Theory of Dimensionless Products: Some Concepts of Linear Algebra

Some Concepts of Linear Algebra

A linear combination of vectors

A vector w is called a linear combination of the vectors v₁, v₂, …, v_r if w can be expressed in the form w = k₁ v₁ + k₂ v₂ + … + k_r v_r where k₁, k₂, …, k_r are scalars.

Given, vectors u = [1, 2, −1]^T and v = [6, 4, 2]^T. Is w = [9, 2, 7]^T a linear combination of u and v? What about w' = [4, −1, 8]^T?

Tackling the question "Is w = [9, 2, 7]^T a linear combination of u and v?"

From definition we know that if w is a linear combination of u and v then there must exist scalars k₁ and k₂ such that

w = k₁ u + k₂ v Substituting the values for the vectors we get $w = k1*u + k2*v$ This system corresponds to k₁ + 6k₂ = 9
2k₁ + 4k₂ = 2
−k₁ + 2k₂ = 7 To solve the system let us add the top and bottom equations k₁ + 6k₂ = 9
−k₁ + 2k₂ = 7

     0 + 8k₂ = 16
         ⇒ k₂ = 2 Substituting k₂ = 2 into the middle equation returns 2k₁ + 4⋅2 = 2
            ⇒ k₁ = −3 Substituting k₁ = −3 and k₂ = 2 into the system returns k₁ + 6k₂ = −3 + 6⋅2 = 9
2k₁ + 4k₂ = 2(−3) + 4⋅2 = 2
−k₁ + 2k₂ = −(−3) + 2⋅2 = 7 The equations do not contradict the system equations. Thus the solution of the system are the scalars k₁ = −3 and k₂ = 2. That is, w can be expressed in the form w = −3u + 2v Hence, w is a linear combination of u and v.

Tackling the question "Is w' = [4, −1, 8]^T a linear combination of u and v?"

For w' to be a linear combination of u and v there must exist (some other) scalars k₁ and k₂ such that

w' = k₁ u + k₂ v or $w' = k1*u + k2*v$ or   k₁ + 6k₂ = 4
   2k₁ + 4k₂ = −1
−k₁ + 2k₂ = 8 To solve the system let us add the top and bottom equations   k₁ + 6k₂ = 4
−k₁ + 2k₂ = 8

     0 + 8k₂ = 12
             ⇒ k₂ = 3/2 Substituting k₂ = 3/2 into the middle equation returns 2k₁ + 4⋅3/2 = −1
                 ⇒ k₁ = −7/2 Substituting k₁ = −7/2 and k₂ = 3/2 into the system returns k₁ + 6k₂ = −7/2 + 6⋅3/2 = −7/2 + 9 ≠ 4
2k₁ + 4k₂ = 2(−7/2) + 4⋅3/2 = −7 +6 = −1
−k₁ + 2k₂ = −(−7/2) + 2⋅3/2 = 7/2 + 3 ≠ 8 Because the equations contradict the system equations k₁ = −7/2 and k₂ = 3/2 are the scalars that solves the system. In other words, the system has no solution. Therefore w' cannot be expressed in the form w' = k₁ u + k₂ v Hence, w' is not a linear combination of u and v.

❶

Span for vector space V

For any vector space V containing the vectors v₁, v₂, …, v_r if every vector in V can be expressed as a linear combination of v₁, v₂, …, v_r, then the vectors v₁, v₂, …, v_r are said to span V.

Given the vector space ℜ³ containing the vectors i = [1, 0, 0]^T, j = [0, 1, 0]^T and k = [0, 0, 1]^T, do the vectors i, j and k span ℜ³?

From definition we know that for the vectors i, j and k to span the vector space any vector in the vector space should be expressible as their linear combination.

Let us consider an arbitrary vector [a, b, c]^T in ℜ³. Picking one element of this vector and multiplying it to one of the span in question, say ai, and doing this for other elements we get

a i + b j + c k Substituting the values for the vectors we get $ai + bj + ck = [a b c]^T$ This is the same as [a, b, c]^T = a i + b j + c k Therefore, the vectors i, j and k span ℜ³ because any arbitrary vector in the said vector space can be represented as a linear combination of i, j and k.

What if the arbitrary vector is specified to be one of the vectors that spans ℜ³, say [a, b, c]^T = [1, 0, 0]^T = i?

The linear combination 1 i + 0 j + 0 k returns

$1i + 0j + 0k = [1 0 0]^T$ Thus, vector i can be expressed as the lienar combination; i = 1 i + 0 j + 0 k.

What this tells us is that when the definition states "every vector in the vector space should be expressible as a linear combination" it includes vectors that are said to span the vector space.

❷

Linearly Dependent and Indenpendent Set (of vectors)

For any vector space V containing the finite set of vectors S = { v₁, v₂, …, v_r } such that the vector equation k₁ v₁ + k₂ v₂ + … + k_r v_r = 0 has at least one solution, namely k₁ = 0, k₂ = 0, …, k_r = 0

If this is the only solution, S is called a linearly independent set.
If there are other solutions, S is called a linearly dependent set.

Given the vector space ℜ³ containing the vectors i = [1, 0, 0]^T, j = [0, 1, 0]^T and k = [0, 0, 1]^T, are the vectors i, j and k linearly independent?

From definition we know that for the vectors i, j and k to be independent its vector equation

k₁ i + k₂ j + k₃ k = 0 must have just one solution; k₁ = 0, k₂ = 0, k₃ = 0.

Substituting the values for the vectors in the vector equation we get

$k1i + k2j + k3k = [0 0 0]^T$ Thus k₁ = 0, k₂ = 0, k₃ = 0 is the only solution to the equation; the set S = { i, j, k } is linearly independent.

Given the vector space V containing the set S = { v₁, v₂, v₃ } such that vectors v₁ = [2, −1, 0, 3]^T, v₂ = [1, 2, 5, −1]^T and v₃ = [7, −1, 5, 8]^T. Are v₁, v₂ and v₃ linearly dependent?

From definition we have the vector equation

k₁ v₁ + k₂ v₂ + k₃ v₃ = 0 Substituting the values for the vectors in the vector equation we get $k1v1 + k2v2 + k3v3 = [0 0 0]^T$ This system corresponds to 2k₁ + k₂ + 7k₃ = 0
−k₁ + 2k₂ − k₃ = 0
         5k₂ + 5k₃ = 0
3k₁ − k₂ + 8k₃ = 0 The third equation says k₂ = −k₃. So if we let k₂ = 1 then k₃ = −1. Substituting these values to any of the remaining three equations yields k₃ = 3. Below shows the result for subsitution done on the first equation. 2k₁ + 1 − 7k₃ = 0
  ⇒    2k₁ − 6 = 0
or   k₁ = 6/2 = 3 Substituting k₁ = 3, k₂ = 1 and k₃ = −1 into the system 2k₁ + k₂ + 7k₃ = 2⋅3 + 1 + 7(−1) = 6 − 6 = 0
−k₁ + 2k₂ − k₃ = −3 + 2 − (−1)= −3 + 3 = 0
5k₂ + 5k₃ = 5 + 5(−1) = 5 − 5 = 0
3k₁ − k₂ + 8k₃ = 3⋅3 − 1 + 8(−1) = 9 − 9 = 0 we observer that the equations do not contradict each other. Therefore, in addition to the solution k₁ = 0, k₂ = 0, k₃ = 0 another solution is k₁ = 3, k₂ = 1, k₃ = −1 Therefore, the set S = { v₁, v₂, v₃ } is linearly dependent.

❸

Linearly Dependent Set and Linear combination of vectors

The term linearly dependent suggests that vectors in the set in some way depend on each other. This statement can be illustrated with an example.

Let the set S = { v₁, v₂, …, v_r } be a linearly dependent set. Then by definition the vector equation

k₁ v₁ + k₂ v₂ + … + k_r v_r = 0 will have a solution other that k₁ = k₂ = … = k_r = 0.

Furthermore, assume k₁ ≠ 0. Multiplying both sides by 1/k₁ and solving for v₁ yields

v₁ = (−k₂/k₁) v₂ + (−k₃/k₁) v₃ + … + (−k_r/k₁) v_r Therefore, v₁ can be expressed as a linear combination of the remaining vectors.

A set with two or more vectors is linearly dependent if and only if at least one of the vectors is a linear combination of the remaining vectors.

Two vectors v₁ and v₂, form a linearly dependent set if and only if one of them is a scalar multiple of the other.

For a linearly dependent set S = { v₁, v₂ } and the vector expression k₁ v₁ + k₂ v₂ = 0, if k₁ ≠ 0 then v₁ = (−k₂/k₁) v₂. That is v₁ is a scalar multiple of v₂.

Therefore, if S = { v₁, v₂ } is linearly independent then geometrically the vectors v₁ and v₂ do not lie along the same line. ♻

If a vector space ℜ² has a set that contains more than two vectors then the set is linearly dependent.

Let S = { v₁, v₂, …, v_r } be a set of vectors in ℜⁿ. If r > n, then S is linearly dependent.

Assume v₁ = [v₁₁, v₁₂, …, v_1n]
v₂ = [v₂₁, v₂₂, …, v_2n]
⋮
v_r = [v_r1, v_r2, …, v_rn] For the vector equation k₁ v₁ + k₂ v₂ + … + k_r v_r = 0 Substituting the values for the vectors we get $k1v1 + k2v2 + ... + k3v3 = [0 0 0]$ This system corresponds to v₁₁ k₁ + v₂₁ k₂ + … + v_r1 k_r = 0
v₁₂ k₁ + v₂₂ k₂ + … + v_r2 k_r = 0
⋮
v_1n k₁ + v_2n k₂ + … + v_rn k_r = 0 That is, a homogeneous system of n equations with r unknowns k₁, k₂, …, k_r.

Invoking the fundamental theorem of homogeneous system of linear equations

A homogeneous system of linear equations with more unknowns than equations always has infinitely many solutions.

We can conclude that since the above vector equation for the set S = { v₁, v₂, …, v_r } in the vector space ℜⁿ is a homogeneous system with r > n the system has nontrivial solutions, i.e. many solutions.

Therefore, for r > n the set S = { v₁, v₂, …, v_r } in ℜⁿ is a linearly dependent set.∎

❹

Basis for vector space V

For any vector space V the finite set of vectors in V, S = { v₁, v₂, …, v_r } is called a basis for V if

S spans V;
S is linearly independent.

Given the vector space ℜ³ and its set of vectors S = { i, j, k } such that i = [1, 0, 0]^T, j = [0, 1, 0]^T and k = [0, 0, 1]^T, is S a basis for ℜ³?

Does S span ℜ³?

We showed in the above example that for an arbitrary vector [a, b, c]^T in ℜ³ the vector equation

a i + b j + c k becomes $ai + bj + ck = [a b c]^T$ which is [a, b, c]^T = a i + b j + c k Therefore, the vectors i, j and k span ℜ³ because any arbitrary vector in the said vector space can be represented as a linear combination of i, j and k.

Is S a linearly independent set?

We showed in another example that for some scalars k₁, k₂ and k₃ which form the coefficients of the vector equation

k₁ i + k₂ j + k₃ k = 0 or $k1i + k2j + k3k = [0 0 0]^T$ We see that k₁ = 0, k₂ = 0, k₃ = 0 is the only solution to the equation; the set S = { i, j, k } is linearly independent.

Thus, the set of vectors S spans its vector space and is a linearly independent set. Thus S is a basis for the vector space ℜ³.

Notice that any vector v = [v₁, v₂, v₃]^T in ℜ³ can be written as a linear combination of all the vectors in the set S.

$v1i + v2j + v3k = [v1 v2 v3]^T$ That is v = v₁ i + v₂ j + v₃ k S is a special kind of basis called the standard basis for ℜ³.

If the set of vectors S = { v₁, v₂, …, v_r } spans the vector space V and any vector v = [v₁, v₂, …, v_r] in V can be expressed as v = v₁ v₁ + v₂ v₂ + … + v_r v_r, then S is called the standard basis for V.

❺

Dimension of vector space V

Given a finite vector space V, the number of vectors in a basis for V is called the dimension of V. Furthermore, the zero vector is defined to have dimension zero.

Given the homogeneous system 2x₁ + 2x₂ − x₃            + x₅ = 0
−x₁ − x₂ + 2x₃ − 3x₄ + x₅ = 0
   x₁ + x₂ − 2x₃            − x₅ = 0
                     x₃ + x₄ + x₅ = 0 What is the dimension of its solution space?

We must first find the set of vectors that span the solution space. But to this we need to know the solutions for the homogeneous system.

The augmented matrix of the homogeneous system is

$[2 2 -1 0 1 0; -1 -1 2 -3 1 0; 1 1 -2 0 -1 0; 0 0 1 1 1 0]$ whose reduced-echelon form is $[1 1 0 0 1 0; 0 0 1 0 1 0; 0 0 0 1 0 0; 0 0 0 0 0 0]$ The corresponding system of equation is x₁ + x₂ +               x₅ = 0
                x₃       + x₅ = 0
                     x₄          = 0 Solving for the leading variables we get              x₁ = −x₂ − x₅
    x₃ = −x₅
x₄ = 0 Setting x₂ and x₅ to some arbitrary constants, say, x₂ = s and x₅ = t, then the solution space (or solution set) for the homogeneous system is x₁ = −s − t,      x₂ = s,      x₃ = −t,      x₄ = 0,      x₅ = t The solution space can be written as the vector [x₁, x₂, x₃, x₄, x₅]^T = [−s − t, s, −t, 0, t]^T. Taking out the unknown terms s and t in the vector we get $[x1 x2 x3 x4 x5] = s [-1 1 0 0 0] + t [-1 0 -1 0 1]$ If we define v₁ = [−1, 1, 0, 0, 0]^T and v₂ = [−1, 0, −1, 0, 1]^T then the solution space can be expressed as the solution vector [x₁, x₂, x₃, x₄, x₅]^T = s v₁ + t v₂ which as we can see is a linear combination of vectors in the set S = { v₁, v₂ } where s and t are scalars. Furthermore we know that the vector equation is consistent for any arbitrary s and t values; that is, the vectors of the set span the solution space.

To check if the set S = { v₁, v₂ } is linearly independent or dependent we need to solve the vector equation

s v₁ + t v₂ = 0 Substituting the values for the vectors we get $sv1 + tv2 = [0 0 0 0 0]^T$ showing that s = 0 and t = 0 is the only solution. Thus, the set is independent and because it also span the solution space we can conclude that the set S = { v₁, v₂ } is a basis for the solution space.

The number of vectors in the basis S = { v₁, v₂ } is two so the solution space is two dimensional.

Finding the dimension of a vector space V requires first finding its basis which by definition requires determining if the set in question spans V and is linearly independent. But, for a vector space V known to be n-dimensional finding if some set of n vectors is a basis for V requires determining if it is either spanning or linearly independent (theorem below).

If S = { v₁, v₂, …, v_n } is a set of n linearly independent vectors in a n-dimensional space V, then S is a basis for V.

If S = { v₁, v₂, …, v_n } is a set of n vectors that spans an n-dimensional space V, then S is a basis for V.

If S = { v₁, v₂, …, v_n } is a set of n linearly independent vectors in a n-dimensional space V and r < n, then S can be enlarged to a basis for V; that is, there are vectors v_r+1, …, v_n such that { v₁, v₂, …, v_r, v_r+1, …, v_n } is a basis for V.

❻

Row space and column space

Given the m × n $A = [a11 a12 ... a1n; a21 a22 ... a2n; ...; am1 am2 ... amn]$ such that vectors r₁ = [a₁₁, a₁₂, …, a_1n]
r₂ = [a₂₁, a₂₂, …, a_2n]
⋮
r_m = [a_r1, a_r2, …, a_mn] from rows of A are called row vectors of A and the vectors c₁ = [a₁₁, a₂₁, …, a_m1]^T
c₂ = [a₁₂, a₂₂, …, a_m2]^T
⋮
c_n = [a_1n, a_2n, …, a_mn]^T formed from columns of A are called column vectors of A.

The subspace (set of vectors) of ℜⁿ spanned by the row vectors of the m × n matrix A is called the row space of A.

The subspace (set of vectors) of ℜⁿ spanned by the column vectors of the m × n matrix A is called the column space of A.

If the row space or column space is taken as a vector space how can its basis be found? To help us find the basis below are some theorems.

Elementary row operations do not change the row space of a matrix A.

Let r₁, r₂, …, r_m represent the row vectors of a m × n matrix A. Also, suppose matrix B is the result of performing elmentary row operations on A.

Then to show A and B have the same row space we need to show that

1. every row vector in B is also in row space of A
2. every row vector in A is in the row space of B

Consider the following possible elementary row operations

• row interchange
B and A will have the same row vectors and hence the same row space.

• scalar multiplication of row or addition of a multiple of one row to another
       The row vectors r₁', r₂', …, r_m' of B are linear combinations of r₁, r₂, …, r_m. Therefore, row vectors of B lie in the row space of A.
       What about the linear combinations of r₁', r₂', …, r_m', do they lie in the row space of A?
       Because of the theorem that says that a vector space is closed under addition and scalar multiplication
If W is a set of one or more vectors from a vector space V, then W is a subspace of V if and only if the following conditions hold.
- W is closed under addition; If u and v are any vectors in W, then u + v is in W.
- W is closed under scalar multiplication; If k is any scalar and u is any vectors in W, then ku is in W.
and since the row vectors of B lie in the row space of A (consequence of the row vectors of B being linear combinations of row vectors of A) we can conclude that all linear combinations of r₁', r₂', …, r_m' will also lie in the row space of A.
Thus, every vector in the row space of B is in the row space of A.∎¹

Since matrix B was obtained from A by performing row operations $A to B by elementary row operations$ we can obtain A by doing the inverse operations on B $A from B by inverse row operations$ Therefore, using the above argument we can show that vectors r₁, r₂, …, r_m lie in the row space of B and all linear combinations of r₁, r₂, …, r_m lie in the row space of B.

Hence, every vector in the row space of A is in the row space of B.∎²

Nonzero row vectors in a row-echelon form of a matrix A form a basis for the row space of A.

Let r₁, r₂, …, r_m represent the row vectors of a m × n matrix A whose row-echelon form is B containing the row vectors r₁', r₂', …, r_m' Let us also denote the zero row vectors r_i' = 0 of B as 0₁, 0₂, …, 0_p and the nonzero row vectors r_i' ≠ 0 of B as 0₁, 0₂, …, 0_q Hence, we need to show that

1. the set { 0₁, 0₂, …, 0_p } is not a basis for the row space of A
2. the set { 0₁, 0₂, …, 0_q } is a basis for the row space of A

By definition a basis S of space V must span V and S is linearly independent.

Because of the theorem

Elementary row operations do not change the row space of a matrix A.

we know that the set { r₁', r₂', …, r_m' } span the row space of A, consequently both { 0₁, 0₂, …, 0_p } and { 0₁, 0₂, …, 0_q } wil span the row space of A.

Expressing set { 0₁, 0₂, …, 0_p } as the vector equation

k₁ 0₁ + k₂ 0₂ + … + k_p 0_p = 0 we see that it has infinite solution since any arbitray values of k₁, k₂, …, k_r is the solution. Hence, the set of zero row vectors are linearly dependent. Therefore, the set is not a basis for the row space of A.∎¹

For the set { 0₁, 0₂, …, 0_q } with the vector equation

k₁ 0₁ + k₂ 0₂ + … + k_q 0_q = 0 the only solution is k₁ = 0, k₂ = 0, …, k_r = 0 The set { 0₁, 0₂, …, 0_q } is linearly independent. Therefore, the set is a basis for the row space of A.∎²

If the space is spanned by the vectors v₁ = [1, −2, 0, 0, 3]
v₂ = [2, −5, −3, −2, 6]
v₃ = [0, 5, 15, 10, 0]
v₄ = [2, 6, 18, 8, 6] then by definition the set S = { v₁, v₂, v₃, v₄ } span the vector space. The set can be viewed as a row space $[1 -2 0 0 3;0 1 3 2 0;0 0 1 1 0;0 0 0 0 0]$ whose row-echelon form returns $[1 -2 0 0 3;2 -5 -3 -2 6;0 5 15 10 0;2 6 18 8 6]$ From this matrix the nonzero row vectors are w₁ = [1, −2, 0, 0, 3]
w₂ = [0, 1, 3, 2, 0]
w₃ = [0, 0, 1, 1, 0] The above theorem tells us that the set { w₁, w₂, w₃ } span the row space. And since the set of row vectors S span the vector space, the set { w₁, w₂, w₃ } span the same vector space.

Therefore, { w₁, w₂, w₃ } is the basis for the vector space.

If A is any matrix, then the row space and column space of A have the same dimension.

Given the m × n matrix $A = [a11 a12 ... a1n; a21 a22 ... a2n; ...; am1 am2 ... amn]$ with row vectors denoted by r₁, r₂, …, r_m, suppose its row space has k dimension with basis of the row space S = { b₁, b₂, …, b_k }, where b_i = [b_i1, b_i2, …, b_in].

Since S is a basis for the row space, each row vector can be expressed as a linear combination of the basis vectors b₁, b₂, …, b_k.

r₁ = c₁₁b₁ + c₁₂b₂ + … + c_1kb_k
r₂ = c₂₁b₁ + c₂₂b₂ + … + c_2kb_k
⋮
r_m = c_m1b₁ + c_m2b₂ + … + c_mkb_k
Because two vectors in a vector space ℜⁿ are equal if and only if corresponding components are equal, equating the j^th component in r_i with the corresponding components of the vectors b₁, b₂, …, b_k on the right hand side of the above expression we get a_1j = c₁₁b_1j + c₁₂b_2j + … + c_1kb_kj
a_2j = c₂₁b_1j + c₂₂b_2j + … + c_2kb_kj
⋮
a_mj = c_m1b_1j + c_m2b_2j + … + c_mkb_kj This is equivalent to $[a1j, a2j, ..., amj]^T = b1j [c11, c21, ..., cm1]^T + b2j [c12, c22, ..., cm2]^T + ... + bkj [c1k, c2k, ..., cmk]^T$ with the left hand side of the equation representing the j^th column vector of A for j = 1, 2, …, n and the right hand side showing that each column vector of A lies in the vector space spanned by the k vectors; [c₁₁, c₂₁, …, c_m1]^T, [c₁₂, c₂₂, …, c_m2]^T, …, [c_1k, c_2k, …, c_mk]^T.

Therefore, the set { [c₁₁, c₂₁, …, c_m1]^T, [c₁₂, c₂₂, …, c_m2]^T, …, [c_1k, c_2k, …, c_mk]^T } spans the column space of A. Hence, the column space has dimension ≤ k.

Since we assumed k to be the dimension of the row space

k = dim(row space of A) thus dim(column space of A) ≤ dim(row space of A) Also, because matrix A is arbitrary, the same conclusion applies to A^T dim(column space of A^T) ≤ dim(row space of A^T) Transposing a matrix converts rows to columns and columns to rows $A = [a11 a12 ... a1n; a21 a22 ... a2n; ...; am1 am2 ... amn] and A^T = [a11 a21 ... am1; a12 a22 ... am2; ...; a1n a2n ... amn]$ row space of A = column space of A^T
column space of A = row space of A^T Thus, dim(row space of A) ≤ dim(column space of A) But, dim(column space of A) ≤ dim(row space of A) therefore dim(column space of A) = dim(row space of A)∎

Given the matrix A $[1 0 1 1;3 2 5 1;0 4 4 -4]$ its row-echelon form is $[1 0 1 1;0 1 1 -1;0 0 0 0]$ Since there are two nonzero row vectors in the row-echelon form that span the row space of A, the row space is two dimensional.

The transpose of matrix A, A^T is

$[1 0 1 1;3 2 5 1;0 4 4 -4]$ whose row-echelon form is $[1 3 0;0 1 2;0 0 0;0 0 0]$ There are two nonzero row vectors therefore, the row space of A^T is two dimensional.

Since the column vectors of A are the row vectors of A^T the basis for the row space of A^T

{ [1, 3, 0]^T, [0, 1, 2]^T } is the basis for the column space of A. Thus, the column space of A is two dimensional; the row space and column space of A are two dimensional.

If A is a matrix that is not square, then either the row vectors of A or the column vectors of A are linearly dependent.

Consider the consistent linear system Ax = b such that A is a m × n matrix, x a n × 1 vector and b a m × 1 column vector. [A b] is the augmented matrix.

From the theorem^{(ibid. 5)}

Nonzero row vectors in a row-echelon form of a matrix A form a basis for the row space of A. we know that the number of the resulting nonzero vectors gives us the dimension of the column or row space of A and consequently the rank of matrix A.

The preceeding theorem states that Ax = b is consistent if and only if b is in the column space of A. Thus, for a consistent system since the only unique column vector, b between matrices A and [A b] must be in the column space of A, the number of nonzero vectors following row operations on the A and [A b] will be equal.

Therefore, for a consistent system Ax = b, matrices A and [A b] will have equal ranks.

❼

Rank of a matrix

The dimension of the row and column space of a matrix A is called the rank of A.

From theorem

If A is any matrix, then the row space and column space of A have the same dimension.

we know that column space of A is two dimensional.

The rank of A is 2.

If A_{m × n} is a matrix and m ≠ n, then the largest possible rank is ≤ m if m < n
≤ n if m > n

From the lemma The dimension of the row space and column space for any A_{m × n} matrix is the same. and the lemma The nonzero row vectors in a row-echelon form of any A_{m × n} matrix is a basis for the the row space. Thus, if m < n, then
dimension of row space = dimension of column space ≤ m
if m > n, then
dimension of row space = dimension of column space ≤ n Then from the definition of rank it follows if m < n, then
the largest possible rank ≤ m
if m > n, then
the largest possible rank ≤ n∎

Since the definition of rank refers to the number of vectors in a set that form a basis and since a basis must span (row/column space) and be linearly independent, then

If A_{m × n} is a matrix and m ≠ n, then if m < n, then
the column vectors of A are linearly dependent
if m > n, then
the row vectors of A are linearly dependent∎

From the preceeding theorem it follows

If A is a matrix that is not square, then either the row vectors of A or the column vectors of A are linearly dependent.

Summary

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Dimensionless Products, π (p:6) ➽

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