Vector Basis And Coordinates In R3: A Step-by-Step Guide
Hey guys! Today, we're diving into a cool math problem involving vector systems, bases, and coordinate transformations in R3. It might sound a bit intimidating at first, but don't worry, we'll break it down step by step so that everyone can follow along. Let's get started!
Understanding the Vector System B
First, let's consider the vector system B = {v1 = (−1, 1, 0), v2 = (1, 2,−1), v3 = (2,−1, 1)} which is a subset of R3. Our main goal here is to show that this system B actually forms a basis for R3 and then figure out how to switch between this basis and the standard one we're all familiar with. Sounds like fun, right? The crucial part here is understanding what it means for a set of vectors to form a basis. Essentially, it means that these vectors are linearly independent and they span the entire space R3. Linear independence ensures that no vector in the set can be written as a linear combination of the others, while spanning means that any vector in R3 can be expressed as a linear combination of the vectors in B.
Proving B is a Basis in R3
To prove that B is a basis in R3, we need to demonstrate two things:
- Linear Independence: The vectors v1, v2, and v3 are linearly independent.
- Spanning: The vectors v1, v2, and v3 span R3.
Linear Independence
To check for linear independence, we set up a linear combination of the vectors equal to the zero vector and see if the only solution is the trivial one (where all coefficients are zero). So, we have:
a * v1 + b * v2 + c * v3 = (0, 0, 0)
a * (−1, 1, 0) + b * (1, 2,−1) + c * (2,−1, 1) = (0, 0, 0)
This gives us the following system of equations:
- -a + b + 2c = 0
- a + 2b - c = 0
- -b + c = 0
We can write this system as a matrix:
[ -1 1 2 ]
[ 1 2 -1 ]
[ 0 -1 1 ]
To determine if the vectors are linearly independent, we calculate the determinant of this matrix:
det = -1 * (2*1 - (-1)*(-1)) - 1 * (1*1 - (-1)*0) + 2 * (1*(-1) - 2*0)
= -1 * (2 - 1) - 1 * (1) + 2 * (-1)
= -1 - 1 - 2
= -4
Since the determinant is non-zero (-4 ≠ 0), the vectors v1, v2, and v3 are indeed linearly independent. This is a crucial step because it confirms that none of these vectors can be expressed as a combination of the others, which is a fundamental requirement for forming a basis.
Spanning R3
Now, we need to show that these vectors span R3. Since we have three linearly independent vectors in R3 (which is a 3-dimensional space), they automatically span R3. This is a theorem in linear algebra: if you have n linearly independent vectors in an n-dimensional space, they form a basis and thus span the entire space. Think of it like this: in a 3D world, if you have three directions that aren't just different ways of pointing along the same line or plane, you can reach any point in that world by combining movements in those three directions. Therefore, B spans R3.
Conclusion: Since B is linearly independent and spans R3, B is a basis in R3. Awesome, we've proved the first part!
Writing the Transition Matrix from B to the Canonical Basis in R3
The transition matrix from B to the canonical basis (also known as the standard basis) in R3 is formed by using the vectors of B as columns. The canonical basis in R3 consists of the vectors e1 = (1, 0, 0), e2 = (0, 1, 0), and e3 = (0, 0, 1). So, the transition matrix, let's call it P, is:
P = [ -1 1 2 ]
[ 1 2 -1 ]
[ 0 -1 1 ]
This matrix P transforms the coordinates of a vector expressed in the basis B to its coordinates in the canonical basis. In other words, if you have a vector's coordinates in basis B, multiplying it by P will give you its coordinates in the standard basis. This matrix is super useful for converting between different representations of the same vector!
Determining the Coordinates of Vector u
Now, let's move on to the second part of the problem. We need to determine the coordinates of a vector u (the problem statement is incomplete, but let’s assume u is given). Let's say the problem gives us a vector u = (x, y, z) in the canonical basis, and we want to find its coordinates in the basis B. To do this, we need to find scalars a, b, and c such that:
u = a * v1 + b * v2 + c * v3
(x, y, z) = a * (−1, 1, 0) + b * (1, 2,−1) + c * (2,−1, 1)
This gives us the following system of equations:
- -a + b + 2c = x
- a + 2b - c = y
- -b + c = z
To find a, b, and c, we need to solve this system of linear equations. We can use various methods like substitution, elimination, or matrix inversion.
Solving the System of Equations
Let's use the matrix method. We can write the augmented matrix:
[ -1 1 2 | x ]
[ 1 2 -1 | y ]
[ 0 -1 1 | z ]
We want to transform this matrix into row-echelon form or reduced row-echelon form. First, let's add the first row to the second row:
[ -1 1 2 | x ]
[ 0 3 1 | x+y ]
[ 0 -1 1 | z ]
Next, let's multiply the third row by 3 and add it to the second row:
[ -1 1 2 | x ]
[ 0 0 4 | x+y+3z ]
[ 0 -1 1 | z ]
From the second row, we get:
4c = x + y + 3z => c = (x + y + 3z) / 4
From the third row, we get:
-b + c = z => b = c - z = (x + y + 3z) / 4 - z = (x + y - z) / 4
From the first row, we get:
-a + b + 2c = x => a = b + 2c - x = (x + y - z) / 4 + 2 * (x + y + 3z) / 4 - x
a = (x + y - z + 2x + 2y + 6z - 4x) / 4 = (-x + 3y + 5z) / 4
So, the coordinates of vector u in the basis B are:
u_B = (a, b, c) = (((-x + 3y + 5z) / 4), ((x + y - z) / 4), ((x + y + 3z) / 4))
Example
Let’s take an example. Suppose u = (1, 2, 3). Then,
a = (-1 + 32 + 53) / 4 = (-1 + 6 + 15) / 4 = 20 / 4 = 5 b = (1 + 2 - 3) / 4 = 0 / 4 = 0 c = (1 + 2 + 3*3) / 4 = (1 + 2 + 9) / 4 = 12 / 4 = 3
So, u_B = (5, 0, 3). This means that u = 5 * v1 + 0 * v2 + 3 * v3.
Conclusion
Alright, guys! We've successfully shown that the vector system B is a basis in R3, found the transition matrix from B to the canonical basis, and determined the coordinates of a vector u in the basis B. This involves understanding linear independence, spanning, and solving systems of linear equations. Keep practicing, and you'll become a pro at these types of problems in no time!