Exercise 05 solutions
🔹 must know, 🔸 ideally should know, ⛑ challenging, 🔫 very challenging
Exercises from the book
🔹 4.4.41
Determine whether the following set is linearly independent:
\[S=\left\{2-x, 2 x-x^2, 6-5 x+x^2\right\}\]Solution
First of all, I will rewrite the vectors into an actual vector form:
\[S=\left\{[2, -1, 0], [0, 2, -1], [6, -5, 1]\right\}\]Now, the question is whether if we linearly combine these vectors, we can get to zero:
\[c_1v_1 + c_2v_2 + c_3v_3 = 0\]where \(v_1, v_2, v_3 \in S\). Well, there is certainly a trivial solution \(c_1 = c_2 = c_3 = 0\). But is there also some non-trivial solution? We can find this out by solving \(Ax = 0\) where \(A\) is a matrix whose columns are formed by \(v_1, v_2, v_3\). Since the system has infinitely many solutions, we conclude that the set is linerly dependent, i.e., any of the vectors in the set can be written as a linear combination of other vectors.
🔹 4.5.40
Determine if \(S=\{(1,2),(1,-1)\} \text { forms basis for } R^2\).
Solution
First of all, notice that we know that basis for \(R^2\) will have two vectors, i.e., \(n = 2\). And \(S\) has also two vectors. This means that it is enough to prove either of the properties of basis - vector independence or span.
Since in the previous exercise, I showed how to prove independence, we can show that these two vectors also span \(R^2\). This means to prove that any vector \(b \in R^2\) can be obtained by linear combination of the two vectors in \(S\): \(Ax = b\) where \(x = [c_1, c_2]\).
Therefore, it is enough to just show that there exists unique solution, or in other words, that \(det(A) \neq 0\). Since \(A\) is a \(2 \times 2\) matrix, we can use closed formula \(det(A) = 1(-1) - 2(1) = -3\). Thus, the conclusion is that vectors in \(S\) form basis of \(R^2\).
🔹 4.5.51
Determine whether the following set
\[S=\left\{\left[\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right],\left[\begin{array}{ll} 1 & 4 \\ 0 & 1 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \\ 3 & 2 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \\ 2 & 0 \end{array}\right]\right\}\]forms basis of \(M_{2, 2}\).
Solution
First, again realize that \(n\) for vector space \(M_{2, 2}\) is \(4\) and \(S\) has 4 vectors. Therefore, it is enough to prove either of the properties. Let’s prove that the vectors in \(S\) are linearly independent.
We start by writing:
\[c_1\left[\begin{array}{ll} 2 & 0 \\ 0 & 3 \end{array}\right]+c_2\left[\begin{array}{ll} 1 & 4 \\ 0 & 1 \end{array}\right]+c_3\left[\begin{array}{ll} 0 & 1 \\ 3 & 2 \end{array}\right]+c_4\left[\begin{array}{ll} 0 & 1 \\ 2 & 0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]\]This can be then translated into the following system of linear equations:
\[\begin{aligned} 2 c_1+c_2 &=0 \\ 4 c_2+c_3+c_4 &=0 \\ 3 c_3+2 c_4 &=0 \\ 3 c_1+c_2+2 c_3 &=0 \end{aligned}\]The solution to this system is only trivial solution, thus we can conclude that given set \(S\) forms basis of \(R^2\).
🔹 4.6.11
Find basis of the row space and rank of the following matrix:
\[A = \left[\begin{array}{rrrr} -2 & -4 & 4 & 5 \\ 3 & 6 & -6 & -4 \\ -2 & -4 & 4 & 9 \end{array}\right]\]Solution
To obtain basis of \(A\)’s row space, we need to first reduce it to row echelon form:
\[\text{... some row transformations} \Rightarrow \left[\begin{array}{rrrr} 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right]\]Now, we know that all rows with leading 1 form basis of the row space, therefore:
\[B = \{(1,2,-2,0),(0,0,0,1)\}\]Since there are 2 vectors in \(B\), we can write that \(rank(A) = 2\).
I would like to also highlight one interesting thing about \(A\), it is non-square. So far, we have mostly talked about square matrices for which we actually find a determinant and also it is easy to understand what the linear transformation in their case means. But linear transformation does \(A\) represent?
Well, in the input space, we have some \(4 \times 1\) vector \(x\). This vector is then mapped into some \(3 \times 1\) vector \(b\). Therefore, we map the original vector from 4D space into 3D space. But we just found out that \(rank(A)\) is \(2\), meaning the transformed vector space has only 2 basis vectors. So this means that the original vector \(x\) is mapped into 3D space onto some plane which is spanned by the two basis vectors.
Why do I mention this? Because I wanted to highlight that given vector space can be 2 dimensional but the actual vector representation has 3 coordinates (3 dimensions). Therefore, when you want to know whether the input space has been squashed by the linear transformation, you compare the input vector space dimensions (number of variables) and transformed vector space dimension (rank of the given transformation matrix).
🔹 4.6.37
Find the null-space of the following matrix:
\[A=\left[\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 0 & 1 & -1 & 2 \\ -2 & -6 & 4 & -8 \end{array}\right]\]Solution
To find null-space of \(A\), we need to find solution to \(Ax = 0\). Using Gauss-Jordan elimination we arrive at:
where \(s, t \in R\). Therefore the null-space set can be described as:
\[\text{nullspace} = \left\{t \cdot\left(\begin{array}{c} -1 \\ 1 \\ 1 \\ 0 \end{array}\right)+s \cdot\left(\begin{array}{c} 2 \\ -2 \\ 0 \\ 1 \end{array}\right)\right\}\]🔹 4.6.51
Determine whether the nonhomogeneous system \(Ax = b\) is consistent. If it is, write the solution in the form \(x = x_p + x_h\), where \(x_p\) is a particular solution of \(Ax = b\) and \(x_h\) is a solution of \(Ax = 0\).
Solution
To check whether the system is consistent, we simply need to attempt to solve it. Therefore, we use again Gauss-Jordan elimination to obtain:
where \(t \in R\). Now, we need to write it as \(x = x_p + x_h\). This is should be easy since \(x_p = [3, 5, 0]\) and \(x_h = t[2, -4, 1]\). Therefore, we can write:
\[\mathbf{x}=\left[\begin{array}{l} 3 \\ 5 \\ 0 \end{array}\right]+t\left[\begin{array}{r} 2 \\ -4 \\ 1 \end{array}\right]\]Exercises from the Rasmus’ notes
⛑ Little proofs exercise
Suppose \(A\) and \(B\) are matrices such that \(B A\) is defined.
- Prove that \(N(A) \subseteq N(B A)\)
- Prove that if \(B\) is invertible then \(N(B A) \subseteq N(A)\). Can you prove this using what you proved in part 1 ? Remember that part 1 applies to all \(A\) and \(B\).
- Conclude that \(B\) invertible implies \(N(A)=N(B A)\)
Solution (based on Rasmus’ official solution)
Proof 1. We have to prove that \(N(A) \subseteq N(B A)\), i.e., nullspace of \(A\) is a subset of nullspace of \(BA\). In other words, any vector \(x\) which is part of \(N(A)\), must also be part of \(N(BA)\). If this is true, then it must be true that:
\[(BA)x = 0\]Since we assume \(x \in N(A)\), then \(Ax = 0\), therefore using associativity axiom with just derive fact, we can write:
\[(BA)x = B(Ax) = B0 = 0\]Thus we have proved \((BA)x = 0\) which proofs that \(N(A) \subseteq N(B A)\).
Proof 2. Essentially what we proved in 1 was:
\[N(C) \subseteq N(DC)\]Therefore, if \(C = BA\) and \(D = B^{-1}\), then it is true that:
\[N(BA) \subseteq N(B^{-1}BA)\]And since we know that \(N(B^{-1}BA) = N(A)\), then we proved that:
\[N(BA) \subseteq N(B^{-1}BA) = N(A)\]given \(B\) is invertible.
Proof 3. In the previous two exercises we have proved that
(1) \(N(A) \subseteq N(B A)\) (the fact that \(B\) is invertible does not change anything thus we did not even need it here)
(2) \(N(BA) \subseteq N(A)\)
which put together naturally implies \(N(A) = N(BA)\) given \(B\) is invertible.
🔍 About the page
- Last updated: 29/09/2022
- 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.