One-to-One & Onto - Linear Algebra

1. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation that orthogonally projects each vector onto the $xy$-plane. Which of the following statements is true?
   Answer: $T$ is neither one-to-one nor onto. The transformation is given by $T(x, y, z) = (x, y, 0)$. To check if it is one-to-one, we find its kernel (null space), which is the set of vectors that map to $\vec{0}$. $T(x, y, z) = (0, 0, 0)$ implies $x=0$ and $y=0$. The $z$ component can be any real number. So, the kernel consists of all vectors of the form $(0, 0, z)$, which is the z-axis. Since the kernel is not just the zero vector, $T$ is not one-to-one. To check if it is onto, we consider its range. The range is the set of all possible output vectors, which are of the form $(x, y, 0)$. This set is the $xy$-plane, which is a 2-dimensional subspace of the codomain $\mathbb{R}^3$. Since the range is not the entire codomain, $T$ is not onto.
2. Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation with standard matrix $A = \begin{pmatrix} 3 & -6 \\ -1 & k \end{pmatrix}$. For which value of $k$ is the transformation $T$ NOT one-to-one?
   Answer: $k = 2$ A linear transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$ is not one-to-one if and only if its standard matrix is not invertible. A $2 \times 2$ matrix is not invertible if and only if its determinant is zero. The determinant of $A$ is $\det(A) = (3)(k) - (-6)(-1) = 3k - 6$. Setting the determinant to zero to find when $T$ is not one-to-one: $3k - 6 = 0$, which gives $3k = 6$, so $k = 2$. When $k=2$, the second column is -2 times the first column, so they are linearly dependent.
3. Consider the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^2$ represented by the matrix $A = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{bmatrix}$. A student claims that $T$ is one-to-one because "the rows are linearly dependent, so the transformation compresses the space nicely." What is wrong with this reasoning?
   Answer: The student should examine column dependence, not row dependence, to determine if $T$ is one-to-one To determine if T is one-to-one, we need to check if ker(T) = {0}, which is equivalent to checking if the columns of A are linearly independent (since Ax = 0 only when x = 0). Row dependence/independence relates to whether the rows span the codomain (onto property), not whether T is one-to-one. The student is examining the wrong vectors. Choice A incorrectly accepts the reasoning. Choice B is too broad - row properties do relate to transformation properties, just not one-to-one. Choice D incorrectly describes what row dependence affects.
4. Let $T_A: \mathbb{R}^3 \to \mathbb{R}^5$ and $T_B: \mathbb{R}^5 \to \mathbb{R}^3$ be linear transformations. Which of the following statements is always true?
   Answer: $T_B$ cannot be one-to-one. Let's analyze each option based on the dimensions. $T_A$ has a $5 \times 3$ matrix $A$. $T_B$ has a $3 \times 5$ matrix $B$. (A) For $T_A$ to be onto, the columns of $A$ must span $\mathbb{R}^5$. But there are only 3 columns, and 3 vectors cannot span a 5-dimensional space. So $T_A$ can never be onto. (B) For $T_B$ to be one-to-one, the 5 columns of $B$ (which are vectors in $\mathbb{R}^3$) must be linearly independent. Any set of 5 vectors in $\mathbb{R}^3$ must be linearly dependent. So $T_B$ can never be one-to-one. (C) $T_A$ does not have to be one-to-one. Its 3 columns could be linearly dependent (e.g., if $A$ is the zero matrix). (D) As explained for (B), the 5 columns of the $3 \times 5$ matrix for $T_B$ must be linearly dependent, so $T_B$ cannot be one-to-one. This statement is always true.
5. Let $T: \mathbb{R}^3 \to \mathbb{R}^2$ be a linear transformation with matrix representation $A = \begin{bmatrix} 2 & -1 & 3 \\ 4 & -2 & 6 \end{bmatrix}$. Which statement about the properties of $T$ is correct?
   Answer: $T$ is neither one-to-one nor onto because $\text{rank}(A) = 1$ The rank of matrix A is 1 (the second row is twice the first row). For T to be one-to-one, we need rank(A) = dim(domain) = 3, but rank(A) = 1 < 3. For T to be onto, we need rank(A) = dim(codomain) = 2, but rank(A) = 1 < 2. Therefore, T is neither one-to-one nor onto. Choice A incorrectly states T is one-to-one. Choice B incorrectly states T is onto. Choice D incorrectly claims both properties hold.
6. A linear transformation $S: \mathbb{R}^5 \to \mathbb{R}^3$ is defined such that its matrix $A$ satisfies $\text{rank}(A) = 3$. For the equation $S(\mathbf{x}) = \mathbf{b}$, which statement correctly describes the solution behavior for different choices of $\mathbf{b}$?
   Answer: Every equation $S(\mathbf{x}) = \mathbf{b}$ has infinitely many solutions because $\text{nullity}(A) = 2$ Since rank(A) = 3 and the codomain is R³, S is onto, so every equation S(x) = b is solvable for any b ∈ R³. By rank-nullity theorem, nullity(A) = 5 - 3 = 2, meaning ker(S) is 2-dimensional. When a solution exists, the general solution is a particular solution plus all vectors in the 2-dimensional kernel, giving infinitely many solutions. Choice A is wrong because nullity > 0 means solutions aren't unique. Choice C correctly notes solvability but the phrasing about affine subspaces is less direct than choice B. Choice D incorrectly suggests some equations are unsolvable.
7. Consider the linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ defined by the matrix $A = \begin{pmatrix} 1 & 2 & 1 \\ -1 & 0 & 1 \\ 0 & 2 & 2 \end{pmatrix}$. Which statement accurately describes $T$?
   Answer: $T$ is neither one-to-one nor onto because the columns of $A$ are linearly dependent. To determine if $T$ is one-to-one or onto, we analyze its standard matrix $A$. We can perform row reduction on $A$. Adding the first row to the second gives $\begin{pmatrix} 1 & 2 & 1 \\ 0 & 2 & 2 \\ 0 & 2 & 2 \end{pmatrix}$. Subtracting the second row from the third gives $\begin{pmatrix} 1 & 2 & 1 \\ 0 & 2 & 2 \\ 0 & 0 & 0 \end{pmatrix}$. This matrix has only two pivots, so its rank is 2. For a $3 \times 3$ matrix, a rank of 3 is required for the transformation to be one-to-one or onto. Since the rank is less than 3, the columns of $A$ are linearly dependent, the null space is non-trivial, and the columns do not span $\mathbb{R}^3$. Therefore, $T$ is neither one-to-one nor onto.
8. Consider the linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^3$ defined by $T(x_1, x_2, x_3, x_4) = (x_1 + 2x_2 - x_3, 2x_1 + x_2 + x_4, x_1 - x_2 + x_3 + 2x_4)$. If $T$ maps the vector $(a, b, c, d)$ to $(1, 3, 2)$, what can be concluded about the uniqueness of this vector?
   Answer: The vector $(a, b, c, d)$ is not unique because the null space of $T$ is nontrivial Since T maps from R⁴ to R³, and dim(domain) = 4 > 3 = dim(codomain), T cannot be one-to-one by the rank-nullity theorem. The null space has dimension at least 4-3=1, so it's nontrivial. This means if T(v) = (1,3,2) for some vector v, then T(v + w) = (1,3,2) for any w in the null space, so the preimage is not unique. Choice A incorrectly assumes larger domain implies one-to-one. Choice C confuses rank with one-to-one property. Choice D mentions onto property which is irrelevant to uniqueness.
9. Consider a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ with matrix $A$. If $\text{rank}(A) = r$, which condition is both necessary and sufficient for $T$ to be one-to-one?
   Answer: $r = n$ and $\text{nullity}(A) = 0$ T is one-to-one if and only if ker(T) = {0}, which means nullity(A) = 0. By the rank-nullity theorem, rank(A) + nullity(A) = n, so nullity(A) = 0 if and only if rank(A) = n. Both conditions in choice B are equivalent and necessary/sufficient. Choice A gives conditions for onto (r = m), not one-to-one. Choice C describes when A has maximal possible rank but doesn't specifically characterize one-to-one. Choice D mixes onto conditions (r = m) with one-to-one conditions incorrectly.
10. Consider the linear transformation $T: \mathbb{R}^4 \to \mathbb{R}^3$ whose matrix has rank 2. If we know that $T(\mathbf{v}_1) = T(\mathbf{v}_2)$ for two distinct vectors $\mathbf{v}_1, \mathbf{v}_2 \in \mathbb{R}^4$, what additional information can be determined?
    Answer: Both properties from choices A and B hold since $\text{rank}(T) = 2$ Since T(v₁) = T(v₂) with v₁ ≠ v₂, we know T is not one-to-one. With rank(T) = 2, by rank-nullity theorem: dim(ker(T)) = 4 - 2 = 2, confirming choice A. Since rank(T) = 2 < 3, T is not onto R³, and the range of T is a 2-dimensional subspace, confirming choice B. Choice C correctly identifies that both properties hold. Choice D is incorrect because the given condition T(v₁) = T(v₂) already proves T is not one-to-one.
11. Let $S: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $S\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x + 2y - z \\ 2x + y + z \\ 3x + 3y \end{bmatrix}$. To determine if $S$ is onto, which approach provides the most direct verification?
    Answer: Show that the columns of the matrix span $\mathbb{R}^3$ by computing their determinant For a linear transformation from R³ to R³, being onto is equivalent to the matrix having full rank (rank 3), which occurs when the column vectors span R³. The most direct check is computing the determinant: if det ≠ 0, the columns span R³. Choice B tests one-to-one property, not onto. Choice C is correct in principle but less direct than the determinant test. Choice D mentions cross products which don't directly test linear independence in this context, and linear independence of columns is equivalent to the determinant test anyway.
12. Let $T: \mathbb{R}^3 \to \mathbb{R}^4$ be a linear transformation whose standard matrix is $A$. If the columns of $A$ are linearly independent, which of the following is true?
    Answer: $T$ is one-to-one but not onto. The transformation is from $\mathbb{R}^3$ to $\mathbb{R}^4$, so its standard matrix $A$ is $4 \times 3$. The condition that the columns of $A$ are linearly independent is the definition of $T$ being a one-to-one transformation. The range of $T$ is the span of its columns. Since there are 3 columns, their span forms a subspace of $\mathbb{R}^4$ with dimension 3. For $T$ to be onto, its range must be all of $\mathbb{R}^4$, which has dimension 4. Since the dimension of the range (3) is less than the dimension of the codomain (4), $T$ is not onto.
13. Let $T: \mathbb{R}^4 \to \mathbb{R}^6$ be a linear transformation. If $T$ is one-to-one, what is the dimension of the range of $T$?
    Answer: 4 The dimension of the range of $T$ is equal to the rank of its standard matrix $A$. The matrix $A$ has dimensions $6 \times 4$. The transformation $T$ is one-to-one, which means that the columns of $A$ are linearly independent. The rank of a matrix is the number of linearly independent columns (which is also the number of pivot positions). Since $A$ has 4 columns and they are all linearly independent, the rank of $A$ must be 4. Therefore, the dimension of the range of $T$ is 4.
14. Let $A$ be an $n \times n$ matrix. All of the following are equivalent to the statement that the transformation $T(\vec{x}) = A\vec{x}$ is one-to-one, EXCEPT:
    Answer: The columns of $A$ form an orthonormal set. For a square matrix $A$, the Invertible Matrix Theorem provides many equivalent conditions for invertibility. A transformation $T(\vec{x}) = A\vec{x}$ on $\mathbb{R}^n$ is one-to-one if and only if $A$ is invertible. Choice (A) says $T$ is onto, which is equivalent to one-to-one for a square matrix. Choice (B) says $A$ is row equivalent to $I_n$, which is equivalent to $A$ being invertible. Choice (C) is the definition of a one-to-one transformation. Choice (D) states the columns are orthonormal. An orthonormal set of vectors is always linearly independent, which implies $A$ is invertible and $T$ is one-to-one. However, the converse is not true; a matrix can be invertible (and thus define a one-to-one transformation) without having orthonormal columns (e.g., $A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$). Thus, having orthonormal columns is a sufficient but not necessary condition, so it is not equivalent.
15. Let $T: \mathbb{R}^n \to \mathbb{R}^n$ be a linear transformation. If $T$ is one-to-one, which of the following must also be true?
    Answer: $T$ is onto, and $T^{-1}$ exists as a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^n$ For a linear transformation T: Rⁿ → Rⁿ, if T is one-to-one, then rank(T) = n (since nullity = 0). Because the codomain also has dimension n, rank(T) = n means T is onto as well. Therefore T is bijective and has an inverse. Choice B is wrong - T need not preserve angles/lengths (e.g., scaling by 2). Choice C is wrong because when domain and codomain have the same finite dimension, one-to-one implies onto. Choice D is wrong for the same reason - T must be onto when one-to-one from Rⁿ to Rⁿ.
16. A linear transformation $S: \mathbb{R}^5 \to \mathbb{R}^5$ has a null space with dimension 2. Which statement correctly describes the properties of $S$?
    Answer: $S$ is neither one-to-one nor onto. A transformation is one-to-one if and only if its null space (kernel) is the zero vector, meaning its dimension is 0. Since the null space of $S$ has dimension 2, $S$ is not one-to-one. By the Rank-Nullity Theorem, for a transformation from $\mathbb{R}^n$, we have $\text{rank}(S) + \text{nullity}(S) = n$. Here, $n=5$ and $\text{nullity}(S) = 2$, so $\text{rank}(S) = 5 - 2 = 3$. A transformation is onto if its rank equals the dimension of the codomain. Here, the codomain is $\mathbb{R}^5$, so a rank of 5 is required. Since the rank is 3, $S$ is not onto.
17. Let $T: \mathbb{R}^7 \to \mathbb{R}^4$ be a linear transformation. Which of the following statements must be true?
    Answer: $T$ cannot be one-to-one. The standard matrix $A$ for the transformation $T$ has dimensions $4 \times 7$. For $T$ to be one-to-one, the columns of $A$ must be linearly independent. Since there are 7 columns in $\mathbb{R}^4$, and any set of more than 4 vectors in $\mathbb{R}^4$ must be linearly dependent, the columns of $A$ cannot be linearly independent. Therefore, $T$ cannot be one-to-one. It is possible for $T$ to be onto (if the rank is 4) or not onto (if the rank is less than 4), so options A and B are not necessarily true.
18. A linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is known to be onto. Which of the following statements MUST be true?
    Answer: The dimension of the domain is greater than or equal to the dimension of the codomain ($n \ge m$). If a transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is onto, its range is the entire codomain $\mathbb{R}^m$. This means the columns of its $m \times n$ standard matrix must span $\mathbb{R}^m$. To span $\mathbb{R}^m$, there must be a pivot in every one of the $m$ rows. A matrix cannot have more pivots than it has columns. Therefore, the number of columns, $n$, must be at least the number of rows, $m$. So, $n \ge m$. Choices C and D are conditions for being one-to-one, not onto.
19. Let $T_A: \mathbb{R}^3 \to \mathbb{R}^2$ and $T_B: \mathbb{R}^2 \to \mathbb{R}^3$ be linear transformations with standard matrices $A$ and $B$ respectively. Consider the composite transformation $T = T_B \circ T_A$. What can be concluded about $T$?
    Answer: $T$ is neither one-to-one nor onto. The transformation $T$ maps from $\mathbb{R}^3$ to $\mathbb{R}^3$. Its standard matrix is the product $C = BA$. Matrix $A$ is $2 \times 3$, and matrix $B$ is $3 \times 2$, so $C$ is a $3 \times 3$ matrix. The rank of a product of matrices is less than or equal to the minimum of their ranks. The rank of $A$ is at most 2 (since it has only 2 rows). The rank of $B$ is at most 2 (since it has only 2 columns). Therefore, $\text{rank}(C) = \text{rank}(BA) \le \min(\text{rank}(B), \text{rank}(A)) \le 2$. For the transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ to be one-to-one or onto, its $3 \times 3$ matrix $C$ must have rank 3. Since the rank of $C$ is at most 2, $T$ can be neither one-to-one nor onto.
20. Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation that reflects vectors across the plane $x + y + z = 0$. Which property does $T$ possess?
    Answer: $T$ is both one-to-one and onto because reflection is an invertible geometric transformation Reflection across a plane through the origin is an invertible linear transformation (you can reflect again to get back the original vector). Since T: R³ → R³ is invertible, it must be both one-to-one and onto. The reflection matrix has determinant -1 ≠ 0, confirming it's invertible. Choice A is wrong because reflections preserve magnitudes. Choice B is wrong because only the zero vector maps to zero in linear transformations. Choice C is wrong because reflection matrices are non-singular.

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