Download Maths linear algebra and more Essays (university) Mathematical Statistics in PDF only on Docsity!
Linear Algebra
Note: This module is prepared from the text book (Elementary Linear Algebra by S. Andrilli and D. Hecker, 4th Edition, 2012, Elsevier) just to help the students. The study material is expected to be useful but not exhaustive. For detailed study, the students are advised to attend the lecture/tutorial classes regularly, and consult the text book.
Appeal: Please do not print this e-module unless it is really necessary.
Dr. Suresh Kumar, Department of Mathematics, BITS Pilani, Pilani Campus
Chapter 2 (2.1-2.4)
Elementary row operations
There are three elementary row operations: (1) Interchanging two rows Ri and Rj (symbolically written as Ri ↔ Rj ) (2) Multiplying a row Ri by a non-zero number k (symbolically written as Ri → kRi) (3) Adding constant k multiple of a row Rj to a row Ri (symbolically written as Ri → Ri + kRj )
To see how row transformations are applied, consider the matrix
A =
Applying R 1 ↔ R 2 , we obtain
A ∼
Applying R 2 ↔ (1/2)R 2 , we obtain
A ∼
Applying R 2 → R 2 − 2 R 1 and R 3 → R 3 − 3 R 1 , we get
A ∼
Note. The matrices resulting from the row transformation(s) are known as row equivalent matrices. That is why the sign of equivalence ∼ is used after applying row transformations. So we can write
A =
Notice that row equivalent matrices can be obtained from each other by applying suitable row transfor- mation(s), but row equivalent matrices need not be equal.
Applying R 2 → R 2 − 3 R 3 and R 1 → R 1 − 3 R 3 , we get
A ∼
Finally, applying R 1 → R 1 − 2 R 2 , we get
A ∼
the RREF of A.
Useful Tip: From the above example, one may notice that for getting RREF of a matrix we make use of first row to make zeros in the first column, second row to make zeros in the second column and so on. Note: REF of matrix is not unique, but RREF is unique.
Inverse of a Matrix
Let A be a matrix with m rows say R 1 , R 2 ,....,Rm, and B be a matrix with n columns say C 1 , C 2 ,.....,Cn. Then the product matrix AB is of order m × n, and
A.B =
R 1
R 2
Rm
[
C 1 C 2 ... Cn
]
R 1 C 1 R 1 C 2 ... R 1 Cn R 2 C 1 R 2 C 2 ... R 2 Cn ... ... ... ... RmC 1 RmC 2 ... RmCn
=^ AB
If we interchange two rows in A say R 1 ↔ R 2 , then the first two rows of AB also get interchanged. Similarly, it is easy to see that applying any of the other two row operations in A is equivalent to applying the same row operation in AB. Thus, we conclude that applying any elementary row operation in the matrix A is equivalent to applying the same elementary row operation in the matrix AB. Hence, if R is any row operation, then R(AB) = R(A)B. Note that the matrix B is left unchanged. We make use of this fact to find the inverse of a matrix. Let A be a given non-singular matrix of order n × n. To find the inverse of A, first we write A = InA. In this identity, we apply the elementary row operations on the left hand side matrix A in such a way that it transforms to In, the RREF of A. As discussed above, the same row operations apply to the first matrix In on right hand side and suppose it transforms to a matrix B. Then, we have In = BA. Therefore, A−^1 = B. This method for obtaining the inverse of a matrix is called Gauss-Jordan method.
Note: One may use elementary column operations (Ci ↔ Cj , Ci → kCi and Ci → Ci + kCj ) also to find A−^1. In this case, we write A = AIn and apply elementary column operations to obtain In = AB so that A−^1 = B. It may be noted that we can not apply row and column operations together for finding A−^1 in the Gauss-Jordan method.
Ex. Use Gauss-Jordan method to find inverse of the matrix
A =
Sol. We write A = I 3 A and therefore
A.
Applying R 1 ↔ R 2 , we obtain
A.
Applying R 2 → R 2 − 2 R 1 and R 3 → R 3 − 3 R 1 , we get
A.
Applying R 2 ↔ R 3 , we obtain
A.
Applying R 2 → −R 2 and R 3 → −R 3 , we get
A.
Applying R 2 → R 2 − 3 R 3 and R 1 → R 1 − 3 R 3 , we get
A.
Finally, applying R 1 → R 1 − 2 R 2 , we get
A.
∴ A−^1 =
Useful Tip: To find inverse of A, first write A = InA. Then change the left hand side matrix A to its RREF by applying suitable row transformations so that In = BA and A−^1 = B.
Note: You might be familiar that inverse of a square matrix A exists if and only if A is non-singular, that is, |A| 6 = 0. Note that RREF of a non-singular matrix is always a unit matrix.
Note: You know that when a system of n linear equations in n variables is represented in the matrix form AX = B, then A is n square matrix of the coefficients of the variables, and solution of the system reads as X = A−^1 B provided A−^1 exists. In case, if number of equations is not equal to the number of variables, then A is not a square matrix, and therefore A−^1 is not defined. In what follows, we present a general strategy for solving a system of linear equations. First we introduce the concept of rank of a matrix.
which is formed by inserting the column of matrix B next to the columns of A, is known as augmented matrix of the matrices A and B. We shall denote it by [A : B]. The following theorem tells us about the consistency of the system AX = B.
Theorem: The system AX = B of linear equations has a (i) unique solution if rank(A)=rank([A : B]) = n, (ii) infinitely many solutions if rank(A)=rank([A : B]) < n, (iii) no solution if rank(A) 6 =rank([A : B]). From this theorem, we deduce the following.
- If B = O, then obviously rank(A)=rank([A : B]). It implies that the homogeneous system AX = O always has at least one solution. Further, it has the unique trivial solution X = O if rank(A)= n and infinitely many solutions if rank(A)< n.
- To find rank([A : B]), we find REF of the augmented matrix [A : B]. From the REF of [A : B], we can immediately write the rank(A). Then using the above theorem, we decide about the nature of solution of the system. In case the solution exists, it can be derived using the REF of the matrix [A : B] as illustrated in the following example.
Ex. Test the consistency of the following system of equations and find the solution, if exists.
2 x + 3y + 4z = 11 , x + 5y + 7z = 15 , 3 x + 11y + 13z = 25.
Sol. Considering the matrix form AX = B of the given system, we have X =
x y z
(^) and the augmented
matrix
[A : B] =
Applying R 1 ↔ R 2 , we obtain
[A : B] ∼
Applying R 2 → R 2 + (−2)R 1 and R 3 → R 3 + (−3)R 1 , we have
[A : B] ∼
Applying R 2 → (− 1 /7)R 2 , we have
[A : B] ∼
Applying R 3 → R 3 + 4R 2 , we have
[A : B] ∼
Applying R 3 → (− 7 /16)R 3 , we have
[A : B] ∼
This is the REF of [A : B], which contains three non-zero rows. So rank([A : B])= 3. Also, we see that REF of the matrix A contains three non-zero rows. So rank(A)= 3. Further, there are three variables in the given system. So rank(A)=rank([A : B]) = 3. Hence, the given system of equations is consistent and has a unique solution. From the REF of [A : B], the given system of equations is equivalent to x + 5y + 7z = 15 , y + (10/7)z = 19 / 7 , z = 4.
From the third equation, we have z = 4. Inserting z = 4 into second equation, we obtain y = −3. Finally, plugging z = 4 and y = −3 into first equation, we get x = 2. Hence, the solution of the given system is x = 2, y = −3 and z = 4. Note: In the above example, first we have found the REF of the matrix [A : B]. Then we have written the reduced system of equations and found the solution using back substitution. This approach is called Gauss Elimination Method. If we use RREF of [A : B] to obtain the solution, then this approach is called Gauss-Jordan Method. For illustration of this method, we start with the REF of the matrix [A : B] as obtained above. We have
[A : B] ∼
Applying R 2 → R 2 + (− 10 /7)R 3 and R 1 → R 1 + (−7)R 3 , we get
[A : B] ∼
Applying R 1 → R 1 − 5 R 2 , we get
[A : B] ∼
The RREF of [A : B] yields x = 2, y = −3 and z = 4.
Ex. Test the consistency of the following system of equations
x + y + 2z + w = 5, 2 x + 3y − z − 2 w = 2, 4 x + 5y + 3z = 7.
Sol. Here the augmented matrix is
[A : B] =
Ex. Test the consistency of the following system of equations and find the solution, if exists.
6 x 1 − 12 x 2 − 5 x 3 + 16x 4 − 2 x 5 = − 53 , − 3 x 1 + 6x 2 + 3x 3 − 9 x 4 + x 5 = 29 , − 4 x 1 + 8x 2 + 3x 3 − 10 x 4 + x 5 = 33.
Sol. Here the augmented matrix is
[A : B] =
Using suitable row transformations (you can do it), we find
[A : B] ∼
We see that the rank([A : B]) = 3 =rank(A) < 5(the number of variables in the system). So the given system of equations has infinitely many solutions. The reduced system of equations is
x 1 − 2 x 2 + x 4 = − 4 , x 3 − 2 x 4 = 5 , x 5 = 2.
The second and fourth columns in RREF of [A : B] do not carry the leading entries, and correspond to the variables x 2 and x 4 which we consider as independent variables. Let x 2 = b and x 4 = d. So from the reduced system of equations, we get
x 1 = 2b − d − 4 , x 2 = b, x 3 = 2d + 5, x 4 = d, x 5 = 2.
Hence, the complete solution set is
{(x 1 , x 2 , x 3 , x 4 , x 5 ) = (2b − d − 4 , b, 2 d + 5, d, 2) : b, d ∈ R}.
Row vectors, linear combinations and row space
The rows of a matrix are treated as row vectors. For example, consider the matrix
A =
Then there are three row vectors given by [2, 3 , 4], [1, 5 , 7] and [3, 11 , 13]. Sum of any scalar multiples of row vectors is called a linear combination while the set of all linear combinations of the row vectors is called row space of the matrix. For example, if a, b and c are any three real numbers, then the expression
a[2, 3 , 4] + b[1, 5 , 7] + c[3, 11 , 13] = [2a + b + 3c, 3 a + 5b + 11c, 4 a + 7b + 13c]
is a linear combination of the vectors [2, 3 , 4], [1, 5 , 7] and [3, 11 , 13] while the set
{[2a + b + 3c, 3 a + 5b + 11c, 4 a + 7b + 13c] : a, b, c ∈ R}
of all linear combinations of the row vectors of A is the row space of A. Recall that two matrices are row equivalent if one can be derived from the other by applying suitable row transformation(s). The row spaces of two row equivalent matrices are always same. Afterall, row spaces are nothing but the linear combinations of the row vectors of the matrices. Consequently, the interplay of row operations gives rise to the same sets of linear combinations of row vectors, and hence the same row spaces. This fact is quite useful for determining a simple form of row space of a matrix. We know RREF of a matrix is its row equivalent matrix, and is unique as well. So we shall prefer to use RREF of the given matrix to write its row space. For example, consider the matrix
A =
Its RREF is
So row space of A reads as
{a[1, 0 , 0] + b[0, 1 , 0] + c[0, 0 , 1] = [a, b, c] : a, b, c ∈ R}.
Ex. Determine whether the row vector [5, 17 , −20] is in the row space of the matrix
P =
Sol. We need to check whether there exists three real numbers a, b and c such that
[5, 17 , −20] = a[3, 1 , −2] + b[4, 0 , 1] + c[− 2 , 4 , −3].
This gives the following system of linear equations:
3 a + 4b − 2 c = 5 , a + 4c = 17 , − 2 a + b − 3 c = − 20.
Here the augmented matrix is
[A : B] =
So we get a = 5, b = −1, c = 3, and
[5, 17 , −20] = 5[3, 1 , −2] − [4, 0 , 1] + 3[− 2 , 4 , −3].
Thus, [5, 17 , −20] is a linear combination of the row vectors of P , and hence is in the row space of P.
So we get a = 0, b = 0, but c is arbitrary. Thus, the row vectors of the matrix P are not LI. Notice that the third row in P is sum of the first two rows. That is why we got the linear dependence of the row vectors of P.
Note: We can talk about the linear independence of the rows of a matrix by looking at the rank of the matrix as well. If rank of a matrix is equal to the number of its rows, then the rows of the matrix are LI. At this stage, you should understand the interplay of row operations, RREF, rank, linear system of equations and linear independence of rows.
Homework: Do the problems of exercises 2.1 to 2.4 from the textbook.
Chapter 3 (3.4)
Eigenvalues and Eigenvectors
A real number λ is an eigenvalue of an n-square matrix A iff there there exists a non-zero n-vector X such that AX = λX or (A − λIn)X = 0. The non-zero vector X is called eigenvector of A correspond- ing to the eigenvalue λ. Since the non-zero vector X is non-trivial solution of the homogeneous system (A − λIn)X = 0 , we must have |A − λIn| = 0. This equation, known as the characteristic equation of A, yields eigenvalues of A. So to find the eigenvalues of A, we solve the equation |A − λIn| = 0.
The eigenvectors of A corresponding to λ are the non-trivial solutions (X 6 = 0 ) of the homogeneous system (A − λIn)X = 0.
The set Eλ = {X : AX = λX} is known as the eigenspace of λ. Note that Eλ contains all eigenvectors of A corresponding to the eigenvalue λ in addition to the vector X = 0 since A 0 = λ 0. Of course, by definition X = 0 is not an eigenvector of A.
Ex. Find eigenvalues and eigenvectors of A =
[
]
Sol. Here, the characteristic equation of A, that is, |A − λI 2 | = 0 reads as ∣ ∣ ∣∣
[
12 − λ − 51 2 − 11 − λ
]∣∣
This leads to a quadratic equation in λ given by
λ^2 − λ − 30 = 0.
Its roots are λ = 6, −5, the eigenvalues of A. Now, the eigenvectors corresponding to λ = 6, are the non-trivial solutions X of the homogeneous system (A − 6 I 2 )X = 0. So to find eigenvectors of A corresponding to the eigenvalue λ = 6, we need to solve the homogeneous system: [ 6 − 51 2 − 17
] [
x 1 x 2
]
[
]
Applying R 1 → (1/6)R 1 , we get [ 1 − 17 / 2 2 − 17
] [
x 1 x 2
]
[
]
Applying R 2 → R 2 − 2 R 1 , we get [ 1 − 17 / 2 0 0
] [
x 1 x 2
]
[
]
So the system reduces to x 1 − (17/2)x 2 = 0. Letting x 2 = a, we get x 1 = (17/2)a. So [x 1 , x 2 ] = [(17/2)a, a] = (1/2)a[17, 2]
. So the eigenvectors corresponding to λ = 6 are non-zero multiples of the vector [17, 2]. The eigenspace corresponding to λ = 6, therefore, is E 6 = {a[17, 2] : a ∈ R}. Likewise, to find the eigenvectors corresponding to λ = −5, we solve the homogeneous system (A + 5I 2 )X = 0 , that is, [ 17 − 51 2 − 6
] [
x 1 x 2
]
[
]
Now, let us find the eigenvectors of A corresponding to λ = 2. For this, we need to find non-zero solutions X of the homogeneous system (A − 2 I 3 )X = 0 , that is,
x 1 x 2 x 3
Using suitable row operations, we find
x 1 x 2 x 3
So the system reduces to
x 1 − (4/3)x 2 + 2x 3 = 0.
Letting x 2 = a and x 3 = b, we get x 1 = (4/3)a − 2 b. So
[x 1 , x 2 , x 3 ] = [(4/3)a − 2 b, a, b] = [(4/3)a, a, 0] + [− 2 b, 0 , b] = (1/3)a[4, 3 , 0] + b[− 2 , 0 , 1].
Thus, the eigenvectors corresponding to λ = 2 are non-trivial linear combinations of the vectors X 2 = [4, 3 , 0] and X 3 = [− 2 , 0 , 1]. So E 2 = {aX 2 + bX 3 : a, b ∈ R} is the eigenspace corresponding to λ = 2.
Note: The algebraic multiplicity of an eigenvalue is defined as the number of times it repeats. In the above example, the eigenvalue λ = 2 repeats two times. So its algebraic multiplicity is 2. Also, we get two linearly independent eigenvectors X 2 = [4, 3 , 0] and X 3 = [− 2 , 0 , 1] corresponding to λ = 2. The follow- ing example shows that there may not exist as many linearly independent eigenvectors as the algebraic multiplicity of an eigenvalue.
Ex. If A =
(^) , then eigenvalues of A are λ = 0, 0 , 0. The eigenvectors corresponding to λ = 0
are non-zero multiples of the vector X = [1, 0 , 0]. The eigenspace corresponding to λ = 0, therefore, is E 0 = {a[1, 0 , 0] : a ∈ R}. Please try this example yourself. Notice that there is only one linearly indepen- dent eigenvector X = [1, 0 , 0] corresponding to the repeated eigenvalue (repeating thrice) λ = 0.
Note: One can easily prove the following properties of eigen values. (i) Sum of eigenvalues of a matrix A is equal to the trace of A, that is, the sum of diagonal elements of A. (ii) Product of eigenvalues of a matrix A is equal to the determinant of A. It further implies that deter- minant of a matrix is vanishes iff at least one eigenvalue of the matrix is 0. (iii) If λ is eigenvalue of A, then λm^ is eigenvalue of Am^ where m is any positive integer; 1/λ is eigenvalue of inverse of A, that is, A−^1 ; λ − k is eigenvalue of A − kI, where k is any real number.
Diagonalization
A square matrix A is said to be similar to a matrix B if there exists a non-singular matrix P such that P −^1 AP = B. In case, B is a diagonal matrix, we say that A is a diagonalizable matrix. Thus, a square matrix A is diagonalizable if there exists a non-singular matrix P such that P −^1 AP = D, where D is a diagonal matrix.
Suppose an n-square matrix A has n linearly independent eigenvectors X 1 , X 2 , ...... , Xn corresponding to the eigenvalues λ 1 , λ 2 , ...... , λn. Let P = [X 1 X 2 .... Xn]. Then we have
AP = [AX 1 AX 2 .... AXn] = [λ 1 X 1 λ 2 X 2 .... λnXn] = [X 1 X 2 .... Xn]
λ 1 0 ... 0 0 λ 2 ... 0 ... ... ... ... 0 0 ... λn
=^ P D.
This shows that if we construct P from eigenvectors of A, then A is diagonalizable, and P −^1 AP = D has eigenvalues of A at the diagonal places.
Note: If A has n different eigenvalues, then it can be proved that there exist n linearly independent eigenvectors of A and consequently A is diagonalizable. However, there may exist n linearly independent eigenvectors even if A has repeated eigenvalues as we have seen earlier. Such a matrix is also, of course, diagonalizable. In case, A does not have n linearly independent eigenvectors, it is not diagonalizable.
Ex. If A =
[
]
, then P =
[
]
and P −^1 AP =
[
]
. (Verify!)
Ex. If A =
(^) , then P =
(^) and P −^1 AP =
. (Verify!)
Note: If A is a diagonalizable matrix, that is, P −^1 AP = D or A = P DP −^1 , then for any positive integer n, we have An^ = P DnP −^1. For,
A^2 = (P DP −^1 )^2 = P DP −^1 P DP −^1 = P D^2 P −^1.
Likewise, A^3 = P D^3 P −^1. So in general, An^ = P DnP −^1. This result can be utilized to evaluate powers of a diagonalizable matrix easily.
Ex. Determine A^2 , where A =
So. P =
, P −^1 =
(^) and D =
So A^2 = P D^2 P −^1 =
Homework: Do exercise 3.4 from the textbook.
Real Vector Space
A non-empty set V is said to be a real vector space if there are defined two operations called vector addition and scalar multiplication denoted by ⊕ and respectively such that for all u, v, w ∈ V and a, b ∈ R, the following properties are satisfied:
- u ⊕ v ∈ V (Closure property)
- u ⊕ v = v ⊕ u (Commutative property)
- (u ⊕ v) ⊕ w = u ⊕ (v ⊕ w) (Associative property)
- There exists some element 0 ∈ V such that u ⊕ 0 = u = 0 ⊕ u. (Existence of additive identity)
- There exists −u ∈ V such that u ⊕ (−u) = 0 = (−u) ⊕ u. (Existence of additive inverse)
- a u ∈ V
- a (u ⊕ v) = a u ⊕ a v
- (a + b) u = a u ⊕ b u
- (ab) u = a (b u)
- 1 u = u
Note: Elements of the vector space V are called as vectors while that of R are called as scalars. In what follows, a vector space shall mean a real vector space.
Note: (i) Any scalar multiplied with the zero vector gives zero vector, that is, a 0 = 0. For,
a 0 = a 0 ⊕ 0 = a 0 ⊕ a 0 ⊕ (−a 0 ) = a ( 0 ⊕ 0 ) ⊕ (−a 0 ) = a 0 ⊕ (−a 0 ) = 0.
For clarity, let us use + and. symbols in place of ⊕ and respectively. Then we have
a. 0 = a. 0 + 0 = a. 0 + a. 0 + (−a. 0 ) = a.( 0 + 0 ) + (−a. 0 ) = a. 0 + (−a. 0 ) = 0.
(ii) The scalar 0 multiplied with any vector gives zero vector, that is, 0 u = 0. For,
0 u = (0 − 0) u = 0 u − 0 u = 0. 0 .u = (0 − 0).u = 0.u − 0 .u = 0.
(iii) (−1) u gives the additive inverse −u of u. For,
(−1) u ⊕ u = (−1) u ⊕ 1 u = (−1 + 1) u = 0 u = 0.
(−1).u + u = (−1).u + 1.u = (−1 + 1).u = 0.u = 0.
Ex. The set R of all real numbers is a vector space with respect to the following operations:
- u ⊕ v = u + v, (Vector Addition)
- a u = au, (Scalar Multiplication)
for all a, u, v ∈ R.
Sol. In this case, V = R, and all the properties of vector space are easily verifiable using the axioms satisfied by the real numbers.
Note: The set V = { 0 }, carrying only one real number namely 0, is a real vector space with respect to the operations mentioned in the above example. Think! It is easy!
Ex. The set R^2 = R × R = {[x 1 , x 2 ] : x 1 , x 2 ∈ R} of all ordered pairs of real numbers is a vector space with respect to the following operations:
- [x 1 , x 2 ] ⊕ [y 1 , y 2 ] = [x 1 + y 1 , x 2 + y 2 ], (Vector Addition)
- a [x 1 , x 2 ] = [ax 1 , ax 2 ], (Scalar Multiplication)
for all a ∈ R and [x 1 , x 2 ], [y 1 , y 2 ] ∈ R^2.
Sol. Let u = [x 1 , x 2 ], v = [y 1 , y 2 ] and w = [z 1 , z 2 ] be members of V = R^2 , and a, b be any two real numbers. Then we have the following properties:
- Closure Property: u ⊕ v = [x 1 , x 2 ] ⊕ [y 1 , y 2 ] = [x 1 + y 1 , x 2 + y 2 ] ∈ R^2 since x 1 + y 1 and x 2 + y 2 are real numbers.
- Commutative Property: u ⊕ v = [x 1 , x 2 ] ⊕ [y 1 , y 2 ] = [x 1 + y 1 , x 2 + y 2 ], v ⊕ u = [y 1 , y 2 ] + [x 1 , x 2 ] = [y 1 + x 1 , y 2 + x 2 ]. But [x 1 + y 1 , x 2 + y 2 ] = [y 1 + x 1 , y 2 + x 2 ] since real numbers are commutative in addition.
∴ u ⊕ v = v ⊕ u.
- Associative Property: (u⊕v)⊕w = ([x 1 , x 2 ]⊕[y 1 , y 2 ])⊕[z 1 , z 2 ] = ([x 1 +y 1 , x 2 +y 2 ])⊕[z 1 , z 2 ] = [(x 1 +y 1 )+z 1 , (x 2 +y 2 )+z 2 ] u⊕(v⊕w) = [x 1 , x 2 ]⊕([y 1 , y 2 ]⊕[z 1 , z 2 ]) = [x 1 , x 2 ]⊕([y 1 +z 1 , y 2 +z 2 ]) = [x 1 +(y 1 +z 1 ), x 2 +(y 2 +z 2 )]. Since real numbers are associative in addition, we have
[(x 1 + y 1 ) + z 1 , (x 2 + y 2 ) + z 2 ] = [x 1 + (y 1 + z 1 ), x 2 + (y 2 + z 2 )].
It implies that (u ⊕ v) ⊕ w = u ⊕ (v ⊕ w).
- Existence of identity: There exists 0 = [0, 0] ∈ R^2 such that u ⊕ 0 = [x 1 , x 2 ] ⊕ [0, 0] = [x 1 + 0, x 2 + 0] = [x 1 , x 2 ] = u, 0 ⊕ u = [0, 0] ⊕ [x 1 , x 2 ] = [0 + x 1 , 0 + x 2 ] = [x 1 , x 2 ] = u. So u ⊕ 0 = u = 0 ⊕ u. Therefore [0, 0] is additive identity in R^2.
- Existence of inverse: There exists −u = [−x 1 , −x 2 ] ∈ R^2 such that u ⊕ (−u) = [x 1 , x 2 ] ⊕ [−x 1 , −x 2 ] = [x 1 − x 1 , x 2 − x 2 ] = [0, 0] = 0 , (−u) ⊕ u = [−x 1 , −x 2 ] ⊕ [x 1 , x 2 ] = [−x 1 + x 1 , −x 2 + x 2 ] = [0, 0] = 0. So u ⊕ (−u) = 0 = (−u) ⊕ u. This shows that −u = [−x 1 , −x 2 ] is additive inverse of u = [x 1 , x 2 ] in R^2.
- a u = a [x 1 , x 2 ] = [ax 1 , ax 2 ] ∈ R^2
