Exercise 04 solutions
🔹 must know, 🔸 ideally should know, ⛑ challenging, 🔫 very challenging
Exercises from the Rasmus’ notes
🔸 Exercise 1
Prove that any plane through the origin is a subspace of \(\mathbb{R}^3\). A plane through the origin can always be written on the form
\[S = \{(x, y, z) \mid a x+b y+c z=0\}\]Solution
Let me first restate the task given. We are given set of vectors \(S\) such that these vectors live on the plane that goes through the origin.
First, we need to prove that \(S\) is non-empty which is clearly through since it goes through origin, i.e., \((x, y, z) = (0, 0, 0)\).
Second, we need to do closure under addition. Therefore, given two vectors \(u_1, u_2\) from \(S\), we need to prove that also sum of these two vectors \(v \in S\). Therefore:
Now, we need to find out whether \(v\) satisfies the condition:
\[a x + b y + c z=0\]We simply plug in the corresponding values:
\[a(x_1 + x_2) + b(y_1 + y_2) + c(z_1 + z_2) = 0\]If we expand it, we get:
\[ax_1 + by_1 + cz_1 + ax_2 + by_2 + cz_2 = 0\]And since we know that \(u_1, u_2 \in S\), we also know that for both them it holds true that:
\[\begin{aligned} ax_1 + by_1 + cz_1 = 0 \\ ax_2 + by_2 + cz_2 = 0 \end{aligned}\]And therefore, we can write:
\[\begin{aligned} ax_1 + by_1 + cz_1 + ax_2 + by_2 + cz_2 = 0 \\ 0 + 0 = 0 \end{aligned}\]Closure under addition is thus satisfied. Next, in a similar fashion we will tackle closure under scalar multiplication. Again, we are given vector \(u /in S\) and scalar \(d\), now we need to prove that \(v \in S\) where \(v = du\). Therefore again, we need to prove that \(v\) satisfies the following condition:
This can be rewritten as follows:
\[d(ax + by + cz) = 0\]And since \(u \in S\), it holds true that:
\[ax + by + cz = 0\]Therefore, we can write:
\[\begin{aligned} d(ax + by + cz) = 0 \\ d0 = 0 \\ 0 = 0 \end{aligned}\]Thus, we have also proved closure under scalar multiplication. Since all three conditions are met, we can conclude that \(S\) is a subspace of \(R^3\).
Let me reiterate over the most important things:
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Within this course, proving that given set is non-empty will always be possible through zero vector. In next session, I will explain why this is. But of course you are free to come up with another example.
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Both closures follow a simple principle, obtain newly created vector \(v\), put it in the condition that has to be satisfied (the formula left of the \(|\) sign) and then use the assumption that the original vectors come from the set \(S\) which allows you to prove it or not
This is an important skill to learn, therefore if you are unsure about anything in the proof procedure, please let me know and I will try to help. 😊
🔸 Exercise 2
Prove that the set \(S = \{(x, y) \mid x y=0\}\) is not a subspace of \(\mathbb{R}^2\).
Solution
One could argue that since the condition \(xy = 0\) is non-linear, \(S\) already does not meet the criteria to be a subspace. But to actually prove it, we can show that closure under addition is not met. Given that \(u_1, u_2 \in S\), then \(v\) must also live in \(S\):
If we submit it to the condition:
\[\begin{aligned} (x_1 + x_2)(y_1 + y_2) = 0 \\ x_1y_1 + x_2y_1 + x_1y_2 + x_2y_2 = 0\\ \end{aligned}\]If we use our assumption that \(u_1, u_2 \in S\) we can simplify the equation to:
\[\begin{aligned} x_1y_1 + x_2y_1 + x_1y_2 + x_2y_2 = 0\\ 0 + x_2y_1 + x_1y_2 + 0 = 0 \\ x_2y_1 + x_1y_2 = 0 \end{aligned}\]Therefore, if we for instance choose \(u_1 = (0, 1)\) and \(u_2 = (2, 0)\), then we get:
\[2 + 0 = 0\]which clearly is not true. Note that you could literally skipped the whole ‘formal’ process and just directly come up with a concrete example for which the addition does not work.
🔸 Exercise 5
Complete Example 6 by showing that \(V\) is closed under scalar multiplication. Do this as formally as possible, following the style of the example above.
Solution
Let us assume that
\[p(x)=a_2 x^2+a_1 x+a_0\]is part of \(V\). (given subset) In addition, that \(d\) is a scalar. Then, we need to proved that:
\[v = (dp)(x)\]also lives in \(V\). Therefore, we must prove that the following is true:
\[(dp)(2) = d(a_2 4 + a_1 2 + a_0) = 0\]Recall that we know:
\[p(2) = a_2 4 + a_1 2 + a_0 = 0\]Therefore, we can write:
\[\begin{aligned} d(a_2 4 + a_1 2 + a_0) = 0 \\ d(0) = 0 \\ 0 = 0 \end{aligned}\]which is of course true. Therefore we have also shoved that the \(V\) is closed under scalar multiplication.
⛑ Exercise 8
Use Theorem 1 to prove that \((-\mathbf{u})+(-\mathbf{v})=-(\mathbf{u}+\mathbf{v})\). Give another proof using Theorem 4.
Solution (based on Rasmus’ official solution)
By Theorem 1, it is enough to show that:
Therefore to do that, we can start by writing using axioms 2 and 3:
Then using axiom 5 we can write:
Finally, using axiom 4 and 5:
Thus we have proved that:
\[(-\mathbf{u})+(-\mathbf{v})+(\mathbf{u}+\mathbf{v}) = 0\]which implies according to theorem 1 that:
Alternatively, using theorem 4 we can start by writing:
and then using axiom 7, we can write:
which according to the theorem 4 is equal to:
⛑ Exercise 10
Prove that a vector space can have at most one zero vector. In other words, prove that if \(V\) is a vector space and \(\mathbf{v} \in V\) satisfies \(\mathbf{u}+\mathbf{v}=\mathbf{u}\) for all \(\mathbf{u}\) in \(V\), then \(\mathbf{v}=\mathbf{0}\).
Solution (based on Rasmus’ official solution)
We can assume that \(\mathbf{u}+\mathbf{v}=\mathbf{u}\) for all \(\mathbf{u}\) in \(V\). Therefore, for \(u = 0\), we get:
\[0 + v = 0\]Using axioms 2 and 4 we can rewrite this as follows:
Now, recall that we want to prove that \(v = 0\), using this assumption we can write:
\[0 = 0 + v = v\]and thus we have proved that \(v = 0\).
🔫 Exercise 11
Prove that \(c \cdot 0=0\) for any \(c\). Hint: Prove this first for \(c=0\) using Theorem 3 , then for \(c \neq 0\) using the previous exercise.
Solution (based on Rasmus’ official solution)
We can start for the case when \(c = 0\). Then according to theorem 3 we know that \(0v = 0\) for any \(v\) in the vector space \(V\), thus also \(v = 0\). Therefore, according to theorem 3 we know that \(0 \times 0 = 0\).
Now, let’s consider case when \(c \neq 0\). We can prove that \(u + c0 = u\), this can be done as follows:
\[\begin{aligned} \mathbf{u}+c \cdot \mathbf{0} &=\left(c \cdot \frac{1}{c}\right) \cdot \mathbf{u}+c \cdot \mathbf{0} \\ &=c \cdot\left(\frac{1}{c} \cdot \mathbf{u}\right)+c \cdot \mathbf{0} \\ &=c \cdot\left(\left(\frac{1}{c} \cdot \mathbf{u}\right)+\mathbf{0}\right) \\ &=c \cdot\left(\frac{1}{c} \cdot \mathbf{u}\right) \\ &=\left(c \cdot \frac{1}{c}\right) \cdot \mathbf{u} \\ &=1 \cdot \mathbf{u} \\ &=\mathbf{u} \end{aligned}\]Exercises from the book
🔹 4.4.5
For the following matrices:
\[A=\left[\begin{array}{rr} 2 & -3 \\ 4 & 1 \end{array}\right] \text { and } B=\left[\begin{array}{rr} 0 & 5 \\ 1 & -2 \end{array}\right]\]determine if the matrix:
\[C = \left[\begin{array}{rr} 6 & -19 \\ 10 & 7 \end{array}\right]\]is a linear combination of \(A\) and \(B\).
Solution (based on the official solution)
Our goal is to obtain \(c_1, c_2\) such that:
\[c_1\left[\begin{array}{rr} 2 & -3 \\ 4 & 1 \end{array}\right]+c_2\left[\begin{array}{rr} 0 & 5 \\ 1 & -2 \end{array}\right]=\left[\begin{array}{rr} 6 & -19 \\ 10 & 7 \end{array}\right]\]This can be translated into the following system of linear equations:
\[\begin{aligned} 2 c_1 &=6 \\ -3 c_1+5 c_2 &=-19 \\ 4 c_1+c_2=& 10 \\ c_1-2 c_2=& 7 \end{aligned}\]Using Gauss-Jordan elimination we obtain \(c_1 = 3\) and \(c_2 = -2\). Therefore the matrix \(C\) is a linear combination of \(A, B\).
🔹 4.4.19
Determine whether
\[S=\{(4,7,3),(-1,2,6),(2,-3,5)\}\]spans \(R^3\). If the set does not span \(R^3\), then give a geometric description of the subspace that it does not span.
Solution (based on the official solution)
If \(S\) spans \(R^3\), then any vector \(u \in R^3\) can be written as a linear combination of the vectors in \(S\) where \(u = (u_1, u_2, u_3)\). More specifically:
\[c_1(4,7,3)+c_2(-1,2,6)+c_3(2,-3,5)=\left(u_1, u_2, u_3\right)\]where \(c_1, c_2, c_3\) are scalars. Our goal is to determine if there exists such scalars that yield \(u\). Therefore, if we turn it into the system of linear equations:
\[\begin{aligned} &4 c_1-c_2+2 c_3=u_1 \\ &7 c_1+2 c_2-3 c_3=u_2 \\ &3 c_1+6 c_2+5 c_3=u_3 \end{aligned}\]And then compute the determinant of the coefficient matrix, we find out that the system has a unique solution since the determinant is non-zero. Thus we can conclude that \(span(s) = R^3\).
🔍 About the page
- Last updated: 21/09/2022
- Unless othewise stated, exercises come from the book:
Elementary Linear Algebra, International Metric Edition, Ron Larson - Purpose: This page was created as part of a preparation for my teaching assistant session in the course Linear algebra and optimization managed by
Rasmus Ejlers Møgelberg. Please note that unless othewise stated, the solutions on this page are my own, thus it should not be considered as official solutions provided as part of the course.