Linear Transformations And Bases In R³ A Detailed Explanation

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Hey guys! Today, we're diving deep into the fascinating world of linear algebra, specifically focusing on linear transformations and bases within the three-dimensional Euclidean space, R³. We'll be tackling a problem that involves demonstrating a set of vectors forms a basis for R³ and expressing a generic vector in terms of this basis. Let's get started!

Understanding the Problem

Before we jump into the solution, let's break down the problem statement. We're given a set of three vectors, ß = {v₁ = (1, 2, 0), v₂ = (0, 1, 1), v₃ = (2, 0, 1)}, and we're told that this set is a basis for R³. We're also given a linear transformation T: R³ → R² defined by its action on these basis vectors: T(v₁) = (1, 0), T(v₂) = (0, -1), and T(v₃) = (2, 1).

The problem has two main parts:

  • Part a): We need to show that the set ß is indeed a basis for R³.
  • Part b): We need to express a generic vector x = (x, y, z) in R³ as a linear combination of the vectors in ß.

So, let's roll up our sleeves and get to work!

a) Proving ß is a Basis for R³

Okay, so how do we prove that a set of vectors forms a basis for a vector space? Remember, a set of vectors is a basis if it satisfies two crucial conditions:

  1. Linear Independence: The vectors in the set must be linearly independent, meaning no vector in the set can be written as a linear combination of the others.
  2. Spanning: The vectors in the set must span the entire vector space, meaning any vector in the space can be written as a linear combination of the vectors in the set.

Since we're dealing with R³, which is a three-dimensional space, and we have three vectors in our set ß, we can use a shortcut. If we can show that these three vectors are linearly independent, then they automatically span R³ and form a basis. This is because in a vector space of dimension n, any set of n linearly independent vectors will form a basis.

Checking for Linear Independence

To check for linear independence, we'll set up a linear combination of the vectors in ß equal to the zero vector and see if the only solution is the trivial solution (where all the coefficients are zero). Let's write this out:

c₁v₁ + c₂v₂ + c₃v₃ = (0, 0, 0)

Substituting our vectors, we get:

c₁(1, 2, 0) + c₂(0, 1, 1) + c₃(2, 0, 1) = (0, 0, 0)

This gives us the following system of linear equations:

  • c₁ + 2c₃ = 0
  • 2c₁ + c₂ = 0
  • c₂ + c₃ = 0

Now, we need to solve this system. We can do this using various methods, such as Gaussian elimination or substitution. Let's use Gaussian elimination by writing the augmented matrix:

[ 1  0  2 | 0 ]
[ 2  1  0 | 0 ]
[ 0  1  1 | 0 ]

Performing row operations to get the matrix in row-echelon form:

  1. Subtract 2 times row 1 from row 2: R₂ = R₂ - 2R₁
[ 1  0  2 | 0 ]
[ 0  1 -4 | 0 ]
[ 0  1  1 | 0 ]
  1. Subtract row 2 from row 3: R₃ = R₃ - R₂
[ 1  0  2 | 0 ]
[ 0  1 -4 | 0 ]
[ 0  0  5 | 0 ]

From the last row, we have 5c₃ = 0, which implies c₃ = 0. Substituting this into the second equation, c₂ - 4c₃ = 0, we get c₂ = 0. Finally, substituting c₃ = 0 into the first equation, c₁ + 2c₃ = 0, we get c₁ = 0.

So, the only solution to the system is c₁ = c₂ = c₃ = 0, which is the trivial solution. This means the vectors v₁, v₂, and v₃ are linearly independent.

Therefore, since ß is a set of three linearly independent vectors in R³, it forms a basis for R³.

b) Expressing a Generic Vector x = (x, y, z) in Terms of ß

Now, for the second part, we need to express a generic vector x = (x, y, z) as a linear combination of the basis vectors v₁, v₂, and v₃. This means we need to find scalars a, b, and c such that:

x = (x, y, z) = av₁ + bv₂ + cv₃

Substituting the vectors, we get:

(x, y, z) = a(1, 2, 0) + b(0, 1, 1) + c(2, 0, 1)

This gives us another system of linear equations:

  • x = a + 2c
  • y = 2a + b
  • z = b + c

We need to solve this system for a, b, and c in terms of x, y, and z. Let's use substitution again.

From the third equation, we have b = z - c. Substituting this into the second equation, we get:

y = 2a + (z - c)

y = 2a + z - c

Now, let's solve the first equation for a:

a = x - 2c

Substitute this value of a into the previous equation:

y = 2(x - 2c) + z - c

y = 2x - 4c + z - c

y = 2x + z - 5c

Now, solve for c:

5c = 2x + z - y

c = (2x + z - y) / 5

Great! We have c in terms of x, y, and z. Now we can find a and b.

Substitute the value of c back into the equation for a:

a = x - 2c = x - 2((2x + z - y) / 5)

a = x - (4x + 2z - 2y) / 5

a = (5x - 4x - 2z + 2y) / 5

a = (x + 2y - 2z) / 5

And finally, substitute the value of c into the equation for b:

b = z - c = z - (2x + z - y) / 5

b = (5z - 2x - z + y) / 5

b = (-2x + y + 4z) / 5

We've done it! We've found a, b, and c in terms of x, y, and z. So, we can express the generic vector x = (x, y, z) as:

(x, y, z) = ((x + 2y - 2z) / 5)v₁ + ((-2x + y + 4z) / 5)v₂ + ((2x + z - y) / 5)v₃

This is the expression of a generic vector x = (x, y, z) in terms of the basis ß.

Conclusion

Woohoo! We've successfully shown that the set ß is a basis for R³ and expressed a generic vector in terms of this basis. This problem highlights the fundamental concepts of linear independence, spanning, and basis in linear algebra. Understanding these concepts is crucial for further exploration of linear transformations and vector spaces. Keep practicing, and you'll become a linear algebra whiz in no time! Remember, linear algebra is not just about equations and matrices; it's about understanding the underlying structure of vector spaces and the transformations that operate on them.

I hope this explanation was helpful, guys! Let me know if you have any more questions. Keep exploring the amazing world of mathematics!