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1 Department of Aerospace Engineering AE602 Mathematics for Aerospace Engineers Assignment No. 4 4.1 Decide whether or not the following vectors are linearly independent, by solving 1 1 + 2 2 + 3 3 + 4 4 = 0: 1 = 1 1 0 0 , 2 = 1 0 1 0 , 3 = 0 0 1 1 , 4 = 0 1 0 1 . Decide also if they span , by trying to solve 1 1 + + 4 4 = 0, 0, 0, 1. SOLUTION: To check for dependency or independency, we use 1 1 + 2 2 + 3 3 + 4 4 =0 .............(1) If = = = = is the only way to solve the above equation, then the vectors are linearly independent. If any of the ≠ 0, then the given vectors are linearly dependent. Solving the equation (1) 1 1 1 0 0 + 2 1 0 1 0 + 3 0 0 1 1 + 4 0 1 0 1 =0 we get four equations with four unknowns as follows 1 + 2 =0 1 + 4 =0 2 + 3 =0 3 + 4 =0 From the above set of equations we get = = = Hence with = ; = ; = ; = , the equation (1) goes to zero. That is 1 1 + 2 2 + 3 3 + 4 4 = 1 1 1 0 0 + 1 1 0 1 0 + 1 0 0 1 1 + 1 0 1 0 1 =0 1 , 2 , 3 , 4 can take any other values also. Therefore the given vectors , , , are linearly independent.

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Department of Aerospace Engineering

AE602 Mathematics for Aerospace Engineers

Assignment No. 4

4.1 Decide whether or not the following vectors are linearly independent, by solving

𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 + 𝑐4𝑣4 = 0:

𝑣1 =

1100

, 𝑣2 =

1010

, 𝑣3 =

0011

, 𝑣4 =

0101

.

Decide also if they span 𝑹𝟒, by trying to solve 𝑐1𝑣1 + ⋯ + 𝑐4𝑣4 = 0, 0, 0, 1 .

SOLUTION:

To check for dependency or independency, we use

𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 + 𝑐4𝑣4 = 0 .............(1)

If 𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝒄𝟒 = 𝟎 is the only way to solve the above equation, then the

vectors are linearly independent. If any of the 𝒄𝒊 ≠ 0, then the given vectors are linearly

dependent. Solving the equation (1)

𝑐1

1100

+ 𝑐2

1010

+ 𝑐3

0011

+ 𝑐4

0101

= 0

we get four equations with four unknowns as follows

𝑐1 + 𝑐2 = 0

𝑐1 + 𝑐4 = 0

𝑐2 + 𝑐3 = 0

𝑐3 + 𝑐4 = 0

From the above set of equations we get 𝒄𝟐 = 𝒄𝟒 = −𝒄𝟑 = −𝒄𝟏

Hence with 𝒄𝟏 = 𝟏; 𝒄𝟐 = −𝟏; 𝒄𝟑 = 𝟏; 𝒄𝟒 = −𝟏 , the equation (1) goes to zero.

That is 𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 + 𝑐4𝑣4 = 1

1100

+ −1

1010

+ 1

0011

+ −1

0101

= 0

𝑁𝑜𝑡𝑒 𝑡𝑕𝑎𝑡 𝑐1, 𝑐2, 𝑐3, 𝑐4 can take any other values also.

Therefore the given vectors 𝒗𝟏, 𝒗𝟐, 𝒗𝟑, 𝒗𝟒 are linearly independent.

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Now trying to solve

𝑐1𝑣1 + ⋯ + 𝑐4𝑣4 = 0, 0, 0, 1

We have

𝑐1 + 𝑐2 = 0.....(2)

𝑐1 + 𝑐4 = 0.....(3)

𝑐2 + 𝑐3 = 0.....(4)

𝑐3 + 𝑐4 = 1.....(5)

From (2) and (3), we get

𝑐2 = −𝑐1 𝑎𝑛𝑑 𝑐2 = −𝑐3 𝑎𝑛𝑑 𝑕𝑒𝑛𝑐𝑒 𝑐1 = 𝑐3

Substituting 𝑐1 = 𝑐3 in equation (3) , we get

𝑐3 + 𝑐4 = 0.....(6)

Equations (5) and (6) contradict each other and hence the vectors do not span 𝑹𝟒 .

4.2 Decide the dependence or independence of

(a) 1, 1, 2 , 1, 2, 1 , 3, 1,1 ;

SOLUTION:

Equation to be solved is,

𝑐1𝑣1 + 𝑐2𝑣2 + 𝑐3𝑣3 = 0

𝑐1 112 + 𝑐2

121 + 𝑐3

311 = 0

This is like solving Ac = 0

Solving Ac = 0 is equivalent to solving Uc = 0 where "U" is the upper triangular matrix form of

"A"

Let A = 1 1 31 2 12 1 1

R2→R2 - R1 ; R3→R3 - 2*R1

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→ 1 1 30 1 −20 −1 −5

R3→R3 + R2

→ 1 1 30 1 −20 0 −7

= U

Now solving Uc = 0

𝑐1 100 + 𝑐2

110 + 𝑐3

3−2−7

= 0

-7. 𝑐3 = 0

𝒄𝟑 = 𝟎

𝑐2 − 2𝑐3 = 0

𝒄𝟐 = 𝟎

𝑐1 + 𝑐2 + 3𝑐3 = 0

𝒄𝟏 = 𝟎

𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝟎 and hence the given vectors are independent.

(b) 𝑣1 − 𝑣2 , 𝑣2−𝑣3, 𝑣3 − 𝑣4, 𝑣4 − 𝑣1 for any vectors 𝑣1 , 𝑣2,𝑣3, 𝑣4;

SOLUTION:

Solving the equation,

𝑐1 𝑣1 − 𝑣2 + 𝑐2 𝑣2−𝑣3 ) + 𝑐3(𝑣3 − 𝑣4 + 𝑐4( 𝑣4 − 𝑣1) = 0.........(1)

If 𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝒄𝟒 = 𝟏, then equation (1) gets satisfied

That is 𝑣1 − 𝑣2 + 𝑣2−𝑣3 + 𝑣3 − 𝑣4 + 𝑣4 − 𝑣1 = 0

Hence the vectors 𝒗𝟏 − 𝒗𝟐, 𝒗𝟐−𝒗𝟑, 𝒗𝟑 − 𝒗𝟒, 𝒗𝟒 − 𝒗𝟏 are linearly dependent for any vectors

𝒗𝟏, 𝒗𝟐,𝒗𝟑, 𝒗𝟒

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(c) 1, 1, 0 , 1, 0, 0 , 0, 1,1 , 𝑥, 𝑦, 𝑧 , for any numbers 𝑥, 𝑦, 𝑧.

Solution:

To find dependency or independency we have to solve

𝑐1 110 + 𝑐2

100 + 𝑐3

011 + 𝑐4

𝑥𝑦𝑧 = 0

𝑐1 + 𝑐2 + 𝑐4𝑥 = 0.........(1)

𝑐1 + 𝑐3 + 𝑐4𝑦 = 0.........(2)

𝑐3 + 𝑐4𝑧 = 0.........(3)

Case 1: When x=y=z=0.

Then from equations (1), (2) and (3), we have 𝒄𝟏 = 𝒄𝟐 = 𝒄𝟑 = 𝟎 but 𝒄𝟒 can assume

any value and hence the vectors are not linearly independent.

Case 2: When either x or y or z not equal to zero (or) x, y & z all are not equal to zero

From equations (1), (2) and (3),

𝑧 =−𝑐3

𝑐4 .........(4)

𝑦 =−(𝑐1+𝑐3)

𝑐4 ...(5)

𝑥 =−(𝑐1+𝑐2)

𝑐4 ....(6)

If either x or y or z is not equal to zero, then for x or y or z to have a finite value (which is not

equal to zero), 𝒄𝟒 must not be equal to zero. If it is zero then the value tends to infinity.

Hence for the case 2 also the given vectors are not linearly dependent.

Hence the given set of vectors are always linearly dependent for any existing non-infinite

𝒙, 𝒚, 𝒛.

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4.3 Prove that if any diagonal element of

𝑇 = 𝑎 𝑏 𝑐0 𝑑 𝑒0 0 𝑓

is zero, the rows are linearly dependent.

SOLUTION:

The equation to be solved for determining dependency or independency is,

𝑐1 𝑎𝑏𝑐 + 𝑐2

0𝑑𝑒 + 𝑐3

00𝑓 = 0

𝑐1𝑎 = 0.........(1)

𝑐1𝑏 + 𝑐2𝑑 = 0.........(2)

𝑐1𝑐 + 𝑐2𝑒 + 𝑐3𝑓 = 0.........(3)

CASE 1: a = 0, d ≠ 0, f ≠ 0

From equation (1),

If a = 0, 𝒄𝟏 need not be zero (it can assume any value). Hence the

rows are linearly dependent.

CASE 2: a ≠ 0, d = 0, f ≠ 0

From (1), we have 𝒄𝟏 = 𝟎

From (2), we have 𝑐2𝑑 = 0

Now since d = 0, 𝒄𝟐 need not be zero which proves that the rows are again linearly

dependent.

CASE 3: a ≠ 0, d ≠ 0, f = 0

From (1), we have 𝒄𝟏 = 𝟎

From (2), since d ≠ 0 , we have 𝒄𝟐 = 𝟎

From (3), since f = 0, 𝒄𝟑 can assume any value.

Hence for the present case the rows are again linearly dependent.

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4.4 Is it true that if 𝑣1 , 𝑣2 ,𝑣3 are linearly dependent, then also the vectors 𝑤1 = 𝑣1 + 𝑣2, 𝑤2 =

𝑣1 + 𝑣3, 𝑤3 = 𝑣2 + 𝑣3, are linearly independent? (Hint: Assume some combination 𝑐1𝑤1 +𝑐2𝑤2 + 𝑐3𝑤3 = 0, and find which 𝑐𝑖 are possible.)

SOLUTION:

If 𝑣1, 𝑣2 ,𝑣3 are linearly dependent, then in the equation

𝑐1𝑣1 + 𝑐2𝑣 + 𝑐3𝑣3 = 0

there is a 𝒄𝒊 which is not equal to zero...........(1)

To check the dependency of vectors 𝑤1 = 𝑣1 + 𝑣2, 𝑤2 = 𝑣1 + 𝑣3, 𝑤3 = 𝑣2 + 𝑣3 , we consider

the equation

𝑐1𝑤1 + 𝑐2𝑤2 + 𝑐3𝑤3 = 0

𝑐1(𝑣1 + 𝑣2) + 𝑐2(𝑣1 + 𝑣3) + 𝑐3(𝑣2 + 𝑣3) = 0

(𝑐1 + 𝑐2)𝑣1 + (𝑐1 + 𝑐3)𝑣2 + (𝑐2 + 𝑐3)𝑣3 = 0

𝐶1𝑣1 + 𝐶2𝑣 + 𝐶3𝑣3 = 0

From (1), if there is a non-zero 𝒄𝒊 , then there should definitely be a non-zero 𝑪𝒊 since 𝑪𝒊 is

composed of 𝒄𝒊 . If there is such a non-zero 𝑪𝒊 , then vectors 𝒘𝟏, 𝒘𝟐, 𝒘𝟑 are also linearly

dependent.

Hence it is not true that if 𝑣1, 𝑣2 ,𝑣3 are linearly dependent, then the vectors 𝑤1 = 𝑣1 + 𝑣2,

𝑤2 = 𝑣1 + 𝑣3, 𝑤3 = 𝑣2 + 𝑣3, are linearly independent.

4.5 Describe geometrically the subspace of 𝑹𝟑 spanned by

(a) 0, 0, 0 , 0, 1, 0 , 0, 2,0 ;

SOLUTION:

Here the x and z components are zero and only the y component is present. Hence the given

vectors span a line (y-axis) in 𝑹𝟑.

(b) 0, 0, 1 , 0, 1, 1 , 0, 2,1 ;

SOLUTION:

Here x component is zero but y and z components are present. Hence these vectors span the yz

-plane in 𝑹𝟑.

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(c) all six of these vectors. Which two form a basis?

SOLUTION:

In all the six vectors, the x-component is zero. Only the y and z components are present. Hence

these six vectors span the yz -plane in 𝑹𝟑.

The vectors (0,1,0) and (0,0,1) form the basis because the other vectors in the column space

can be expressed as a linear combination of these two vectors.

For example 2 010 + 1

001 =

021

(d) all vectors with positive components.

SOLUTION:

The vectors with positive components can be multiplied by any scalar (positive or negative)

so that they span the whole of 𝑹𝟑.

4.6 To decide whether 𝑏 is in the subspace spanned by 𝑤1,… , 𝑤𝑙 , let the vectors 𝑤 be the

columns of 𝐴 and try to solve 𝐴𝑥 = 𝑏 . What is it the result for

(a) 𝑤1 = 1, 1, 0 , 𝑤2 = 2, 2, 1 , 𝑤3 = 0, 0, 2 , 𝑏 = 3, 4, 5 ;

SOLUTION:

For the above system 𝑨𝒙 = 𝒃 to have a solution, the vector "b" must lie in the column

space of "A". Otherwise the system will not have a solution. "A" matrix is formed by the

vectors "w" as follows

A = 1 2 01 2 00 1 2

Hence 𝑨𝒙 = 𝒃 is 1 2 01 2 00 1 2

𝑥1

𝑥2

𝑥3

= 345

Applying Gauss elimination to the "A" matrix

R2 → R2-R1

→ 1 2 00 0 00 1 2

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Interchanging rows 2 and 3

→ 1 2 00 1 20 0 0

= U

Here the third pivot is equal to zero and hence the system does not have a solution. As already

stated if there is no solution then it means that the vector "b" does not lie in the column space of

"A". Hence 𝒃 is not in the subspace spanned by 𝒘𝟏, 𝒘𝟐, 𝒘𝟑.

(b) 𝑤1 = 1, 2, 0 , 𝑤2 = 2, 5, 0 , 𝑤3 = 0, 0, 2 , 𝑤4 = 0, 0, 0 , and any 𝑏?

SOLUTION:

Ax = 120

2 0 05 0 00 2 0

𝑥1

𝑥2

𝑥3

𝑥4

=

𝑏1

𝑏2

𝑏3

Applying Gauss elimination

R2 → R2-2*R1

→ 100

2 0 01 0 00 2 0

𝑥1

𝑥2

𝑥3

𝑥4

=

𝑏1

𝑏2

𝑏3

− 2𝑏1

There are pivots in column 1, 2 and 3. Hence 𝒙𝟏, 𝒙𝟐𝒂𝒏𝒅 𝒙𝟑 are basic variables whereas 𝒙𝟒

is a free variable.

2𝑥3 = 𝑏3

𝒙𝟑 = 𝒃𝟑/𝟐

𝒙𝟐 = 𝒃𝟐 − 𝟐𝒃𝟏

𝑥1 + 2𝑥2 = 𝑏1

𝒙𝟏 = 𝟓𝒃𝟏 − 𝟐𝒃𝟐

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𝑥1

𝑥2

𝑥3

𝑥4

=

5𝑏1 − 2𝑏2

𝑏2 − 2𝑏1

𝑏3/20

Whatever be the values of 𝑏1, 𝑏2, 𝑏3, there is definitely going to be a solution available as we can

clearly see from the above equation. Hence definitely vector "b" is in the subspace spanned

by vectors 𝒘𝟏, 𝒘𝟐, 𝒘𝟑, 𝒘𝟒 (which is the column space in 𝑹𝟑).

4.7 By locating the pivots, find a basis for the column space of

𝑈 =

0000

1000

4200

3200

Express each column that is not the basis as a combination of the basic columns. Find also a

matrix 𝐴 with this echelon form 𝑈, but a different column space.

SOLUTION:

Given

𝑈 =

0000

1000

4200

3200

which is in the echelon form. Here the columns 2 and 3 contain the pivots (1 and 2) respectively.

Hence columns 2 and 3 are the basis for the column space of "A".

Basis for column space are

1000

and

4200

....................(1)

Now columns that are not the basis can be expressed as a linear combination of the basis

columns as follows:

𝒗𝟒 = 𝒗𝟑 − 𝒗𝟐

That is

3200

=

4200

1000

Also 𝒗𝟏 = 𝟎𝒗𝟑 − 𝟎𝒗𝟐

By looking into "U" we can see that A =

𝟎𝟎𝟎𝟎

𝟏𝟎𝟎𝟐

𝟒𝟐𝟒𝟖

𝟑𝟐𝟒𝟔

for which when we apply the Gaussian

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elimination, we get back the given "U" matrix. Again here columns 2 and 3 are the basis

for the column space as the remaining two columns (1 and 4) can be expressed as a linear

combination of the basis columns.

Now the basis for column space are

1002

and

4248

which are different from the basis obtained in

(1). Hence the above matrix 𝐴 is a matrix with this echelon form 𝑈, but a different column space.

4.8 Find the dimension and construct a basis for the four subspaces associated with each of the

matrices

𝐴 = 0 1 40 2 8

00 and 𝑈 =

0 1 40 0 0

00 .

SOLUTION :

First let us consider the matrix 𝐴 = 0 1 40 2 8

00

Here m=2; n=4; r=1 (where r is the rank of the matrix A)

Column space R(A):

Dimension of R(A) is "r" which is equal to 1.

Basis for R(A) is 12 because in the echelon form of "A", only the second

column contains the pivot (1 is the pivot).

Row space R(𝐴𝑇):

Dimension of R(𝐴𝑇) is "r" which is equal to 1.

Basis for R(𝐴𝑇) is 0 1 4 0 because it is the only non-zero row in

the echelon form of "A".

Null space N(A):

Dimension of N(A) is "n-r" which is equal to 3.

Basis for null space are found by solving Ax=0 where the solution vector

"x" forms the null space of "A".

Solving Ax = 0 is the same as solving Ux = 0

0 1 40 0 0

00

𝑥1

𝑥2

𝑥3

𝑥4

= 00

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𝑥2 + 4𝑥3 = 0

𝑥2 = −4𝑥3

𝑥1

𝑥2

𝑥3

𝑥4

=

𝑥1

−4𝑥3

𝑥3

𝑥4

where 𝒙𝟏, 𝒙𝟑, 𝒙𝟒 are the free variables

When 𝒙𝟏 = 𝟏, 𝒙𝟑 = 𝟎, 𝒙𝟒 = 𝟎

𝑥1

𝑥2

𝑥3

𝑥4

=

1000

When 𝒙𝟏 = 𝟎, 𝒙𝟑 = 𝟎, 𝒙𝟒 = 𝟏

𝑥1

𝑥2

𝑥3

𝑥4

=

0001

When 𝒙𝟏 = 𝟎, 𝒙𝟑 = 𝟏, 𝒙𝟒 = 𝟎

𝑥1

𝑥2

𝑥3

𝑥4

=

0−410

Basis of null space of "A" are

1000

,

0001

and

0−410

Left Null space N(𝐴𝑇):

Dimension of N(𝐴𝑇) is "m-r" which is equal to 1

For finding the left null space of "A", we solve 𝑨𝑻𝒚 = 𝟎

0140

0280

𝑦1

𝑦2 =

0000

𝒚𝟏 = −𝟐𝒚𝟐

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Here 𝒚𝟐 is the free variable because in the echelon form of "𝑨𝑻", only the first column

contains a pivot.

U =

1000

2000

is the echelon form

When 𝒚𝟐 = 𝟏, 𝒚𝟏 = −𝟐

Therefore basis for left null space is −21

Now considering the matrix 𝑈 = 0 1 40 0 0

00

Column space:

Dimension is 1.

Basis is the column containing the pivots which is 10

Row space:

Dimension is 1.

Basis is the non-zero row which is 0 1 4 0

Null space:

Dimension is 3.

Basis are

1000

,

0001

and

0−410

Left Null space:

Dimension is 1.

Basis is obtained by solving 𝑼𝑻𝒚 = 𝟎

0140

0000

𝑦1

𝑦2 =

0000

Here again 𝒚𝟐 is the free variable because in the echelon form of "𝑼𝑻", only the first column

contains a pivot.

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When 𝒚𝟐 = 𝟏, basis of left null space is 01

4.9 Find the dimension and a basis for the four fundamental subspaces for both

𝐴 = 1 2 00 1 11 2 0

101 and 𝑈 =

1 2 00 1 10 0 0

100 .

SOLUTION:

First considering the matrix 𝐴 = 1 2 00 1 11 2 0

101

Here m=3; n=4; r=2 (since in the echelon form, the last row becomes zero

and there are two non-zero rows)

Column space:

Dimension is 2.

Basis are 101 𝑎𝑛𝑑

212 since these are the columns which contains pivots in

"U" matrix.

Row space:

Dimension is 2.

Basis are 1 2 0 1 and 0 1 1 0

Null space:

dimension is 2.

Basis are obtained by solving Ax = 0 or Ux= 0

1 2 00 1 10 0 0

100

𝑥1

𝑥2

𝑥3

𝑥4

= 000

Here 𝒙𝟏 & 𝑥𝟐𝒂𝒓𝒆 𝒃𝒂𝒔𝒊𝒄 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔 𝒂𝒏𝒅 𝒙𝟑 & 𝒙𝟒 𝒂𝒓𝒆 𝒕𝒉𝒆 𝒇𝒓𝒆𝒆 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔

𝒙𝟐 = −𝒙𝟑 and 𝒙𝟏 = 𝟐𝒙𝟑 − 𝒙𝟒

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𝑥1

𝑥2

𝑥3

𝑥4

=

2𝑥3 − 𝑥4

−𝑥3

𝑥3

𝑥4

= 𝑥3

2−110

+ 𝑥4

−1001

When 𝒙𝟑 = 𝟎, 𝒙𝟒 = 𝟏

𝑥1

𝑥2

𝑥3

𝑥4

=

−1001

When 𝒙𝟑 = 𝟏, 𝒙𝟒 = 𝟎

𝑥1

𝑥2

𝑥3

𝑥4

=

2−110

Therefore the basis are

−𝟏𝟎𝟎𝟏

𝒂𝒏𝒅

𝟐−𝟏𝟏𝟎

Left null space:

dimension is 1.

Solving 𝑨𝑻𝒚 = 𝟎

1 0 12 1 20 1 01 0 1

𝑦1

𝑦2

𝑦3

=

0000

Here 𝑦3 is the free variable and 𝒚𝟏 = −𝒚𝟑 & 𝒚𝟐 = 𝟎

When 𝒚𝟑 = 𝟏, the basis becomes −101

Now considering the matrix 𝑈 = 1 2 00 1 10 0 0

100

Column space:

Dimension is 2.

Basis are 100 𝑎𝑛𝑑

210

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Row space:

Dimension is 2.

Basis are 1 2 0 1 and 0 1 1 0

Null space:

Dimension is 2.

Basis are

−1001

𝑎𝑛𝑑

2−110

Left null space:

Dimension is 1.

Solving 𝑼𝑻𝒚 = 𝟎

1 0 02 1 00 1 01 0 0

𝑦1

𝑦2

𝑦3

=

0000

Here 𝑦3 is the free variable ; 𝒚𝟏 = 𝟎, 𝒚𝟐 = 𝟎,

Hence assigning 𝑦3 = 1, 𝑤𝑒 𝑔𝑒𝑡 𝑡𝑕𝑒 𝑏𝑎𝑠𝑖𝑠 𝑎𝑠 001

4.10 Describe the four subspaces in 3-dimensional space associated with

𝐴 = 0 1 00 0 10 0 0

.

SOLUTION:

Here m=3; n=3; r=2

Column space:

Dimension is 2.

Basis are 100 𝑎𝑛𝑑

010

The column space spans the x-y plane in the space 𝑹𝟑.

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Row space:

Dimension is 2.

Basis are 0 1 0 and 0 0 1

The row space spans the y-z plane in the vector space 𝑹𝟑

Null space:

Dimension is 1.

Solving Ax = 0

0 1 00 0 10 0 0

𝑥1

𝑥2

𝑥3

= 000

Basis for null space is 100

The null space spans the line x-axis in the vector space 𝑹𝟑.

Left null space:

Dimension is 1.

Solving 𝑨𝑻𝒚 = 𝟎

0 0 01 0 00 1 0

𝑦1

𝑦2

𝑦3

= 000

𝑦1 = 0, 𝑦2 = 0 (𝑦3 𝑖𝑠 𝑡𝑕𝑒 𝑓𝑟𝑒𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒)

Therefore Basis is 001

The left null space spans the line (z-axis) in the vector space 𝑹𝟑

4.11 Find the rank of 𝐴 and write the matrix as 𝐴 = 𝑢𝑣𝑇:

𝐴 = 1 0 00 0 02 0 0

306 and 𝐴 =

2 −22 −2

.

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SOLUTION:

a) 𝐴 = 1 0 00 0 02 0 0

306

R3 → R3-2*R1

→ 1 0 00 0 00 0 0

300 which is the echelon form

Rank of "A" is 1. All matrices with rank 1 can be expressed in the

form 𝑨 = 𝒖𝒗𝑻

Therefore A = 102 1 0 0 3

b) 𝐴 = 2 −22 −2

= 2 1 −11 −1

R2 → R2-R1

→ 1 −10 0

which is the echelon form.

Hence rank is 1. All matrices with rank 1 can be expressed in the

form 𝑨 = 𝒖𝒗𝑻

A = (2) 11 1 −1

4.12 Find a left-inverse and/or a right –inverse (when they exist) for

𝐴 = 1 1 00 1 1

and 𝑀 = 1 01 10 1

and 𝑇 = 𝑎 𝑏0 𝑎

.

SOLUTION:

a) 𝐴 = 1 1 00 1 1

Here r=2; m=2; n=3

r=m implies right inverse exists for "A"

AC = I where C is the right inverse and C = 𝑨𝑻(𝑨𝑨𝑻)−𝟏

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𝐴𝐴𝑇 = 2 11 2

(𝑨𝑨𝑻)−𝟏 = 2

3 −1

3

−13

23

C = 𝑨𝑻(𝑨𝑨𝑻)−𝟏 =

23

13

−13

−13

13

23

is the right inverse of "A"

b) 𝑀 = 1 01 10 1

R2 → R2-R1

→ 1 00 10 1

R3 → R3-R2

→ 1 00 10 0

which is the echelon form

Here r=2; m=3; n=2

r=n implies there exists a left inverse for "M"

BM = I where B is the left inverse and is equal to

B = (𝑴𝑻𝑴)−𝟏𝑴𝑻

𝑀𝑇𝑀 = 2 11 2

(𝑀𝑇𝑀)−1 = 2

3 −1

3

−13

23

B = (𝑴𝑻𝑴)−𝟏𝑴𝑻 = 2

3 1

3 −1

3

−13

13

23 is the left

inverse of "M"

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c) 𝑇 = 𝑎 𝑏0 𝑎

Here r=m=n=2. Hence both left and right inverse exist

and they both are the same.

det(T) = 𝒂𝟐

𝑻−𝟏 = 𝟏𝒂𝟐

𝑎 −𝑏0 𝑎

𝑻−𝟏 = 1

𝑎 −𝑏𝑎2

0 1𝑎

is the left and right inverse of "T"