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LU decomposition of a matrix

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1) Given that A is a non singular matrix.
Consider the non homogeneous system
A IAI=
Observe that the middle A between the identity matrices is just a variable and is not a matrix.
Apply elementary row operations on A and the same operations on the pre multiple I and
the column operations on the post multiple I to reduce the A on the left side of the equation
to the normal form.
That is,
N LAU
=
where we can see that the operations used to reduce A into row reduced
echelon form allows the lower triangular matrix L formed from the identity matrix.
Similarly, the column operations on A to reduce it to the normal form N will be applied on the
post multiple and thus, it becomes the upper triangular matrix.
Further, the inverse of a lower triangular matrix is again a lower triangular and similarly, the
inverse of the upper triangular matrix is upper triangular.
Therefore,
( ) ( )
1 1 1 1
1 1
1 1
1 1
1 1
{ }
is also an identity matrix of same order
L NU L LAU U
L NU L L A UU
LU IAI N
LU A
=
=
=
=
Q
This way, we could reduce the given coefficient matrix A in to the product of lower and upper
triangular matrices. Conveniently, let these matrices be Q and R
( )
where
Ax QRx
Q Rx
Qy Rx y
=
=
= =
Since
Ax b
=
is the given system, now it looks like
Qy b=
where
y =
the column vector of
new variables
1
n
y
y
.
The system
Qy b=
looks like the row reduced echelon form and so, solving from below, we
get the values of
1 2
, ,...,
n
y y y
readily.
Now, substitute
y
in
, we once again readily get the values of the required variables
1 2
, ,...,
n
x x x
readily.
Therefore,
1
n
x
x
is easily solved.
Further, A is non singular allows that all the entries of the principal diagonal of L are non zero
and just the eigen values of A.
So, we can write the lower triangular matrix as the product of a lower triangular matrix
1
Q
and a diagonal matrix D whose diagonal entries are nothing but the diagonal entries of the

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lower triangular matrix Q. Further, the diagonal entries of the new lower triangular matrix
1
Q
are 1’s and the actual diagonal entries are supplied into D.
Therefore,
1
Ax Q DR=
where
1
Q
is the lower triangular, D is the diagonal and R is the upper
triangular matrix.
Supplying the Eigen value and Eigen vector properties, we can see that
A
λ
=
x x
where
λ
is
an Eigen value and x is the respective Eigen vector.
So, we can easily see that
( ) ( )
k k
A
λ
=
x x
where
λ
is the diagonal entry of D in the present case.
In other words,
( ) ( )
k k
A D=x x
which has the rows of
'
i
se
and 1 is multiplied with
λ
in each
row. Therefore, it requires n operations which are nothing but the order of A.
a) Observe that whether zero structure of
'
i
e s
is taken into account, then there would be no
change in the computation procedure while the diagonal entries are the only to be effected.
Till now, the entire procedure requires
2
n
computations at the most to change A into the
product of lower and upper triangular matrices.
Also, writing the diagonal matrix, by splitting it from L, it requires
2
n
operations by
considering the row entries of
'
i
e s
also in the other than principal diagonal, for each iteration
of
( )
k
x
, it requires n computations and so, in total there are nk computations for this part.
Thus, the total number of computations required is
( )
2 2
2n n nk n n k
+ + = +
b) This same is true for when the zero structure of
'
i
e s
is not taken into account also.
Then the number of computations required is
( )
2 2
1n n nk n k n
+ + = + +
2a) suppose
AC B
=
is the product of matrices in which A is non-singular. Also, suppose
these are n
th
order matrices.
Then considering
A IA=
where I is the identity matrix of same order and apply the row
operations on left hand side matrix and I, we can reduce the equation into
and thus,
we easily see that E is the inverse A.
Further, A is non-singular allows that no diagonal entry of E is zero.
Since there are
2
n
entries in the matrix A, it requires at the most
2
n
operations to reduce A into
I.
Thus, we get
1
A E
=
Now,
( )
1 1
AC B A AC A B
= =
Or simply
( )
1 1
A A C IC C A B
= = =
See that each row of
1
A
is to be multiplied with each column of B and while there are n rows
in
1
A
and n columns in B, it requires
2
n
computations.
Among each of these
2
n
computations, there is n number of real number multiplications and n
– 1 additions.
So, the arithmetic computations in each row and column multiplication requires
2n – 1 arithmetic computations.
In total, the number of required computations is
( )
2
2 1n n
.

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Including the finding of E =
1
A
, it requires
( )
2 2 3
2 1 2n n n n
+ =
operations in total.
2b) on the other hand, if each row
j
c
of C is to be multiplied to construct the columns of B,
then to each column, it the n entries of the row
j
c
are multiplied with all the entries of
1
A
,
then the action requires
( )
2
2 1n n
×
operations. For all the rows, it requires
( )
2 4 3
2 1 2n n n n n
=
operations.
Including the finding of inverse of A, the number of computations required is
4 3 2
2n n n +
operations.
Note that, to construct the LU decomposition of A, there are very many methods, through
which the number of computations varies.
Naturally, the end course requires different number of computations. So, take a precaution
keeping this in view.

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