Linear Combinations: When Vectors Span R^2 And R^3

by Andrew McMorgan 51 views

Hey guys, ever wondered when a couple of vectors can actually cover everything in a space, like our good old 2D plane or even the 3D world? Today, we're diving deep into the fascinating world of linear combinations and figuring out exactly what conditions need to be met for this to happen. We'll tackle a couple of cool problems that will make this concept crystal clear. So, grab your favorite beverage, and let's get our math on!

Part (a): Spanning R^2 with Two Vectors

Alright, let's kick things off with a classic scenario. We're given two vectors, u = (βˆ’3, p) and v = (2, 3), and we want to know for which real values of p can we create any point in R^2 (that's our entire 2D plane, folks!) by just adding multiples of u and v together? This is what we call the span of u and v. For the set of all linear combinations of u and v to be all of R^2, these two vectors need to be linearly independent. What does that mean, you ask? It means that neither vector can be written as a scalar multiple of the other. If one vector was just a stretched or shrunk version of the other, they'd be pointing in the same or opposite directions, and you'd only be able to create points along a single line, not the whole plane.

So, how do we check for linear independence in R^2? A super handy trick is to look at the determinant of the matrix formed by these vectors. If the determinant is non-zero, the vectors are linearly independent and they will span R^2. Let's set up that matrix with u and v as our columns (or rows, it doesn't matter for the determinant):

(βˆ’32p3) \begin{pmatrix} -3 & 2 \\ p & 3 \end{pmatrix}

Now, we calculate the determinant:

det⁑(βˆ’32p3)=(βˆ’3)(3)βˆ’(2)(p)=βˆ’9βˆ’2p \det \begin{pmatrix} -3 & 2 \\ p & 3 \end{pmatrix} = (-3)(3) - (2)(p) = -9 - 2p

For our vectors to span R^2, this determinant must not be equal to zero. So, we set:

βˆ’9βˆ’2pβ‰ 0 -9 - 2p \neq 0

Solving for p:

βˆ’2pβ‰ 9 -2p \neq 9

pβ‰ βˆ’92 p \neq -\frac{9}{2}

So, the only value of p that doesn't work is -9/2. This means that for all other real values of p, the vectors u = (βˆ’3, p) and v = (2, 3) will be linearly independent and their linear combinations will cover the entire R^2 plane. Pretty neat, huh? It’s like finding the secret ingredients that make sure your vectors are perfectly positioned to paint the whole canvas of R^2 without any gaps!

Justification for R^2

To justify this answer rigorously, let's think about what it means for the set of linear combinations to be all of R^2. We're saying that for any arbitrary vector w = (a, b) in R^2, there must exist scalars c1 and c2 such that:

c1u+c2v=w c_1 \mathbf{u} + c_2 \mathbf{v} = \mathbf{w}

c1(βˆ’3,p)+c2(2,3)=(a,b) c_1 (-3, p) + c_2 (2, 3) = (a, b)

This gives us a system of linear equations:

βˆ’3c1+2c2=a -3c_1 + 2c_2 = a

pc1+3c2=b pc_1 + 3c_2 = b

This system has a solution for c1 and c2 for all values of a and b if and only if the coefficient matrix of this system has a non-zero determinant. The coefficient matrix is precisely the one we formed earlier:

egin{pmatrix} -3 & 2 \\ p & 3 \end{pmatrix}

As we calculated, its determinant is βˆ’9βˆ’2p-9 - 2p. This system has a unique solution for c1c_1 and c2c_2 (and thus spans R^2) if and only if βˆ’9βˆ’2peq0-9 - 2p eq 0, which means peqβˆ’9/2p eq -9/2. If p=βˆ’9/2p = -9/2, the determinant is zero, indicating that the vectors are linearly dependent, and their span is only a line, not the entire R^2. Therefore, p can be any real number except -9/2.

Part (b): Spanning R^3 with Three Vectors

Now, let's level up to R^3, the 3D space we live in! The question here is a bit more complex, involving three vectors: u = (1, p, βˆ’1), and v = (2, 1, 3), and w = (1, 0, q). We need to find the real values of p and q such that the set of all linear combinations of u, v, and w forms the entire R^3 space. For three vectors to span R^3, they must be linearly independent. In R^3, if you have three vectors, they span the entire space if and only if they are linearly independent. If they are linearly dependent, they will only span a plane or a line, or just the origin.

To check for linear independence of three vectors in R^3, we can again use a determinant. This time, we form a 3x3 matrix using our vectors as rows or columns. Let's use them as rows:

(1pβˆ’121310q) \begin{pmatrix} 1 & p & -1 \\ 2 & 1 & 3 \\ 1 & 0 & q \end{pmatrix}

For these vectors to span R^3, the determinant of this matrix must be non-zero. Let's compute the determinant. We can expand along the third row (because of the zero):

det⁑=1β‹…det⁑(pβˆ’113)βˆ’0β‹…det⁑(1βˆ’123)+qβ‹…det⁑(1p21) \det = 1 \cdot \det \begin{pmatrix} p & -1 \\ 1 & 3 \end{pmatrix} - 0 \cdot \det \begin{pmatrix} 1 & -1 \\ 2 & 3 \end{pmatrix} + q \cdot \det \begin{pmatrix} 1 & p \\ 2 & 1 \end{pmatrix}

det⁑=1β‹…((p)(3)βˆ’(βˆ’1)(1))βˆ’0+qβ‹…((1)(1)βˆ’(p)(2)) \det = 1 \cdot ((p)(3) - (-1)(1)) - 0 + q \cdot ((1)(1) - (p)(2))

det⁑=(3p+1)+q(1βˆ’2p) \det = (3p + 1) + q(1 - 2p)

det⁑=3p+1+qβˆ’2pq \det = 3p + 1 + q - 2pq

For the vectors u, v, and w to span R^3, this determinant must not be equal to zero:

3p+1+qβˆ’2pqβ‰ 0 3p + 1 + q - 2pq \neq 0

This inequality tells us the condition on p and q for the vectors to span R^3. Any pair of (p, q) that satisfies this inequality will work. It's a bit more complex than the R^2 case because now we have a relationship between two variables, p and q, instead of just one. It means that if you pick a p, there’s likely only one specific value of q (or possibly none) that would make the determinant zero, thus making the vectors linearly dependent. All other pairs (p, q) will result in linear independence and span R^3.

Justification for R^3

Let's justify this by considering the system of equations. For any arbitrary vector b = (b1, b2, b3) in R^3, we want to know if there exist scalars c1, c2, and c3 such that:

c1u+c2v+c3w=b c_1 \mathbf{u} + c_2 \mathbf{v} + c_3 \mathbf{w} = \mathbf{b}

c1(1,p,βˆ’1)+c2(2,1,3)+c3(1,0,q)=(b1,b2,b3) c_1 (1, p, -1) + c_2 (2, 1, 3) + c_3 (1, 0, q) = (b_1, b_2, b_3)

This translates into the following system of linear equations:

c1+2c2+c3=b1 c_1 + 2c_2 + c_3 = b_1

pc1+c2=b2 pc_1 + c_2 = b_2

βˆ’c1+3c2+qc3=b3 -c_1 + 3c_2 + qc_3 = b_3

This system has a solution for c1,c2,c3c_1, c_2, c_3 for every vector b in R^3 if and only if the coefficient matrix of this system is invertible. The coefficient matrix is:

egin{pmatrix} 1 & 2 & 1 \\ p & 1 & 0 \\ -1 & 3 & q \end{pmatrix}

Notice that this matrix is the transpose of the matrix we used earlier (with vectors as rows). The determinant of a matrix is equal to the determinant of its transpose, so its determinant is also 3p+1+qβˆ’2pq3p + 1 + q - 2pq. The system has a solution for all b if and only if this determinant is non-zero. Therefore, the set of all linear combinations of u, v, and w spans R^3 if and only if 3p+1+qβˆ’2pqeq03p + 1 + q - 2pq eq 0. This means p and q can be any real numbers as long as they satisfy this condition. For example, if p=0, then 1+qeq01+q eq 0, so qeqβˆ’1q eq -1. If q=0, then 3p+1eq03p+1 eq 0, so peqβˆ’1/3p eq -1/3. They are linked, but generally, any pair (p,q) works as long as they don't make the determinant zero.

So there you have it! We’ve explored how linear independence dictates whether a set of vectors can span an entire space. Whether it's R^2 or R^3, the key lies in ensuring our vectors are pointing in sufficiently different directions. Keep practicing these concepts, guys, and you'll be mastering linear algebra in no time!