Linear Algebra Homework

This document presents solutions to a linear algebra homework assignment. The problems cover a range of topics, including proof by induction, determinant calculations, matrix factorization (LU decomposition), linear independence and rank of matrices, subspaces, properties of linear transformations, eigenvalues and eigenvectors, orthogonal bases (Gram-Schmidt process), and solving systems of linear equations. Each problem is worked out step-by-step, showing the detailed calculations and reasoning involved. The solutions demonstrate a strong understanding of fundamental linear algebra concepts and techniques.

Problem 1
Proof by induction
n ≥ 0 , then2n ≥ n where n is a natural number
p ( n ) ; 2n> n , for n=1 we hane p ( 1 ) ;21 >1 hence p (1) is true
Suppose p ( k )=true for k >+0 then p ( k +1 )=true whenever pk=true
2k >k [n=k ]
2∗2k >2 k
2k +1> 2k
but 2 k thats (k +k )
Being that ( k + k ) >k +1
Then 2k +1> k +1 therefore p ( k +1 ) istrue whenever p ( k ) istrue
Hence by principal of mathematical induction p ( n ) istrue for all n ≥ 0
Problem 2
A=
[1 2 3 2
0 1 4 1
5 6 0 6
0 0 1 1 ]
a) A−1
A−1= 1
det ( A ) ∗adj( A)
det ( A ) = ( 1∗1∗0∗1 ) + ( 1∗4∗6∗0 ) + ( 1∗1∗6∗1 ) + ( 2∗0∗6∗1 ) + ( 2∗4∗5∗1 ) + ( 2∗1∗0∗0 )∗ ( 3∗0∗6∗1 ) + ( 3∗1∗
This gives
6+ 40+10−6−24−10−15=1
Hence the determinant is 1
Adj ( A )=
[b11 b12 b13 b14
b21 b22 b23 b24
b31 b32 b33 b34
b41 b42 b43 b44
]now
b11= ( 1∗0∗1 ) + ( 4∗1∗0 ) + ( 1∗6∗1 ) − ( 1∗6∗1 ) − ( 4∗6∗1 ) − ( 1∗0∗0 ) =−23
b12= ( 2∗1∗1 ) + ( 3∗6∗1 ) + ( 0∗0∗0 ) − ( 2∗0∗1 ) − ( 3∗6∗0 ) − ( 2∗6∗1 ) =8

b13= ( 2∗4∗1 )+ ( 3∗1∗0 ) + ( 2∗1∗1 )− (2∗1∗1 ) − ( 3∗1∗1 )− ( 2∗4∗0 )=5
b14= ( 2∗1∗0 ) + ( 3∗1∗6 ) + ( 2∗4∗6 ) − ( 2∗4∗6 ) − ( 3∗1∗6 ) − ( 2∗1∗0 ) =0
b21= ( 0∗6∗1 ) + ( 4∗5∗1 )+ ( 1∗0∗0 )− ( 0∗0∗1 )− ( 4∗6∗0 ) − ( 1∗5∗1 )=15
b22= (1∗0∗1 ) + ( 3∗6∗0 ) + ( 2∗5∗1 )− ( 1∗6∗1 )− ( 3∗5∗1 ) − ( 2∗0∗0 )=−11
b23= ( 1∗1∗1 )+ (3∗0∗1 ) + ( 2∗4∗0 )− ( 1∗4∗1 )− ( 3∗1∗0 )− ( 2∗0∗1 )=−3
b24= ( 1∗4∗6 ) + ( 3∗1∗5 )+ ( 2∗0∗0 ) − ( 1∗1∗0 )− (3∗0∗6 )− ( 2∗4∗5 )=−1
b31= ( 0∗6∗1 ) + ( 1∗6∗0 ) + ( 1∗5∗0 ) − ( 0∗6∗0 ) − ( 1∗5∗1 ) − ( 1∗6∗0 ) =−5
b32= (1∗6∗0 ) + ( 2∗5∗1 ) + ( 2∗6∗0 )− (1∗6∗1 )− ( 2∗6∗0 )− (2∗5∗0 )=4
b33= ( 1∗1∗1 )+ (2∗1∗0 ) + ( 2∗0∗0 )− ( 1∗1∗0 ) − ( 2∗0∗1 )− ( 2∗1∗0 )=1
b34= ( 1∗1∗6 ) + ( 2∗0∗6 ) + ( 2∗1∗5 )− (1∗1∗6 )− ( 2∗1∗5 ) − ( 2∗0∗0 )=0
b41=(0∗0∗0) ¿+(1∗5∗1)+( 4∗6∗0)−(0∗6∗1)−(1∗0∗0)−( 4∗5∗0)=5
b42= ( 1∗6∗1 ) + ( 2∗0∗0 ) + ( 3∗5∗0 )− ( 1∗0∗0 )− ( 2∗5∗1 ) − ( 3∗6∗0 )=−4
b43= ( 1∗4∗0 ) + ( 2∗0∗1 ) + ( 3∗1∗1 ) − ( 1∗1∗1 ) − ( 2∗4∗0 ) − ( 3∗0∗0 ) =2
b44= ( 1∗1∗0 )+ ( 2∗4∗5 )+ ( 3∗0∗6 )− ( 1∗4∗6 )− ( 2∗0∗0 )− ( 3∗1∗5 ) =1
Now the
Adj ( A )=
[−23 8 5 0
15 −11 −3 −1
−5 4 1 0
5 −4 2 1 ]
since thedeterminant of A=1 thentheinverse of A=adj( A)
which is
[−23 8 5 0
15 −11 −3 −1
−5 4 1 0
5 −4 2 1 ]
b) det ( A ) =1 as calculated in part (a) above
The determinant of A−1
A−1=
[ −23 8 5 0
15 −11 −3 −1
−5 4 1 0
5 −4 2 1 ]
det ( A−1 ) = (−23∗−11∗0 )+ (−23∗−3∗0∗4 ) + (−23∗−1∗4∗2 ) + ( 8∗15∗0∗2 )+ ( 8∗−3∗−5∗1 ) + ( 8∗0∗1∗5

This gives
184+120+ 300+100−276−92−120−80−275+100=−39
Which is the determinant of A−1
c) LU = A
Where L is the lower triangular matrix and U is the upper triangular matrix
[1 0 0 0
a 1 0 0
b c 1 0
d e f 1 ]∗
[ g h i j
0 k l m
0 0 n o
0 0 0 p ]=
[1 2 3 2
0 1 4 1
5 6 0 6
1 1 1 1 ]
multiplying bthe LU matrixes we get
[ g h i j
ag ah+ k ai+l aj+ m
bg bh+ck bi+cl+n bj+cm+o
dg dh+ck di+¿+ fn dj+em+of +ip ]=A
Comparing the elements of LU with the corresponding elements of A we obtain the
values of the unknowns as
g=1, h=2 ,i=3 , j=2
a=0 , b=5 , d=0 , k=1 , c=−4
e=0 , l=4 , n=1, f =1 , o=14 , p=−3
Now replacing the values of the unknowns in the matrixes L and U we get
L=
[1 0 0 0
0 1 0 0
5 −4 1 0
0 0 1 1 ]while U=
[1 2 3 2
0 1 4 1
0 0 1 14
0 0 0 −3 ]
d) Number of linearly independent columns of A
When the matrix is reduced to reduced row echelon form we obtain the matrix
[1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1 ]here therefore theequation Ax=0 will have a trivial solution . A therefore have 4
linearly independent columns.
e) Rank of A
f) Based on the matrix reduced to row echelon form above. there are 4 non-zero rows in
the matrix. Therefore, rank of A is 4.
As indicated on the row echelon matrix the kernel of A is {0,0,0,0}

Which is 0
g) The system has at least one solution if the vector be being non-zero and have at least
one non-zero element. That’s b=[0,0,02 ,]
h) Solution of the system Ax=b
The reduced row echelon form of vector A is
[1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1 ]
now Ax=b=
[1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1 ]∗
[ x1
x2
x3
x4
] therefore, the value of x’s is
x1=0 , x2=0 , x3 =0∧x4 =2
Problem 3
a) v=R3∗3 , w={ A ∈ v : a=− AT }
wϵv isa subspace of v then
w ≠ 0
Since A ∈ v∧− AT is also ∈ R3∗3 then − AT ϵW meaning w is a subspace of v
b) For any value of x xϵ π5 then wϵ π5 hence w is a subspace of v the span of W is
3
Problem 4
a) True; by definition a set of vectors {
v1 , v2 , … v p }∈real space is linearly independent if the vector x 1 v 1+ x 2 v 2 … xpvp h
as only the trivial solution.
b) True; homogeneous systems are linear system in the form Ax=0where 0 is the 0
vector
c) True’ the function is one to one as it maps R3 ¿ R3
d) True; given A2=I 2 then A=± I2 which is I 2∨−I 2
Problem 5
1) if N =p (−1 ) DP then this means
N=D as themultiplication of D by p−1 ∧P is same as mulplicationby an identity ¿since
N=d then there determinant is also equal satisfying det ( N )=det ( D)
2) a) the complex eigen value are a+ib which is λi
b) the complex eigen value are algebraically multiplied based on their conjugate
c) c is not diagonalizable as some of the eigen values are not real numbers
d) the det of C will be zero as it is not diagonalizable

3) the eigen vectors are composed form the matrix m hence the matrix is
1 1
−1 −1
Problem 6(a)
a) the orthogonal basis will be
[ 1 8 0
2 1 0
0 −6 1 ] by the gram-Schmidt
b) Normalizing means we divide the basis by the magnitude hence we will have vector
[0.5 4 0
1 0.5 0
0 −3 0.5 ]
Problem 6(a)
a) The matrix which is orthogonal to B is the identity matrix
[1 0 0
0 1 0
0 0 1 ]
b) The matrix Bx=b
First, we reduce the matrix to row echelon form and obtain
[1 0 3
0 1 2
0 0 7 ]Hence the values of x are
x1=3 , x2 =2∧x3=7
Which is the solution.