Topic 2 Vector Spaces and Subspaces

This topic focuses on the abstract structure of vector spaces, including tests for linear independence, determining spanning sets, finding bases for Null spaces and Column spaces, understanding the Rank Theorem, and proving whether a given set constitutes a valid subspace.

Question 1: Linear Independence of Vectors

Source: Practice Exam 4, Question 1
Exact Question: Are the vectors ${\mathbf{v}}_1=\left[\begin{array}{c}1\\ -3\\ 3\end{array}\right]$, ${\mathbf{v}}_2=\left[\begin{array}{c}2\\ 6\\ -1\end{array}\right]$ and ${\mathbf{v}}_3=\left[\begin{array}{c}-7\\ -3\\ -7\end{array}\right]$ linearly independent?

Methodology:

1. Objective: Determine if the only solution to the vector equation $x_1{\mathbf{v}}_1+x_2{\mathbf{v}}_2+x_3{\mathbf{v}}_3=\mathbf{0}$ is the trivial solution ($x_1=x_2=x_3=0$).
2. Matrix Setup: Form a matrix $A$ where the given vectors are the columns.
3. Logic: Row reduce the matrix to echelon form.
   • If every column has a pivot, there are no free variables, meaning the trivial solution is the only solution. The vectors are linearly independent.
   • If there is a column without a pivot, free variables exist, meaning there are non-trivial solutions. The vectors are linearly dependent.

Solution:

• Step 1: Construct the matrix.$$A=\left[\begin{array}{ccc}1& 2& -7\\ -3& 6& -3\\ 3& -1& -7\end{array}\right]$$
• Step 2: Row reduce to echelon form.
  $R_2\to R_2+3R_1$
  $R_3\to R_3-3R_1$$$\sim \left[\begin{array}{ccc}1& 2& -7\\ 0& 12& -24\\ 0& -7& 14\end{array}\right]$$
• Step 3: Eliminate the entry in row 3, column 2.
  $R_3\to R_3+\frac{7}{12}{R}_2$$$\sim \left[\begin{array}{ccc}1& 2& -7\\ 0& 12& -24\\ 0& 0& 0\end{array}\right]$$
• Conclusion: The $3\times 3$ matrix only has 2 pivots (in columns 1 and 2). Column 3 lacks a pivot, meaning the equation $A\mathbf{x}=\mathbf{0}$ has a non-trivial solution.
  Final Answer: No, the vectors are linearly dependent.

Question 2: Spanning a Space (${\mathbb{R}}^3$)

Source: Exam 2, Question 1
Exact Question: Consider the vectors ${\mathbf{v}}_1=\left[\begin{array}{c}1\\ -2\\ 4\end{array}\right]$, ${\mathbf{v}}_2=\left[\begin{array}{c}3\\ 0\\ -2\end{array}\right]$, ${\mathbf{v}}_3=\left[\begin{array}{c}-4\\ -4\\ 12\end{array}\right]$ and ${\mathbf{v}}_4=\left[\begin{array}{c}6\\ -6\\ h\end{array}\right]$. For which value(s) of $h$ do we have $\text{Span}\{{\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3,{\mathbf{v}}_4\}={\mathbb{R}}^3$?

Methodology:

1. Objective: Find the condition on $h$ such that any vector in ${\mathbb{R}}^3$ can be written as a linear combination of the given vectors.
2. Logic: A set of vectors spans ${\mathbb{R}}^m$ if and only if their corresponding matrix has a pivot position in every row.
3. Process: Place the vectors as columns in a matrix and row reduce to echelon form. Analyze the bottom row to find the values of $h$ that guarantee a pivot exists there.

Solution:

• Step 1: Construct the matrix and begin row reduction.$$\left[\begin{array}{cccc}1& 3& -4& 6\\ -2& 0& -4& -6\\ 4& -2& 12& h\end{array}\right]$$
• Step 2: $R_2\to R_2+2R_1$, and $R_3\to R_3-4R_1$.$$\sim \left[\begin{array}{cccc}1& 3& -4& 6\\ 0& 6& -12& 6\\ 0& -14& 28& h-24\end{array}\right]$$
• Step 3: To make the math match the provided exam solution exactly (which pre-simplified rows), let's scale row 2 by $1/6$:$$\sim \left[\begin{array}{cccc}1& 3& -4& 6\\ 0& 1& -2& 1\\ 0& -14& 28& h-24\end{array}\right]$$
• Step 4: $R_3\to R_3+14R_2$.$$\sim \left[\begin{array}{cccc}1& 3& -4& 6\\ 0& 1& -2& 1\\ 0& 0& 0& h-10\end{array}\right]$$
• Conclusion: For the vectors to span ${\mathbb{R}}^3$, there must be a pivot in the third row. The only potential pivot is the entry $(h-10)$. Thus, it must not be zero.
  Final Answer: $h-10\ne 0⟹h\ne 10$.

Question 3: Finding a Basis for Column Space ($\text{Col }A$)

Source: Exam 2, Question 4
Exact Question: Let $A=\left[\begin{array}{cccc}1& 4& 2& 5\\ 2& 8& -4& 2\\ -1& -4& 7& 4\end{array}\right]$. Find a basis for $\text{Col}(A)$.

Methodology:

1. Definition: The column space ($\text{Col }A$) is the set of all linear combinations of the columns of $A$.
2. Logic: The pivot columns of a matrix form a basis for its column space.
3. Process:
   • Row reduce $A$ to echelon form to identify which columns contain the pivots.
   • Crucial Rule: Always select the columns from the original matrix $A$, not the echelon form matrix. Row operations change the column space, but they preserve the linear dependence relations between columns.

Solution:

• Step 1: Row reduce $A$.
  $R_2\to R_2-2R_1$, $R_3\to R_3+R_1$.$$\sim \left[\begin{array}{cccc}1& 4& 2& 5\\ 0& 0& -8& -8\\ 0& 0& 9& 9\end{array}\right]$$
• Step 2: Scale rows to simplify (optional but helpful), $R_2\to -1/8R_2$, $R_3\to 1/9R_3$.$$\sim \left[\begin{array}{cccc}1& 4& 2& 5\\ 0& 0& 1& 1\\ 0& 0& 1& 1\end{array}\right]$$
• Step 3: $R_3\to R_3-R_2$.$$\sim \left[\begin{array}{cccc}1& 4& 2& 5\\ 0& 0& 1& 1\\ 0& 0& 0& 0\end{array}\right]$$
• Conclusion: The pivots are in the 1st and 3rd columns. Therefore, the 1st and 3rd columns of the original matrix $A$ form the basis.
  Final Answer:$$\{\left[\begin{array}{c}1\\ 2\\ -1\end{array}\right],\left[\begin{array}{c}2\\ -4\\ 7\end{array}\right]\}$$

Question 4: Finding a Basis for a Subspace Defined by an Equation ($\text{Nul }A$)

Source: Practice Exam 3, Question 2
Exact Question: Let $H$ be the subspace of ${\mathbb{R}}^4$ described by $-3x_1-5x_2+4x_3=0$. Find a basis $\mathcal{B}$ for $H$.

Methodology:

1. Objective: Find a set of linearly independent vectors that span the solutions to the given equation. This is equivalent to finding a basis for the null space ($\text{Nul }A$) of the matrix representing the equation.
2. Process:
   • Write the equation as a matrix system $A\mathbf{x}=0$. Here, $A=\left[\begin{array}{cccc}-3& -5& 4& 0\end{array}\right]$. Note the 0 for $x_4$ since it's missing from the equation but the space is ${\mathbb{R}}^4$.
   • Identify the basic variable (pivot) and the free variables.
   • Solve for the basic variable in terms of the free variables.
   • Write the general solution in parametric vector form ($\mathbf{x}=x_2{\mathbf{v}}_1+x_3{\mathbf{v}}_2+x_4{\mathbf{v}}_3$).
   • The vectors attached to the free variables form the basis.

Solution:

• Step 1: Matrix representation is $\left[\begin{array}{cccc}-3& -5& 4& 0\end{array}\right]$. The pivot is on $x_1$. The free variables are $x_2,x_3,x_4$.
• Step 2: Solve for $x_1$.
  $-3x_1-5x_2+4x_3=0⟹3x_1=-5x_2+4x_3⟹x_1=-\frac{5}{3}{x}_2+\frac{4}{3}{x}_3$
• Step 3: Write in parametric vector form $\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]$:$$\left[\begin{array}{c}-\frac{5}{3}{x}_2+\frac{4}{3}{x}_3\\ x_2\\ x_3\\ x_4\end{array}\right]=x_2\left[\begin{array}{c}-\frac{5}{3}\\ 1\\ 0\\ 0\end{array}\right]+x_3\left[\begin{array}{c}\frac{4}{3}\\ 0\\ 1\\ 0\end{array}\right]+x_4\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]$$Final Answer:$$\mathcal{B}=\{\left[\begin{array}{c}-\frac{5}{3}\\ 1\\ 0\\ 0\end{array}\right],\left[\begin{array}{c}\frac{4}{3}\\ 0\\ 1\\ 0\end{array}\right],\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]\}$$

Question 5: The Rank Theorem and Subspace Dimensions

Source: Exam 2, Question 3
Exact Question: Suppose $A$ is a matrix with 4 pivot positions. In addition, suppose that the solution set to the equation $A\mathbf{x}=\mathbf{b}$ for a certain vector $\mathbf{b}$ takes the form $\mathbf{x}={\mathbf{v}}_1+s{\mathbf{v}}_2+t{\mathbf{v}}_3$, where the set $\{{\mathbf{v}}_1,{\mathbf{v}}_2,{\mathbf{v}}_3\}$ is linearly independent. How many columns does $A$ have?

Methodology:

1. Formulas / Logic:
   • Rank Theorem: $\text{Rank}(A)+\dim (\text{Nul }A)=n$, where $n$ is the number of columns.
   • $\text{Rank}(A)$ is strictly equal to the number of pivot positions.
   • The structure of the solution to $A\mathbf{x}=\mathbf{b}$ is $\mathbf{x}={\mathbf{x}}_p+{\mathbf{x}}_h$, where ${\mathbf{x}}_p$ is a particular solution (here, ${\mathbf{v}}_1$) and ${\mathbf{x}}_h$ is the general solution to the homogeneous equation $A\mathbf{x}=\mathbf{0}$ (here, $s{\mathbf{v}}_2+t{\mathbf{v}}_3$).
   • The vectors attached to the parameters in the homogeneous part span $\text{Nul }A$.
2. Process: Identify $\text{Rank}(A)$ and $\dim (\text{Nul }A)$ from the prompt, and add them together.

Solution:

• Step 1: The matrix has 4 pivot positions, so $\text{Rank}(A)=4$.
• Step 2: The homogeneous solution set is spanned by ${\mathbf{v}}_2$ and ${\mathbf{v}}_3$. Since they are linearly independent, they form a basis for $\text{Nul }A$. Therefore, there are 2 basis vectors, so $\dim (\text{Nul }A)=2$.
• Step 3: Apply the Rank Theorem. Number of columns = $\text{Rank}(A)+\dim (\text{Nul }A)=4+2=6$.
  Final Answer: 6 columns.

Question 6: Coordinate Vectors

Source: Practice Exam 4, Question 3
Exact Question: Let $H$ be the subspace of ${\mathbb{R}}^3$ with basis $\mathcal{B}=\{\left[\begin{array}{c}-1\\ -3\\ 1\end{array}\right],\left[\begin{array}{c}-1\\ -6\\ 10\end{array}\right]\}$ and let $\mathbf{x}=\left[\begin{array}{c}1\\ 0\\ 8\end{array}\right]$. Is $\mathbf{x}$ a vector in $H$? If so, compute the $\mathcal{B}$-coordinate vector of $\mathbf{x}$.

Methodology:

1. Objective: Determine if $\mathbf{x}$ can be written as a linear combination of the basis vectors ($c_1{\mathbf{b}}_1+c_2{\mathbf{b}}_2=\mathbf{x}$). If yes, the scalars $(c_1,c_2)$ form the coordinate vector $[\mathbf{x}{]}_{\mathcal{B}}$.
2. Process: Set up the augmented matrix $[{\mathbf{b}}_1{\mathbf{b}}_2\mid \mathbf{x}]$ and row reduce.
   • If the system is inconsistent (a row of $0=\text{non-zero}$), $\mathbf{x}$ is not in $H$.
   • If it is consistent, solve for $c_1$ and $c_2$. The coordinate vector is $[\mathbf{x}{]}_{\mathcal{B}}=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$.

Solution:

• Step 1: Set up the augmented matrix.$$\left[\begin{array}{ccc}-1& -1& 1\\ -3& -6& 0\\ 1& 10& 8\end{array}\right]$$
• Step 2: Row reduce. $R_3\to R_3+R_1$, $R_2\to R_2-3R_1$.$$\sim \left[\begin{array}{ccc}-1& -1& 1\\ 0& -3& -3\\ 0& 9& 9\end{array}\right]$$
• Step 3: $R_3\to R_3+3R_2$.$$\sim \left[\begin{array}{ccc}-1& -1& 1\\ 0& -3& -3\\ 0& 0& 0\end{array}\right]$$
• Step 4: There is no contradictory row, so $\mathbf{x}$ is in $H$. Solve the system.
  Row 2: $-3c_2=-3⟹c_2=1$.
  Row 1: $-c_1-c_2=1⟹-c_1-1=1⟹-c_1=2⟹c_1=-2$.
  Final Answer: Yes, it is in $H$. $[\mathbf{x}{]}_{\mathcal{B}}=\left[\begin{array}{c}-2\\ 1\end{array}\right]$.

Question 7: Proving or Disproving a Subspace

Source: Exam 2, Question 10
Exact Question: Consider the subset $S$ of ${\mathbb{R}}^3$ of all vectors $\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]$ with $|x_1|=|x_2+x_3|$. In order for $S$ to be a subspace of ${\mathbb{R}}^3$, it needs to satisfy the three defining properties of a subspace.
a. Verify for each property whether it is satisfied by this particular set $S$. Motivate your answers.
b. Conclude whether $S$ is a subspace of ${\mathbb{R}}^3$.

Methodology:

1. Objective: Test a subset against the strictly defined rules of a vector subspace.
2. The Three Properties:
   • (i) Contains the Zero Vector: Plug in $\mathbf{0}=(0,0,0)$ and see if the condition holds.
   • (ii) Closed under Vector Addition: Take two arbitrary vectors $\mathbf{u}$ and $\mathbf{v}$ that exist in $S$. Test if their sum $\mathbf{u}+\mathbf{v}$ satisfies the condition. Tip: For non-linear conditions like absolute values, look for a counterexample where addition breaks the rule.
   • (iii) Closed under Scalar Multiplication: Take an arbitrary vector $\mathbf{u}$ in $S$ and a scalar $c$. Test if $c\mathbf{u}$ satisfies the condition.

Solution:
Part a:

• (i) Zero Vector: Check $\mathbf{0}=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$. The rule requires $|0|=|0+0|$. Since $0=0$, the property is satisfied.
• (ii) Closure under Addition: We suspect this fails because $|a+b|\ne |a|+|b|$ generally. Let's find a counterexample.
  Let $\mathbf{u}=\left[\begin{array}{c}-1\\ 2\\ -1\end{array}\right]$. Check: $|-1|=|2+(-1)|⟹1=1$. So $\mathbf{u}\in S$.
  Let $\mathbf{v}=\left[\begin{array}{c}1\\ 1\\ 0\end{array}\right]$. Check: $|1|=|1+0|⟹1=1$. So $\mathbf{v}\in S$.
  Now check $\mathbf{u}+\mathbf{v}=\left[\begin{array}{c}0\\ 3\\ -1\end{array}\right]$.
  Check rule: Does $|0|=|3+(-1)|$? No, $0\ne 2$.
  Thus, $\mathbf{u}+\mathbf{v}$ is NOT in $S$. The property is not satisfied.
• (iii) Closure under Scalar Multiplication: Let $\mathbf{u}\in S$, so $|u_1|=|u_2+u_3|$. Let $c\in \mathbb{R}$. We test $c\mathbf{u}=\left[\begin{array}{c}c{u}_1\\ c{u}_2\\ c{u}_3\end{array}\right]$.
  Check rule: Does $|c{u}_1|=|c{u}_2+c{u}_3|$?
  $|c{u}_1|=|c||u_1|$. We know $|u_1|=|u_2+u_3|$, so:
  $|c||u_1|=|c||u_2+u_3|=|c(u_2+u_3)|=|c{u}_2+c{u}_3|$.
  This is true. The property is satisfied.

Part b:

• Conclusion: Because property (ii) fails, $S$ is not a subspace.