Suppose \(\vec x = \begin{bmatrix} c \\ 1 \\ 0 \end{bmatrix}\), \(\vec y = \begin{bmatrix} 1 \\ c \\ 1 \end{bmatrix}\), and \(\vec z = \begin{bmatrix} 0 \\ 1 \\ c \end{bmatrix}\), where \(c \in \mathbb{R}\) is a constant.
a)
8 pts Find a positive value of \(c\) such that \(\vec x\), \(\vec y\), and \(\vec z\) are linearly dependent. Show your work, and \(\boxed{\text{circle}}\) your final answer, which should be a positive number with no variables.
Solution
\(c = \sqrt{2}\).
For \(\vec x\), \(\vec y\), and \(\vec z\) to be linearly dependent, there must exist scalars \(a\), \(b\), and \(c\) such that
$$ a \vec x + b \vec y + \vec z $$
(or equivalently, \(a \vec x + b \vec y + d\vec z = \vec 0\), but the former approach involves one fewer variable to solve for).
Substituting in the given vectors, we have
$$ a \begin{bmatrix} c \\\\ 1 \\\\ 0 \end{bmatrix} + b \begin{bmatrix} 1 \\\\ c \\\\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\\\ 1 \\\\ c \end{bmatrix} $$
As a system of equations, we have
$$ \begin{align*} c a + b &= 0 \\\\ a + c b &= 1 \\\\ b &= c \end{align*} $$
The third equation gives us \(b = c\), and the second gives us \(a = 1 - cb = 1 - c^2\). Substituting these into the first equation gives us
$$ c(1 - c^2) + c = 0 \implies c - c^3 + c = 0 \implies c(2 - c^2) = 0 $$
This equation has three solutions for \(c\): \(c = 0\), \(c = \sqrt{2}\), and \(c = -\sqrt{2}\). We’re asked to find a positive value of \(c\), so \(c = \sqrt{2}\) for this part, and either \(0\) or \(-\sqrt{2}\) for the next part.
b)
3 pts Provide one other value of \(c\) (that is, not your answer from the previous part) such that \(\vec x\), \(\vec y\), and \(\vec z\) are linearly dependent. Your answer should be a number with no variables.
other value of \(c =\) ______
Solution
other value of \(c =\)\(\boxed{\text{0 or }-\sqrt{2}}\)
Suppose \(\vec x_1, \vec x_2, \ldots \vec x_{12}\) are 12 non-zero vectors in \(\mathbb{R}^{7}\). Furthermore, suppose:
\(\vec x_1\), \(\vec x_2\), and \(\vec x_3\) span a 2-dimensional subspace of \(\mathbb{R}^{7}\).
\(\vec x_4\), \(\vec x_5\), and \(\vec x_6\) span the same 2-dimensional subspace of \(\mathbb{R}^{7}\) as \(\vec x_1\), \(\vec x_2\), and \(\vec x_3\), i.e.
4 pts Let \(r\) be the dimension of the subspace of \(\mathbb{R}^{7}\) spanned by \(\vec x_1, \vec x_2, \ldots \vec x_{12}\). What are the smallest and largest possible values of \(r\)? Your answers should be integers with no variables.
smallest possible value of \(r =\) ______ largest possible value of \(r =\) ______
Solution
smallest possible value of \(r =\)\(\boxed{2}\)
largest possible value of \(r =\)\(\boxed{7}\)
b)
4 pts Which of the following could form a basis for \(\mathbb{R}^{7}\)? Select all that apply. Blank answers will receive no credit.
The first choice only includes 6 vectors, but since the span of \(\vec x_1, \vec x_2, \ldots \vec x_{12}\) is 7-dimensional, it must include at least 7 vectors. So, the first choice is not a valid basis.
The fourth choice includes 7 vectors, but we know that \(\vec x_1, \vec x_2, \vec x_5\) are a linearly dependent set since they all lie on the same 2-dimensional subspace of \(\mathbb{R}^7\) (and you only need 2 vectors to uniquely define a 2-dimensional subspace), so the fourth choice is not a valid basis.
The other options all include 7 vectors that could be linearly independent, and so they could form a basis for \(\mathbb{R}^7\).
c)
4 pts Suppose the intersection of \(\text{span}(\lbrace\vec x_1, \vec x_2\rbrace)\) and \(\text{span}(\lbrace \vec x_4, \vec x_5 \rbrace)\) is a line (i.e. a 1-dimensional subspace) in \(\mathbb{R}^{7}\). Which of the following must be true? Select all that apply. Blank answers will receive no credit.
Hint: Don’t forget the assumptions introduced at the start of the problem.
\(\vec x_2\), \(\vec x_4\), and \(\vec x_5\) can all be written as scalar multiples of \(\vec x_1\).
The set \(\lbrace \vec x_2, \vec x_4 \rbrace\) is linearly independent.
The set \(\lbrace \vec x_3, \vec x_4 \rbrace\) is linearly independent.
The set \(\lbrace \vec x_3, \vec x_6 \rbrace\) is linearly independent.
None of the above.
Solution
None of the above.
The intended answer to the problem was options 1 and 3. The scenario we had in mind was that \(\operatorname{span}(\lbrace\vec x_1, \vec x_2\rbrace) = \operatorname{span}(\lbrace\vec x_4, \vec x_5\rbrace) = \text{the same line}\). The two spans can’t both be different planes that happen to intersect in a line, since we’re told that \(\vec x_1, \vec x_2, \vec x_3\) span a 2-dimensional subspace of \(\mathbb{R}^7\) and \(\vec x_4, \vec x_5, \vec x_6\) span the same 2-dimensional subspace of \(\mathbb{R}^7\). So, if the two spans are planes, they’re the same plane, and they would intersect at a plane. Since the two spans intersect at a line, we thought they’d both have to be lines. If that was the case, then \(\vec x_2\), \(\vec x_4\), and \(\vec x_5\) would all be scalar multiples of \(\vec x_1\), and so \(\vec x_3\) would have to not be on that line (for \(\vec x_1, \vec x_2, \vec x_3\) to span a 2-dimensional subspace), which is why Options 1 and 3 were our originally intended answers.
But after releasing exam scores, a student brought up a possibility we hadn’t considered: it’s possible that \(\operatorname{span}(\lbrace\vec x_1, \vec x_2\rbrace)\) is a plane, and \(\operatorname{span}(\lbrace\vec x_4, \vec x_5\rbrace)\) is a line that is contained on that plane. That setup would satisfy all of the assumptions provided in the problem statement, but it would imply that none of the options are true.
So, retroactively, we gave full credit to everyone for this part.
Let \(\vec u\) and \(\vec v\) be as in the previous problem.
a)
4 pts Suppose that for some value of \(c\), \(P\) is the matrix that projects vectors in \(\mathbb{R}^3\) onto \(\text{span}(\lbrace\vec u, \vec v\rbrace)\). Select all true statements below.
\(P^2 = P\)\(P\) is invertible\(P\) is orthogonal\(P\) is symmetric
Solution
\(P^2 = P\)\(P\) is invertible\(P\) is orthogonal\(P\) is symmetric
If we let \(X = \begin{bmatrix} | & | \\ \vec u & \vec v \\ | & | \end{bmatrix}\), then no matter what \(c\) is, \(\text{rank}(X) = 2\), meaning the \(2 \times 2\) matrix \(X^TX\) is invertible. Then,
$$ P = X (X^TX)^{-1}X^T $$
With this in mind:
\(P^2 = P\) is true. This is the defining property of a projection matrix: once a vector has been projected onto the plane, projecting it again does nothing.
Conceptually, \(P\) is not invertible, because multiple different vectors \(\vec y\) can be projected onto the same vector \(\vec p\). The act of multiplying by \(P\) is not one-to-one, so \(P\) is not invertible.
\(P\) is not an orthogonal matrix. Orthogonal matrices preserve lengths, but projection usually shortens vectors unless they already lie in the plane. Also, orthogonal matrices are invertible, but \(P\) is not.
\(P\) is symmetric. This is a standard property of orthogonal projection matrices, and you can also verify it directly from \(P = X(X^TX)^{-1}X^T\) by taking the transpose.
b)
4 pts Now, suppose \(\vec y \in \mathbb{R}^3\). Let \(\vec p \) be the projection of \(\vec y\) onto \(\text{span}(\lbrace\vec u, \vec v\rbrace)\), and let \(\vec e = \vec y - \vec p\).
There is no value of \(c\) that guarantees that the components of \(\vec e\) sum to 0, for every \(\vec y \in \mathbb{R}^3\). That is, it is not guaranteed that \(e_1 + e_2 + e_3 = 0\) for every \(\vec y \in \mathbb{R}^3\).
Give a 1-2 sentence English explanation for why it is not guaranteed that \(e_1 + e_2 + e_3 = 0\) for every \(\vec y \in \mathbb{R}^3\). Hint: What would have to be true about \(\vec u\) and \(\vec v\) to make this guarantee for every \(\vec y\)?
Solution
For \(e_1 + e_2 + e_3\) to always equal 0, every error vector \(\vec e\) would have to be orthogonal to \(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\). Since every error vector is orthogonal to \(\text{span}(\lbrace\vec u, \vec v\rbrace)\), this would require \(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\) to lie in \(\text{span}(\lbrace\vec u, \vec v\rbrace)\), but no value of \(c\) makes that happen.
2 pts True or False: The set \(\lbrace\vec x_1, \vec x_2, \vec x_3, \vec x_4, \vec x_5, \vec x_6\rbrace\) is linearly independent.
True False
Solution
True False
This is false.
If these vectors were linearly independent, they would span a 6-dimensional subspace of \(\mathbb{R}^9\); since they only span a 4-dimensional subspace, they must be linearly dependent, and two of them are “redundant”.
b)
4 pts Consider the statement:
“There exists a vector \(\vec b \in \mathbb{R}^9\) such that the number of ways to write \(\vec b\) as a linear combination of \(\vec x_1, \ldots, \vec x_6\) is .”
In each part below, a possible way to fill in the blank is given. Determine whether the statement that results from filling in the blank is True or False.
zero
True False
exactly one
True False
exactly two
True False
infinite
True False
Solution
True False
(True) zero ways: \(S\), the set of all linear combinations of \(\vec x_1, \ldots, \vec x_6\), is a 4-dimensional subspace of \(\mathbb{R}^9\). Since \(S\) isn’t all of \(\mathbb{R}^9\), there are plenty of vectors \(\vec b \in \mathbb{R}^9\) that are not in \(S\), and therefore can’t be written as a linear combination of \(\vec x_1, \ldots, \vec x_6\). So, it’s true that there are some \(\vec b\)’s such that there are zero ways to write \(\vec b\) as a linear combination of \(\vec x_1, \ldots, \vec x_6\).
(False) exactly one way: Linear combinations are only unique if the spanning vectors are linearly independent. Since \(\vec x_1, \ldots, \vec x_6\) are linearly dependent, there is a non-trivial linear combination of them that equals the zero vector, which could be added to any other existing linear combination that sums to \(\vec b\) to “invent” a new, different-looking linear combination that sums to \(\vec b\). So, it’s false that there is exactly one way to write \(\vec b\) as a linear combination of \(\vec x_1, \ldots, \vec x_6\) — if there is one way, there are infinitely many.
(False) exactly two ways: Same logic as above. If this thinking is a bit confusing, see the solution to part c).
(True) infinite ways: For any vector \(\vec b \in S\), there are infinitely many ways to write \(\vec b\) as a linear combination of \(\vec x_1, \ldots, \vec x_6\).
c)
4 pts Suppose \(\vec b\) is some vector in \(S\) such that both of the following equations are true:
This is not the only possible answer, but it’s probably the easiest one. For example, you could repeat this process with one of the original two \(\vec b\)’s along with the new representation of \(\vec b\) to get another valid representation of \(\vec b\).
You also could have subtracted the two representations of \(\vec b\) to get a linear combination of \(\vec x_1, \ldots, \vec x_6\) that equals the zero vector, which could be added to any other existing linear combination that sums to \(\vec b\) to “invent” a new, different-looking linear combination that sums to \(\vec b\) (as we said in the solution to part b)). If you did this, you’d find that
4 pts Let \(T = \text{span}(\lbrace \vec x_1, \vec x_2, \vec x_3 \rbrace)\) and \(U = \text{span}(\lbrace \vec x_4, \vec x_5, \vec x_6 \rbrace)\). Suppose \(W\) is the intersection of \(T\) and \(U\), i.e. \(W = T \cap U\). \(W\) is also a subspace of \(\mathbb{R}^9\).
What are the smallest and largest possible values of \(\text{dim}(W)\), the dimension of \(W\)? Give your answers as integers.
smallest possible value \(=\) ______
largest possible value \(=\) ______
Solution
\(T\) and \(U\) are each individually at most 3-dimensional, since they are each spanned by 3 vectors. So, the intersection \(W\) must be at most 3-dimensional. This means the possible dimensions to consider are 3, 2, 1, or 0. Let’s reason about them, starting with 3.
To give examples, we’ll use the standard basis vectors \(\vec e_1, \vec e_2, \ldots, \vec e_9\) of \(\mathbb{R}^9\). In \(\mathbb{R}^2\), \(\vec e_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\) and \(\vec e_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}\), so (for instance) in \(\mathbb{R}^9\),
Could \(\text{dim}(W) = 3\)? No. If \(\text{dim}(W) = 3\), it would mean that \(T\) and \(U\) are both the same 3-dimensional subspace of \(\mathbb{R}^9\), and intersect everywhere. But if that were the case, then \(S = \text{span}(\lbrace \vec x_1, \vec x_2, \vec x_3, \vec x_4, \vec x_5, \vec x_6 \rbrace)\) would be a 3-dimensional subspace of \(\mathbb{R}^9\), which contradicts the problem statement that \(S\) is a 4-dimensional subspace of \(\mathbb{R}^9\). So, \(\text{dim}(W) < 3\), and the maximum possible value is something less than 3.
Could \(\text{dim}(W) = 2\)? Yes, and all smaller values are also possible. \(T\) and \(U\) could overlap in a 2-dimensional subspace of \(\mathbb{R}^9\), while each containing a direction that the other doesn’t.
For example, \(T\) could be the span of \(\lbrace\vec e_1, \vec e_2, \vec e_3\rbrace\) and \(U\) could be the span of \(\lbrace\vec e_2, \vec e_3, \vec e_4\rbrace\). In this case, \(W\) would be the span of \(\lbrace\vec e_2, \vec e_3\rbrace\), which is a 2-dimensional subspace of \(\mathbb{R}^9\).
Could \(\text{dim}(W) = 1\)? Yes. For example, \(T\) could be the span of \(\lbrace\vec e_1, \vec e_2\rbrace\) and \(U\) could be the span of \(\lbrace\vec e_2, \vec e_3, \vec e_4 \rbrace\). In this case, \(W\) would be the span of \(\lbrace\vec e_2\rbrace\), which is 1-dimensional, while \(S\) would still be 4-dimensional. (In this example, we said that \(T\) is the span of \(\vec e_1\) and \(\vec e_2\) though we defined it in the problem statement to be the span of three vectors. No problem — just pick the third vector to be a linear combination of \(\vec e_1\) and \(\vec e_2\). That is, \(\vec x_1 = \vec e_1\), \(\vec x_2 = \vec e_2\), and \(\vec x_3 = \vec e_1 + \vec e_2\) would work as an example.)
Could \(\text{dim}(W) = 0\)? Yes. \(T\) and \(U\) could be two completely disjoint subspaces, except for \(\vec 0\), which is in every subspace.
For example, let \(\vec x_1 = \vec e_1\), \(\vec x_2 = \vec e_2\), \(\vec x_3 = \vec e_1 + \vec e_2\), which makes \(T\) a 2-dimensional subspace of \(\mathbb{R}^9\), and \(\vec x_4 = \vec e_3\), \(\vec x_5 = \vec e_4\), \(\vec x_6 = \vec e_3 + \vec e_4\), which makes \(U\) a 2-dimensional subspace of \(\mathbb{R}^9\). In this case, \(W\) would be the set \(\lbrace\vec 0\rbrace\), while \(S\) would still be 4-dimensional.
So, the smallest possible value of \(\text{dim}(W)\) is \(\boxed{0}\), and the largest possible value is \(\boxed{2}\).
$$ S = \left\{ \begin{bmatrix} x_1\\\\x_2\\\\x_3\\\\x_4\\\\x_5\\\\x_6 \end{bmatrix} \in \mathbb{R}^6 : x_1+x_2+x_3=0 \text{ and } x_4=x_5 \right\} $$
Find \(\dim(S)\). Give your answer as an integer with no variables.
\(\dim(S)=\_\_\_\_\_\_\)
Solution
There are \(6\) variables total. The condition
$$ x_1+x_2+x_3=0 $$
removes one degree of flexibility, and the condition
$$ x_4=x_5 $$
removes one flexibility. So
$$ \dim(S) = 6-2 = \boxed{4} $$
Another way to think about it is to think of what a basis for \(S\) looks like. Every vector in \(S\) is of the form
$$ \begin{bmatrix} a \\\\ b \\\\ -a-b \\\\ c \\\\ c \\\\ d \end{bmatrix} $$
where \(a, b, c, d\) are real numbers. \(a\) and \(b\) (components 1 and 2) can both be anything, but component 3 is automatically determined once \(a\) and \(b\) are chosen. Similarly, \(c\) and \(d\) (components 4 and 6) can both be anything, but once component 4 is chosen, component 5 is automatically determined.
\(S\) is the set of all vectors that fit the template above. But
$$ \begin{bmatrix} a \\\\ b \\\\ -a-b \\\\ c \\\\ c \\\\ d \end{bmatrix} = a \begin{bmatrix} 1 \\\\ 0 \\\\ -1 \\\\ 0 \\\\ 0 \\\\ 0 \end{bmatrix} + b \begin{bmatrix} 0 \\\\ 1 \\\\ -1 \\\\ 0 \\\\ 0 \\\\ 0 \end{bmatrix} + c \begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 1 \\\\ 1 \\\\ 0 \end{bmatrix} + d \begin{bmatrix} 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 0 \\\\ 1 \end{bmatrix} $$
4 pts Fill in the blanks to complete the sentence:
For all values of \(c\), \(\text{span}(\lbrace\vec u,\vec v\rbrace)\) is a __(i)__-dimensional subspace of __(ii)__.
(i): ______ (ii): ______
Solution
The vectors \(\vec u\) and \(\vec v\) are never scalar multiples of each other. If \(\vec v=\lambda\vec u\), then the first entries force \(\lambda=2\), the second entries force \(c=2\), and the third entries force \(-2=2c=4\), which is impossible. Therefore, the span is always a 2-dimensional subspace of \(\mathbb R^3\).
Why \(\mathbb R^3\)? Because both \(\vec u\) and \(\vec v\) live in \(\mathbb R^3\), so their span must also live in \(\mathbb R^3\).
b)
7 pts Suppose the plane spanned by \(\vec u\) and \(\vec v\) is
$$ ax+24y+3z=0 $$
where \(a\) is also a constant. Find the value of \(c\). Show your work in the space provided, and write your final answer in the box provided. Your answer should be a number with no variables.
$$ c = \_\_\_\_\_\_ $$
Solution
There are a few ways to approach this. The first way starts by using the fact that \(\vec u\) and \(\vec v\) lie in the plane, which gives us a system of two equations and two unknowns. Plugging in the coordinates of \(\vec u\) into the plane gives us
$$ 3a+24+3c=0 \implies a + 8 + c = 0 $$
and plugging in the coordinates of \(\vec v\) into the plane gives us
$$ 6a+24c-6=0 \implies a + 4c - 1 = 0 $$
Subtracting the simplified versions of the two equations gives us
Another way to approach this is to find the cross product of \(\vec u\) and \(\vec v\), and try and write it as a scalar multiple of the vector \(\begin{bmatrix} a \\ 24 \\ 3 \end{bmatrix}\).
$$ \vec u \times \vec v = \begin{bmatrix} 3 \\\\ 1 \\\\ c \end{bmatrix} \times \begin{bmatrix} 6 \\\\ c \\\\ -2 \end{bmatrix} = \begin{bmatrix} 1 \cdot (-2) - c \cdot c \\\\ c \cdot 6 - 3 \cdot (-2) \\\\ 3 \cdot c - 1 \cdot 6 \end{bmatrix} = \begin{bmatrix} -2 - c^2 \\\\ 6c + 6 \\\\ 3c - 6 \end{bmatrix} $$
Strictly speaking, this vector, \(\begin{bmatrix} -2 - c^2 \\ 6c + 6 \\ 3c - 6 \end{bmatrix}\), is a scalar multiple of \(\begin{bmatrix} a \\ 24 \\ 3 \end{bmatrix}\), but we don’t know what the scalar is yet. So, we really should try and solve
7 pts Are \(\lbrace\vec p,\vec q,\vec r,\vec s\rbrace\) linearly independent?
Select an answer:
Yes No
Prove your answer using the formal definition of linear independence. Hint: You did something similar in Homework 4, Problem 6.
Solution
Yes No
If \(\vec p,\vec q,\vec r,\vec s\) are linearly independent, then the only solution to the equation \(a \vec p + b \vec q + c \vec r + d \vec s = \vec 0\) is \(a = b = c = d = 0\).
That’s not the case here! Consider the linear combination
$$ \vec p-\vec q+\vec r-\vec s $$
How did I think of this? I noticed that if I start with \(\vec p\), subtracting \(\vec q\) gets rid of all \(\vec v_2\)’s, but makes \(\vec v_3\) negative, so I need a positive \(\vec r\) to cancel that out. Then, \(\vec p - \vec q + \vec r = \vec v_1 + \vec v_4\); subtracting \(\vec s\) then gets rid of both \(\vec v_1\) and \(\vec v_4\), leaving me with \(\vec 0\).
$$ \vec p - \vec q + \vec r - \vec s = (\vec v_1+\vec v_2)-(\vec v_2+\vec v_3)+(\vec v_3+\vec v_4)-(\vec v_4+\vec v_1)=\vec 0 $$
The coefficients \(1,-1,1,-1\) are not all zero, so this proves that \(\lbrace\vec p,\vec q,\vec r,\vec s\rbrace\) is linearly dependent.
b)
3 pts What is the dimension of \(\text{span}(\lbrace\vec p,\vec q,\vec r,\vec s\rbrace)\)? Give your answer as a number with no variables.
Part a) shows that the four vectors are linearly dependent, so the dimension of \(\text{span}(\lbrace\vec p,\vec q,\vec r,\vec s\rbrace)\) is at most\(3\). (For the dimension to be 4, which is the number of vectors in question, they would need to be linearly independent. There’s no way to have a span of 5 or more dimensions using just 4 vectors.)
But just because the dimension of \(\text{span}(\lbrace\vec p,\vec q,\vec r,\vec s\rbrace)\) is at most \(3\) doesn’t mean that the dimension is actually \(3\) — for this span to be 3-dimensional, it needs to be the span of 3 linearly independent vectors.
Fortunately, \(\vec p,\vec q,\vec r\) are linearly independent. If
Since \(\vec v_1,\vec v_2,\vec v_3,\vec v_4\) are linearly independent, we must have
$$ a=0,\qquad a+b=0,\qquad b+c=0,\qquad c=0 $$
This gives \(a=b=c=0\), so \(\vec p,\vec q,\vec r\) are linearly independent. Therefore, among \(\left\lbrace \vec p,\vec q,\vec r,\vec s \right\rbrace\), there are 3 linearly independent vectors, and thus
The subspace \(S\) has dimension \(3\) because the single constraint lets us solve
$$ x_4=x_1-x_2+x_3 $$
This means that components 1, 2, and 3 are free to vary, and component 4 is fully determined by those first three components. So, \(S\) has three “degrees of freedom”, and therefore has dimension \(3\).
So a basis for \(S\) is any set of three linearly independent vectors that all lie in \(S\).
The first and third choices are bases: in both of those choices, the set has 3 vectors that are linearly independent, and all 3 vectors lie in \(S\).
The second choice has 4 vectors in a 3-dimensional subspace, so it cannot be a basis.
The fourth choice has 3 vectors but they are not linearly independent, since at least one of them can be written as a linear combination of the other two:
Suppose \(S = \left\lbrace \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} : x_1 + x_2 + 2x_3 = 0 \text{ and } x_3 = x_4 \right\rbrace\). State one basis for \(S\). Your answer should be a list of vectors with no variables.
\(\text{one basis for } S =\) ______
Solution
The condition \(x_3=x_4\) means we can write \(x_3=x_4=b\). The other condition gives
$$ x_1+x_2+2b=0 $$
so \(x_1=-x_2-2b\). Let \(x_2=a\). Then every vector in \(S\) can be written as
Another way to think about this: since \(\dim(S)=2\) (the subspace has two “degrees of freedom”, or free variables), any two linearly independent vectors in \(S\) span all of \(S\) (see Chapter 4.3). So, we could just play with the numbers until we end up with two vectors that are not scalar multiples of each other that both satisfy the conditions of inclusion in \(S\). For instance,
5 pts Find an equation for the plane \(P\) of the form \(Ax+By+Cz+d=0\). Simplify your answer so that \(A=1\).
Solution
Because \(P\) is a span, it contains the origin, so \(d=0\). With \(A=1\), the equation is \(x+By+Cz=0\). Substituting the two spanning vectors gives
$$ 12-2B+6C=0,\qquad 3+3C=0 $$
The second equation gives \(C=-1\), and then the first gives \(B=3\). Thus,
$$ \boxed{x+3y-z=0} $$
Both vectors lie in this plane and are linearly independent, so they span the entire plane.
For part b), suppose that \(\vec x_3,\vec x_4\in\mathbb R^3\) and that \(\operatorname{span}(\lbrace \vec x_1,\vec x_2,\vec x_3,\vec x_4\rbrace )\) is a plane in \(\mathbb R^3\).
b)
5 pts For each statement below, determine whether it is impossible, possible (but not guaranteed), or guaranteed to be true, given the above assumptions. The first statement has been done for you as an example.
statement
impossible?
possible?
guaranteed?
\(i\)
\(\lVert \vec x_3 \rVert = 5\).
\(ii\)
\(\vec x_3\) and \(\vec x_4\) are scalar multiples of one another.
\(iii\)
\(\vec x_4 = \vec 0\).
\(iv\)
The vectors \(\vec x_1, \vec x_2, \vec x_3\) are linearly independent.
\(\lbrace \vec x_1, \vec x_2\rbrace \) is a basis for \(P\).
Solution
The span of all four vectors contains \(P\) and is a plane, so it must equal \(P\). Thus, \(\vec x_3\) and \(\vec x_4\) both lie in \(P\).
Possible. A vector in \(P\) can have length \(5\), but it doesn’t need to.
Possible. For example, choose \(\vec x_3=\vec x_1\) and \(\vec x_4=2\vec x_1\). They don’t need to be scalar multiples: we could instead choose \(\vec x_3=\vec x_1\) and \(\vec x_4=\vec x_2\).
Possible. We may choose \(\vec x_4=\vec0\) because the first two vectors already span \(P\).
Impossible. Three vectors in a two-dimensional space cannot be linearly independent.
Possible. Choosing \(\vec x_3=\vec x_1\) and \(\vec x_4=\vec x_2\) works. Choosing both to be zero shows that this is not guaranteed.
Guaranteed. By definition, \(\vec x_1\) and \(\vec x_2\) span \(P\), and they are not scalar multiples of one another, so they form a basis.
Additionally, suppose that \(\vec x_3,\vec x_4\in\mathbb R^3\) and that \(\operatorname{span}(\lbrace \vec x_1,\vec x_2,\vec x_3,\vec x_4\rbrace )\) is a plane in \(\mathbb R^3\).
c)
4 pts Give one nonzero vector orthogonal to \(3\vec x_1-7\vec x_2+12\vec x_3-\vec x_4\). Your answer should have no variables. Briefly justify it.
Solution
From part a), \(\vec n=\begin{bmatrix}1\\3\\-1\end{bmatrix}\) is normal to \(P\). All four vectors lie in \(P\), so \(\vec n\cdot\vec x_i=0\) for \(i=1,2,3,4\). By distributivity,
Thus, one valid answer is \(\boxed{\begin{bmatrix}1\\3\\-1\end{bmatrix}}\). Any nonzero scalar multiple of this vector also works.
For part d), suppose that \(\vec x_3,\vec x_4\in\mathbb R^3\) but that \(\operatorname{span}(\lbrace \vec x_1,\vec x_2,\vec x_3,\vec x_4\rbrace ) = \mathbb R^3\).
d)
5 pts Which statements are impossible? Select all that apply.
\(\vec x_3\) and \(\vec x_4\) are on \(P\).
The vectors \(\vec x_3,\vec x_4\) are linearly independent.
The vectors \(\vec x_1,\vec x_2,\vec x_3,\vec x_4\) are linearly independent.
There is a nonzero vector orthogonal to all four vectors.
Every subset containing exactly three of the four vectors is a basis for \(\mathbb R^3\).
Solution
Every subset containing exactly three of the four vectors is a basis for \(\mathbb R^3\).
Select options 1, 3, and 4.
Impossible. If both additional vectors were on \(P\), all four vectors would span only \(P\), not \(\mathbb R^3\).
Possible. Choose \(\vec x_3\) outside \(P\) and \(\vec x_4=\vec x_1\). These two vectors are independent, and all four span \(\mathbb R^3\).
Impossible. Four vectors in \(\mathbb R^3\) are always linearly dependent.
Impossible. A vector orthogonal to all four is orthogonal to their entire span, \(\mathbb R^3\). In particular, it is orthogonal to itself and must be zero.
Possible. Choose \(\vec x_3\) outside \(P\) and \(\vec x_4=\vec x_1+\vec x_2+\vec x_3\). The first three vectors form a basis. Replacing any one with their sum still gives a basis, since the omitted vector can be recovered by subtracting the other two from the sum.
As \(t\) ranges over \(\mathbb R\), so does \(t+2\). Hence \(L\) is the span of \(\begin{bmatrix}3\\1\\-2\end{bmatrix}\), so it is a subspace of \(\mathbb R^3\).
Consider the vectors \(\vec v_1, \vec v_2, \ldots, \vec v_d\in\mathbb R^n\), where \(d\geq 2\). For some scalar \(t\in\mathbb R\), define:
$$ \vec s=t(\vec v_1+\cdots+\vec v_d) $$
Then, for each \(i=1, 2, \ldots,d\), define
$$ \vec u_i = \vec v_i - \vec s $$
a)
5 pts For this part only, suppose \(d=3\). Find an expression for \(\vec u_1 + \vec u_2 + \vec u_3\) in terms of \(t\), \(\vec v_1\), \(\vec v_2\), and \(\vec v_3\). Note that your answer cannot involve \(\vec s\).
Solution
For \(d=3\), \(\vec s=t(\vec v_1+\vec v_2+\vec v_3)\). Therefore,
4 pts Find a value of \(t\) for which \(\vec u_1,\ldots,\vec u_d\) are guaranteed to be linearly dependent, regardless of the original vectors. Give your answer as an expression in terms of \(n\), \(d\), and/or constants.
Solution
As in part a), observe what happens when we sum all of the \(\vec u_i\). This is a linear combination of the \(\vec u_i\), with every coefficient equal to \(1\):
For this value of \(t\), the sum is \(\vec0\). Since the coefficients in this linear combination are all \(1\), they are not all zero. By the definition of linear independence, the \(\vec u_i\) are therefore linearly dependent.
c)
2 pts Suppose \(t>0\). Fill in the \(\boxed{???}\) with the relationship that is guaranteed:
The guaranteed relationship is \(\boxed{\geq}\). Equality can occur, for example, when all the original vectors are the same nonzero vector, so a strict inequality is not guaranteed.