2. DIAGONALIZATION
EIGEN VALUE
EIGEN VECTOR
ORTHOGONAL
CAYLEY HAMILTON THEOREM
kshumENGG2013 2
3.
4. A square matrix M is called
diagonalizable if we can find an
invertible matrix, say P, such that the
product P–1
M P is a diagonal matrix.
A diagonalizable matrix can be raised to
a high power easily.
› Suppose that P–1
M P = D, D diagonal.
› M = PD P–1
.
› Mn
= (PD P–1
) (PD P–1
) (PD P–1
) … (PD P–1
)
= PDn
P–1
.
4
5. Let
A is diagonalizable because we can find
a matrix
such that
5
6. By definition a matrix M is diagonalizable
if
P–1
M P = D
for some invertible matrix P, and diagonal
matrix D.
or equivalently,
6
8. Given a square matrix A, a non-zero
vector v is called an eigenvector of A, if
we an find a real number λ (which may
be zero), such that
This number λ is called an eigen value of
A, corresponding to the eigenvector v.
8
Matrix-vector product Scalar product of a vector
9. If v is an eigenvector of A with eigen
value λ, then any non-zero scalar
multiple of v also satisfies the definition of
eigenvector.
9
k ≠ 0
10. First eigenvalue = 2, with eigenvector
where k is any nonzero real number.
Second eigenvalue = -5, with eigenvector
where k is any nonzero real number.
10
18. The inverse A-1
of a matrix A has the
property:
AA-1
=A-1
A=I
A-1
exists only if
Terminology
› Singular matrix: A-1
does not exist
› Ill-conditioned matrix: A is close to being singular
25. • A is orthonormal if:
• Note that if A is orthonormal, it easy to find its inverse:
Property:
26. Diagonalize the matrix A = by means of an
orthogonal transformation.
Solution:-
Characteristic equation of A is
204
060
402
66,2,λ
0λ)16(6λ)λ)(2λ)(6(2
0
λ204
0λ60
40λ2
−=⇒
=−−−−−⇒
=
−
−
−
26
Example :-
28. 2
2I
0
⇒ −
∴
1
2
3
1
2
3
1 3
1 3
1 3 2
2 2 3
x
whenλ = 6,let X = x betheeigenvector
x
then (A -6 )X =0
-4 0 4 x 0
0 0 x = 0
4 0 -4 x 0
4x +4x =0
4x -4x =0
x = x and x isarbitrary
x must be so chosen that X and X are orthogonal among th
.1
emselves
and also each is orthogonal with X
28
29.
∴
∴
∴
2 3
3 1
3 2
3
1α
Let X = 0 and let X =β
1γ
Since X is orthogonal to X
α- γ = 0 ...(4)
X is orthogonal to X
α+ γ = 0 ...(5)
Solving (4)and(5), we getα = γ = 0 and β is arbitra ry.
0
Takingβ =1, X = 1
0
1 1 0
Modal matrix is M= 0 0 1
-1 1
0
29
33.
11 12 1n
21 22 2n
n1 n2 nn
φ(λ) = A - λI
a -λ a ... a
a a -λ ... a
=
... ... ... ...
a a ... a -λ
| A -λI|= 0
34. ⇒ n n-1 n-2
0 1 2 n
n n-1 n-2
0 1 2 n
We are to prove that
pλ +p λ +p λ +...+p = 0
p A +p A +p A +...+p I= 0 ...(1)
I
⇒
n-1 n-2 n-3 -1
0 1 2 n-1 n
-1 n-1 n-2 n-3
0 1 2 n-1
n
0 =p A +p A +p A +...+p +p A
1
A =- [p A +p A +p A +...+p I]
p