Row, Column, and Space - Linear Algebra

1. Let $A$ be an $m \times n$ matrix. The set of all vectors that are simultaneously in the row space of $A$ and the null space of $A$ is
   Answer: a subspace containing only the zero vector. The row space of $A$ and the null space of $A$ are orthogonal subspaces of $\mathbb{R}^n$. If a vector $x$ is in both subspaces, it must be orthogonal to itself. That is, its dot product with itself must be zero: $x \cdot x = 0$. The only vector in $\mathbb{R}^n$ with this property is the zero vector, $x = \vec{0}$. Therefore, the intersection of these two subspaces contains only the zero vector, forming the zero vector space.
2. Let $A$ be a $3 \times 4$ matrix. The vector $v = \begin{pmatrix} 2 \\ -1 \\ 5 \\ 1 \end{pmatrix}$ is known to be in the row space of $A$. Which of the following vectors CANNOT be in the null space of $A$?
   Answer: $w_4 = \begin{pmatrix} 4 \\ 3 \\ -1 \\ 2 \end{pmatrix}$ The row space of a matrix $A$ is orthogonal to its null space. This means that for any vector $v$ in the row space and any vector $w$ in the null space, their dot product must be zero ($v \cdot w = 0$). We must test the dot product of the given vector $v$ with each of the answer choices. The vector whose dot product with $v$ is non-zero cannot be in the null space. A. $v \cdot w_1 = (2)(-2) + (-1)(1) + (5)(1) + (1)(-1) = -4 - 1 + 5 - 1 = -1$. Wait, there's a calculation error in my thought process. Let me re-calculate. A. $v \cdot w_1 = (2)(-2) + (-1)(1) + (5)(1) + (1)(-1) = -4 - 1 + 5 - 1 = -1$. This is non-zero, so it cannot be in the null space. Let me check the others to ensure only one is correct. I must have designed this problem with a specific answer in mind. Let me correct the choices to fit the intended logic. Re-designing choices: Let $v = \begin{pmatrix} 2 \\ -1 \\ 5 \\ 1 \end{pmatrix}$. Choice A: $w_1 = \begin{pmatrix} -2 \\ -3 \\ 1 \\ 0 \end{pmatrix}$. $v \cdot w_1 = (2)(-2) + (-1)(-3) + (5)(1) + (1)(0) = -4 + 3 + 5 + 0 = 4$. Correct, this cannot be in Nul(A). Choice B: $w_2 = \begin{pmatrix} 3 \\ 6 \\ 0 \\ 0 \end{pmatrix}$. $v \cdot w_2 = (2)(3) + (-1)(6) + (5)(0) + (1)(0) = 6 - 6 = 0$. This could be in Nul(A). Choice C: $w_3 = \begin{pmatrix} 1 \\ 7 \\ 1 \\ 0 \end{pmatrix}$. $v \cdot w_3 = (2)(1) + (-1)(7) + (5)(1) + (1)(0) = 2 - 7 + 5 = 0$. This could be in Nul(A). Choice D: $w_4 = \begin{pmatrix} 0 \\ 5 \\ 1 \\ 0 \end{pmatrix}$. $v \cdot w_4 = (2)(0) + (-1)(5) + (5)(1) + (1)(0) = -5 + 5 = 0$. This could be in Nul(A). My initial choices were flawed. I will use a corrected set. The original intent was for D to be the answer. Let's make it so. Let $v = \begin{pmatrix} 2 \\ -1 \\ 5 \\ 1 \end{pmatrix}$. Let's make A, B, C have dot product zero and D not. A: $w_1 = \begin{pmatrix} -2 \\ 1 \\ 1 \\ 0 \end{pmatrix}$. $v \cdot w_1 = -4-1+5=0$. OK. B: $w_2 = \begin{pmatrix} 3 \\ 6 \\ 0 \\ 0 \end{pmatrix}$. $v \cdot w_2 = 6-6=0$. OK. C: $w_3 = \begin{pmatrix} 1 \\ -3 \\ -1 \\ 6 \end{pmatrix}$. $v \cdot w_3 = 2+3-5+6=6$. OK, this will be the answer. D: $w_4 = \begin{pmatrix} 0 \\ 5 \\ 1 \\ 0 \end{pmatrix}$. $v \cdot w_4 = -5+5=0$. OK. Final question setup with correct choices: Stem: Let $A$ be a $3 \times 4$ matrix. The vector $v = \begin{pmatrix} 2 \\ -1 \\ 5 \\ 1 \end{pmatrix}$ is known to be in the row space of $A$. Which of the following vectors CANNOT be in the null space of $A$?\na: $w_1 = \begin{pmatrix} -2 \\ 1 \\ 1 \\ 0 \end{pmatrix}$\nb: $w_2 = \begin{pmatrix} 3 \\ 6 \\ 0 \\ 0 \end{pmatrix}$\nc: $w_3 = \begin{pmatrix} 1 \\ -3 \\ -1 \\ 6 \end{pmatrix}$\nd: $w_4 = \begin{pmatrix} 0 \\ 5 \\ 1 \\ 0 \end{pmatrix}$ Correct Answer: C. Explanation: The row space and null space are orthogonal complements. Thus, the dot product of any vector in the row space with any vector in the null space must be zero. We compute the dot product of $v$ with each choice: A. $v \cdot w_1 = (2)(-2) + (-1)(1) + (5)(1) + (1)(0) = -4 - 1 + 5 = 0$. B. $v \cdot w_2 = (2)(3) + (-1)(6) + (5)(0) + (1)(0) = 6 - 6 = 0$. C. $v \cdot w_3 = (2)(1) + (-1)(-3) + (5)(-1) + (1)(6) = 2 + 3 - 5 + 6 = 6$. Since the dot product is non-zero, $w_3$ cannot be in the null space of $A$. D. $v \cdot w_4 = (2)(0) + (-1)(5) + (5)(1) + (1)(0) = 0 - 5 + 5 = 0$.
3. Let $A$ be an $m \times n$ matrix. Which condition ensures that the equation $Ax = b$ is consistent for every vector $b \in \mathbb{R}^m$?
   Answer: The null space of $A^T$ contains only the zero vector. The equation $Ax=b$ is consistent for every $b \in \mathbb{R}^m$ if and only if the column space of $A$ spans all of $\mathbb{R}^m$. This means $\text{Col}(A) = \mathbb{R}^m$. By the Fundamental Theorem of Linear Algebra, the column space of $A$ is the orthogonal complement of the null space of $A^T$, i.e., $\text{Col}(A) = (\text{Nul}(A^T))^{\perp}$. For $(\text{Nul}(A^T))^{\perp}$ to be all of $\mathbb{R}^m$, the null space of $A^T$ must contain only the zero vector. A and B are equivalent statements meaning $rank(A)=n$. This only ensures a unique solution if one exists, but does not guarantee a solution for all $b$ unless $n=m$. C is a necessary condition for this to be possible (since $rank(A) \le \min(m,n)$), but it is not sufficient.
4. A basis for the null space of a certain $3 \times 4$ matrix $A$ is $\left\{ \begin{pmatrix} 1 \\ -1 \\ 0 \\ 2 \end{pmatrix}, \begin{pmatrix} 0 \\ 2 \\ 1 \\ -1 \end{pmatrix} \right\}$. Which of the following vectors must be in the row space of $A$?
   Answer: $w = \begin{pmatrix} 1 \\ 3 \\ -2 \\ 1 \end{pmatrix}$ The row space of $A$ is the orthogonal complement of the null space of $A$ in $\mathbb{R}^4$. A vector $w$ is in the row space if and only if it is orthogonal to every vector in the basis for the null space. Let the basis vectors be $v_1$ and $v_2$. We need to find the choice $w$ for which $w \cdot v_1 = 0$ and $w \cdot v_2 = 0$. Let's test choice D: $w = (1, 3, -2, 1)$. $w \cdot v_1 = (1)(1) + (3)(-1) + (-2)(0) + (1)(2) = 1 - 3 + 0 + 2 = 0$. $w \cdot v_2 = (1)(0) + (3)(2) + (-2)(1) + (1)(-1) = 0 + 6 - 2 - 1 = 3$. Oh, calculation error again. Let's fix this. I need a vector orthogonal to both. Let $w=(w_1,w_2,w_3,w_4)$. Then $w_1-w_2+2w_4=0$ and $2w_2+w_3-w_4=0$. Let $w_4=2, w_2=1$. Then $w_3 = w_4-2w_2 = 2-2=0$. And $w_1 = w_2-2w_4 = 1-4=-3$. So $w=(-3, 1, 0, 2)$ is in the row space. Let's make this choice D. Let's test the other choices with the original basis vectors. A: $w=(1,1,-2,0)$. $w \cdot v_1 = 1-1=0$. $w \cdot v_2 = 2-2=0$. So A is correct. Let me check my initial setup. Maybe I made a mistake somewhere else. Oh, I see. I must have intended for only one to work. Let me check the other initial choices with the basis. B: $w=(1,1,2,0)$. $w \cdot v_1 = 1-1=0$. $w \cdot v_2 = 2+2=4$. So B is wrong. C: $w=(2,0,1,-1)$. $w \cdot v_1 = 2-2=0$. $w \cdot v_2 = 1+1=2$. So C is wrong. So my choice A was already correct. Let me re-verify D: $w=(1,3,-2,1)$. $w \cdot v_1 = 1-3+2=0$. $w \cdot v_2 = 6-2-1=3$. So D is wrong. The correct answer is A.
5. Let $A$ be a non-zero $m \times n$ matrix with $rank(A) = r$. If a vector $b$ is in the column space of $A$, and a vector $v$ is in the row space of $A$, which of the following statements must be true about the equation $Ax=b$?
   Answer: The equation has at least one solution, and among all solutions, there is exactly one that lies in the row space of $A$. Since $b$ is in the column space of $A$, the system $Ax=b$ is consistent and has at least one solution. The general solution is of the form $x = x_p + x_h$, where $x_p$ is a particular solution and $x_h$ is any vector in the null space of $A$. A fundamental result states that for any consistent system $Ax=b$, there is a unique solution $x^*$ that lies in the row space of $A$. This unique solution $x^*$ is the projection of any particular solution onto the row space. All other solutions are of the form $x^* + x_h$ where $x_h \in \text{Nul}(A)$ and $x_h \neq 0$. Since $x_h$ is orthogonal to the row space, $x^*+x_h$ is not in the row space. Thus, there is exactly one solution in the row space.
6. A matrix $A$ has a 4-dimensional column space, a 3-dimensional null space, and a 2-dimensional left null space (null space of $A^T$). What are the dimensions $m \times n$ of matrix $A$?
   Answer: $m=6, n=7$ We are given: $\text{dim}(\text{Col}(A)) = 4$, $\text{dim}(\text{Nul}(A)) = 3$, and $\text{dim}(\text{Nul}(A^T)) = 2$. The dimension of the column space is the rank of the matrix, so $rank(A) = 4$. The Rank-Nullity Theorem applied to $A$ states $rank(A) + \text{dim}(\text{Nul}(A)) = n$ (the number of columns). Substituting the given values: $4 + 3 = n$, so $n=7$. The Rank-Nullity Theorem can also be applied to $A^T$. The rank of $A^T$ is the same as the rank of $A$. The number of columns in $A^T$ is $m$. So, $rank(A^T) + \text{dim}(\text{Nul}(A^T)) = m$. Substituting the given values: $4 + 2 = m$, so $m=6$. Therefore, the matrix $A$ is a $6 \times 7$ matrix.
7. Let $A$ be a $5 \times 8$ matrix. If the linear transformation $T(x) = Ax$ maps $\mathbb{R}^8$ onto $\mathbb{R}^5$, what is the dimension of the null space of $A$?
   Answer: 3 The statement that the transformation $T(x) = Ax$ maps $\mathbb{R}^8$ onto $\mathbb{R}^5$ means that the column space of $A$ spans all of $\mathbb{R}^5$. Therefore, the dimension of the column space of $A$, which is the rank of $A$, must be 5. So, $rank(A) = 5$. The Rank-Nullity Theorem states that for an $m \times n$ matrix, $rank(A) + \text{dim}(\text{Nul}(A)) = n$. Here, $A$ is a $5 \times 8$ matrix, so $n=8$. Substituting the known values, we get $5 + \text{dim}(\text{Nul}(A)) = 8$. Solving for the dimension of the null space gives $\text{dim}(\text{Nul}(A)) = 8 - 5 = 3$.
8. The vectors $x_1 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix}$ and $x_2 = \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix}$ are two distinct solutions to the non-homogeneous system of equations $Ax = b$. Which of the following vectors must be in the null space of $A$?
   Answer: $v_3 = \begin{pmatrix} -2 \\ 1 \\ -3 \end{pmatrix}$ When you encounter multiple solutions to a non-homogeneous system $Ax = b$, you're dealing with a fundamental relationship in linear algebra. The key insight is that if $x_1$ and $x_2$ are both solutions to $Ax = b$, then their difference $x_1 - x_2$ must lie in the null space of $A$. Here's why: Since $Ax_1 = b$ and $Ax_2 = b$, we have $A(x_1 - x_2) = Ax_1 - Ax_2 = b - b = 0$. This means $x_1 - x_2$ is in the null space of $A$. Let's calculate $x_1 - x_2 = \begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} - \begin{pmatrix} 3 \\ 1 \\ 4 \end{pmatrix} = \begin{pmatrix} -2 \\ 1 \\ -3 \end{pmatrix}$ This is exactly vector A) $v_3$, which must be in the null space of $A$. Looking at the wrong answers: B) $v_2$ is just $x_1$, which satisfies $Ax_1 = b$, not $Ax_1 = 0$, so it's not in the null space. C) $v_1 = \begin{pmatrix} 4 \\ 3 \\ 5 \end{pmatrix}$ equals $x_1 + x_2$, which would satisfy $A(x_1 + x_2) = 2b \neq 0$. D) $v_4$ has no clear relationship to our given solutions and cannot be determined to be in the null space from the given information. Remember this pattern: when you have two solutions to a non-homogeneous system, their difference always reveals a vector in the null space. This connects the solution structure of non-homogeneous systems to the fundamental concept of null spaces.
9. Let $A$ be a $5 \times 3$ matrix. A basis for the null space of $A^T$ is given by $\left\{ \begin{pmatrix} 1 \\ 0 \\ 2 \\ 0 \\ -1 \end{pmatrix}, \begin{pmatrix} 0 \\ 1 \\ -1 \\ 3 \\ 0 \end{pmatrix} \right\}$. For which vector $b$ is the system $Ax = b$ guaranteed to be consistent?
   Answer: $b = \begin{pmatrix} 2 \\ -4 \\ -1 \\ 1 \\ 5 \end{pmatrix}$ The system $Ax=b$ is consistent if and only if $b$ is in the column space of $A$. The Fundamental Theorem of Linear Algebra states that the column space of $A$ is the orthogonal complement of the null space of $A^T$. Therefore, $b$ must be orthogonal to every vector in the basis of $\text{Nul}(A^T)$. Let the basis vectors be $v_1$ and $v_2$. We need to find the choice for $b$ such that $b \cdot v_1 = 0$ and $b \cdot v_2 = 0$. For choice C: $b = (2, -4, -1, 1, 5)$. $b \cdot v_1 = (2)(1) + (-4)(0) + (-1)(2) + (1)(0) + (5)(-1) = 2 + 0 - 2 + 0 - 5 = -5$. Oh wait, calculation error again. Let me fix the correct answer. I need a vector orthogonal to both $v_1$ and $v_2$. Let $b=(b_1,b_2,b_3,b_4,b_5)$. $b_1+2b_3-b_5=0$ and $b_2-b_3+3b_4=0$. Let $b_3=1, b_4=1, b_5=1$. Then $b_1 = 1-2 = -1$. $b_2 = 1-3 = -2$. So $b=(-1, -2, 1, 1, 1)$ should work. Let's make this choice C. Corrected choices: A. $b = (1, 1, 1, 1, 1)$, $b \cdot v_1 = 1+2-1=2 \neq 0$. B. $b = (-2, 1, 1, -3, 0)$, $b \cdot v_1 = -2+2=0$, but $b \cdot v_2 = 1-1-9=-9 \neq 0$. C. $b = (-1, -2, 1, 1, 1)$, $b \cdot v_1 = -1+2-1=0$, and $b \cdot v_2 = -2-1+3=0$. This is the correct choice. D. $b = (3, -2, 0, 6, -3)$, $b \cdot v_1 = 3+3=6 \neq 0$. My original choice C was wrong. The new C is correct.
10. Let $A$ be the matrix $A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ -3 & -6 & k \end{pmatrix}$ For which value of $k$ is the dimension of the null space of $A$ equal to 2?
    Answer: $k = 3$ When you encounter questions about the dimension of the null space, remember that it's directly connected to the rank of the matrix through the rank-nullity theorem: for an $n \times n$ matrix, $\text{rank}(A) + \text{nullity}(A) = n$. Since we want the null space dimension to be 2, and $A$ is $3 \times 3$, we need $\text{rank}(A) = 1$. To find when the rank equals 1, let's row reduce the matrix. Notice that row 2 is exactly 2 times row 1, and row 3 is $-3$ times row 1 (except for the $k$ entry). Starting with row operations: $\begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \\ -3 & -6 & k \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 2 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & k-3 \end{pmatrix}$ For the rank to be 1, we need the third row to also become all zeros, which happens when $k - 3 = 0$, so $k = 3$. Choice A ($k = 3$) is correct because it makes the rank equal to 1, giving nullity = 2. Choice B ($k = 2$) is wrong because this makes the third row $(0, 0, -1)$, giving rank = 2 and nullity = 1. Choice C ($k = -1$) is wrong because this makes the third row $(0, 0, -4)$, also giving rank = 2 and nullity = 1. Choice D is wrong because only $k = 3$ produces the desired nullity of 2; any other value gives nullity = 1. Study tip: When finding null space dimensions, always use the rank-nullity theorem and look for when rows become linearly dependent through row reduction.
11. Let $A$ be a $3 \times 5$ matrix, and let $R$ be its reduced row echelon form, given by $R = \begin{pmatrix} 1 & 3 & 0 & -2 & 0 \\ 0 & 0 & 1 & 5 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$ Which statement about the matrix $A$ must be true?
    Answer: The dimension of the row space of $A$ is 3. The rank of a matrix is the number of pivots (leading 1s) in its reduced row echelon form. Here, $R$ has 3 pivots, so $rank(A) = 3$. The dimension of the row space is equal to the rank. Therefore, the dimension of the row space of $A$ is 3. A is incorrect because a basis for the column space of $A$ is formed by the columns of $A$ that correspond to the pivot columns of $R$. The pivot columns in $R$ are columns 1, 3, and 5. B is a common misconception; row operations change the column space, so the columns of $R$ do not generally span the column space of $A$. C is incorrect because by the Rank-Nullity Theorem, $\text{dim}(\text{Nul}(A)) = n - rank(A) = 5 - 3 = 2$.
12. Let $A$ be a matrix and $R$ be its reduced row echelon form. Which of the following statements about the fundamental subspaces is generally FALSE?
    Answer: The column space of $A$ is equal to the column space of $R$. Elementary row operations do not change the linear dependencies among the columns. However, they do change the column vectors themselves and thus can change the column space. For example, consider $A = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$. Its RREF is $R = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$. The column space of $A$ is the set of all multiples of $(1, 2)$, while the column space of $R$ is the set of all multiples of $(1, 0)$. These are different subspaces. The other statements are true: row operations preserve the row space, the null space (solutions to $Ax=0$), and the rank (which is the dimension of the column space).

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