A and B are n x n matrices. Investigate if
(A + B)2 = A2 + 2.A.B + B2
|
(A + B)2 = (A + B).(A + B) = A.A + A.B + B.A + B.B
= A2 + A.B + B.A + B2
Since the product is not commutative, we do NOT have
(A + B)2 = A2 + 2.A.B + B2
We say that two matrices A and B commute, if and only if AB = BA.
Show that all matrices like
[a -b]
[b a]
commute.
|
[a -b] [a' -b'] = [aa'-bb' -ab'-ba']
[b a] [b' a'] [ba'+ab' -bb'+aa']
and
[a' -b'] [a -b] = [aa'-bb' -ab'-ba']
[b' a'] [b a] [ba'+ab' -bb'+aa']
Prove that for each n x n matrix A , AT .A = symmetric |
We'll prove that AT .A is equal to its transpose. (AT .A)T = AT .(AT)T = AT .A
Prove that if A is skew-symmetric, then A.A is symmetric. |
A is skew-symmetric <=> A = - AT . So, (A.A)T = AT .AT = (- A ).(- A ) = A.A Hence, A.A is symmetric.
Given X = [x y] and A is a 2 x 2 matrix . All elements of the matrices are real numbers. Examine if X . AT . A . XT can be negative. |
First we see that X . AT . A . XT is a 1 x 1 matrix.
It is a number, so maybe it is negative.
But,
X . AT = is a 1 x 2 matrix, say [a b].
A . XT = (X . AT )T = [a b]T
Then
X . AT . A . XT = (X . AT ).(X . AT )T
= [a b] .[a b]T
= a2 + b2
So, X . AT . A . XT can not be negative.
[ 1 m - 1 2 m - 3 ]
Given A = [ m 2 m - 2 2 ]
[ m + 1 3 m - 3 m.m - 1 ]
Calculate the condition for m such that A is regular.
Assume that m satisfies this condition and consider the system
with A as coefficient matrix.
/ x + (m - 1)y + (2 m - 3)z = 1
|
| m x + (2 m - 2)y + 2 z = 0
|
\ (m + 1)x + (3 m - 3)y + (m.m - 1)z = 0
Now, let x = 1 and calculate the values of m such that the system
has a solution for y and z.
|
2
Det(A) = m (1 - m) (m - 2)
A is regular if and only if m is NOT in { 0, 1, 2 }
With x = 1, the system becomes
/ (m - 1)y + (2 m - 3)z = 0
|
| (2 m - 2)y + 2 z = - m
|
\ (3 m - 3)y + (m.m - 1)z = -(m + 1)
The coefficient matrix of the first and the second equation is
[m - 1 2 m - 3 ]
[2 m - 2 2 ]
The determinant is -4 (m - 1) (m - 2) and since m is not in { 0, 1, 2 },
the determinant is not 0.
The condition is
|m - 1 2 m - 3 0 |
|2 m - 2 2 - m | = 0
|3 m - 3 m.m - 1 - m - 1|
<=>
2
(m + 4) (m - 1) (m - 2) = 0
Since m is not in { 0, 1, 2 }, there are no real values of m such
that the system has a solution for y and z.
The matrix x is a 2 x 2 matrix.
Calculate three solutions of the quadratic equation
x2 - x = 0
|
x2 - x = 0
<=>
x (x-I) = 0
x = 0-matrix and x = Identity matrix are two solutions, but we know
that x (x-I) can be zero without x = 0 or x = I.
To find another solution of the given equation we'll search for a matrix
[ 1 b ]
x = [ ] such that x2 = x
[ c d ]
x2 = x
<=>
[ 1 b ] [ 1 b ] [ 1 b ]
[ ].[ ] = [ ]
[ c d ] [ c d ] [ c d ]
<=>
[ 1 + b c b + b d ] [ 1 b ]
[ ] = [ ]
[ c + d c c b + d2 ] [ c d ]
<=>
[ b c b d ] [0 0 ]
[ ] = [ ]
[ c d b c + d2 - d ] [0 0 ]
Since we need only one solution, we choose d = 0 and c = 0.
[ 1 b ]
x = [ ]
[ 0 0 ]
[-2 -9 ] n [1-3n -9n ]
Given A = [ ] . Prove that A = [ ]
[ 1 4 ] [ n 1+3n ]
|
We'll prove this by complete induction.
a) The property is trivial for n = 1
b) Assume the property is true for n = k.
Then we have to prove that the property is true for n = k+1.
k+1 k [1-3k -9k ] [-2 -9 ] [-2-3k -9-9k]
A = A .A = [ ].[ ] = [ ]
[ k 1+3k ] [ 1 4 ] [k+1 3k+4]
[1-3(k+1) -9(k+1) ]
= [ ]
[ k+1 1+3(k+1) ]
[1 1 0 ]
Given : A = [0 1 0 ]
[0 0 1 ]
Show that A is regular.
n
Calculate A .
Calculate the real numbers a and b such that
A2 + a A + b I = 0 ( I is the 3 x 3 identity matrix)
Show that there are real numbers c0,c1,c2, ... ,cn such that
A-n = c0.I + c1.A2 + c2.A3 + c3.A4 + c4.A + ... + cn.An
|
Det(A) = 1 , so A is regular.
2 [1 2 0 ] 3 [1 3 0 ]
A = [0 1 0 ] A = [0 1 0 ]
[0 0 1 ] [0 0 1 ]
From this we'll prove by complete induction that
n [1 n 0 ]
A = [0 1 0 ]
[0 0 1 ]
This expression is true for n = 1
Assume the property is true for n = k.
Then we have to prove that the property is true for n = k+1.
k+1 k [1 k 0 ][1 1 0 ] [ 1, k + 1, 0 ]
A = A .A = [0 1 0 ][0 1 0 ] = [ 0, 1, 0 ]
[0 0 1 ][0 0 1 ] [ 0, 0, 1 ]
A2 + a A + b I = 0
<=>
[1 2 0 ] [1 1 0 ] [1 0 0 ] [0 0 0 ]
[0 1 0 ] + a [0 1 0 ] + b [0 1 0 ] = [0 0 0 ]
[0 0 1 ] [0 0 1 ] [0 0 1 ] [0 0 0 ]
<=>
1 + a + b = 0
2 + a = 0
<=>
a = -2, b = 1
From this we have
A2 - 2 A + I = 0
<=>
2 A - A2 = I
<=>
A (2 I - A ) = I
From this, we see that (2 I - A ) is the inverse of A
A-1 = 2 I - A
=> A-n = (2 I - A )n
=> A-n = a polynomial in A of degree n, with real coefficients
=> there are real numbers c0,c1,c2, ... ,cn such that
A-n = c0.I + c1.A + c2.A2 + c3.A3 + c4.A4 + ... + cn.An