenvector for λ
= -2 .
For the eigenvalue λ = 4, since
,
Thus, if we choose
= ==
1, then
is an eigenvector for the eigenvalue λ = 4. Thus, the eigenvector for A
are
To make
orthonormal,
we use the Gram-Schmidt process.
Define
We normalize
also
to get
are orthonormal basis of
of
eigenvectors of A. Thus, for
one checks that
.
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10.3.9 Exercises: |
Click here to take a Quiz: Quiz 10.3 |
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