Linear Algebra Quiz: One To One And Onto

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One To One And OntoQuestion 1 of 20

For a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ , the condition $m \ge n$ is:

a necessary, but not sufficient, condition for

T

to be one-to-one.

a sufficient, but not necessary, condition for

T

to be one-to-one.

both a necessary and sufficient condition for

T

to be one-to-one.

a necessary, but not sufficient, condition for

T

to be onto.

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Linear Algebra Quiz

Linear Algebra Quiz: One To One And Onto

Practice One To One And Onto in Linear Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on One To One And Onto, giving you a quick way to practice the rules, question types, and explanations that matter most for Linear Algebra.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ , the condition $m \ge n$ is:

a necessary, but not sufficient, condition for $T$ to be one-to-one. (correct answer)
a sufficient, but not necessary, condition for $T$ to be one-to-one.
both a necessary and sufficient condition for $T$ to be one-to-one.
a necessary, but not sufficient, condition for $T$ to be onto.

Explanation: For

T

to be one-to-one, the

n

columns of its

m \times n

standard matrix must be linearly independent. This requires a pivot position in every column. To have

n

pivot positions, the matrix must have at least

n

rows, so

m \ge n

. Thus, the condition is necessary. However, it is not sufficient. For example, the transformation

T: \mathbb{R}^2 \to \mathbb{R}^3

given by the zero matrix satisfies

3 \ge 2

, but is not one-to-one. Therefore, the condition is necessary but not sufficient for

T

to be one-to-one.

Question 2

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:

The range of $T$ is $\mathbb{R}^n$ .
$A$ is row equivalent to the identity matrix $I_n$ .
The only solution to $A\vec{x} = \vec{0}$ is the trivial solution.
The columns of $A$ form an orthonormal set. (correct answer)

Explanation: For a square matrix

A

, the Invertible Matrix Theorem provides many equivalent conditions for invertibility. A transformation

T(\vec{x}) = A\vec{x}

\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.

Question 3

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 $T$ ?

$T$ is one-to-one because the equation $A\vec{x}=\vec{0}$ has only the trivial solution.
$T$ is onto because the columns of $A$ span $\mathbb{R}^3$ .
$T$ is both one-to-one and onto because $A$ is a square matrix.
$T$ is neither one-to-one nor onto because the columns of $A$ are linearly dependent. (correct answer)

Explanation: 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.

Question 4

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?

$k = -2$
$k = 1/2$
$k = 2$ (correct answer)
$k = -18$

Explanation: A linear transformation from

\mathbb{R}^2

\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

\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.

Question 5

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 $T$ ?

$T$ is one-to-one, but might not be onto.
$T$ could be both one-to-one and onto.
$T$ is onto, but might not be one-to-one.
$T$ is neither one-to-one nor onto. (correct answer)

Explanation: The transformation

T

maps from

\mathbb{R}^3

\mathbb{R}^3

. Its standard matrix is the product

C = BA

. Matrix

A

2 \times 3

, and matrix

B

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.

Question 6

Let $T: \mathbb{R}^7 \to \mathbb{R}^4$ be a linear transformation. Which of the following statements must be true?

$T$ must be onto.
$T$ cannot be onto.
$T$ must be one-to-one.
$T$ cannot be one-to-one. (correct answer)

Explanation: 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.

Question 7

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?

$T$ is one-to-one but not onto because $\text{rank}(A) = 1 < 3$
$T$ is onto but not one-to-one because $\text{rank}(A) = 1 < 2$
$T$ is neither one-to-one nor onto because $\text{rank}(A) = 1$ (correct answer)
$T$ is both one-to-one and onto because the matrix has full row rank

Explanation: 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.

Question 8

The linear transformation $F: \mathbb{R}^2 \to \mathbb{R}^4$ has matrix representation $B = \begin{bmatrix} 1 & 2 \\ 3 & 1 \\ 0 & 5 \\ 2 & 7 \end{bmatrix}$ . After row reducing $B$ to find its rank, what conclusion can be drawn about $F$ ?

$F$ is onto because the matrix has more rows than columns
$F$ cannot be one-to-one because the codomain dimension exceeds the domain dimension
$F$ is one-to-one if and only if $\text{rank}(B) = 4$
$F$ cannot be onto because $\text{rank}(B) \leq 2 < 4$ (correct answer)

Explanation: When analyzing linear transformations, the key relationships to remember are how the rank of the matrix connects to the "onto" and "one-to-one" properties of the transformation. For any linear transformation

F: \mathbb{R}^m \to \mathbb{R}^n

with matrix

A

, two crucial facts apply:

F

is onto if and only if

\text{rank}(A) = n

(the codomain dimension), and

F

is one-to-one if and only if

\text{rank}(A) = m

(the domain dimension). Here,

F: \mathbb{R}^2 \to \mathbb{R}^4

means we're going from 2-dimensional space to 4-dimensional space. Since

B

is a

4 \times 2

matrix, its rank cannot exceed 2 (the number of columns). For

F

to be onto, we'd need

\text{rank}(B) = 4

, but this is impossible since

\text{rank}(B) \leq 2 < 4

. Therefore,

F

cannot be onto, making answer D correct. Looking at the wrong answers: A incorrectly assumes that having more rows than columns guarantees the transformation is onto—this ignores rank entirely. B confuses the conditions: when the codomain dimension exceeds the domain dimension, the transformation cannot be onto (not "cannot be one-to-one"). C states

F

is one-to-one if

\text{rank}(B) = 4

, but since the domain is 2-dimensional, we'd need

\text{rank}(B) = 2

for one-to-one, and rank 4 is impossible anyway. Study tip: Always check rank against both domain and codomain dimensions. For onto: rank must equal codomain dimension. For one-to-one: rank must equal domain dimension.

Question 9

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?

$T$ is one-to-one but not onto because reflection changes vector magnitudes
$T$ is onto but not one-to-one because some vectors map to the zero vector
$T$ is neither one-to-one nor onto because the transformation matrix is singular
$T$ is both one-to-one and onto because reflection is an invertible geometric transformation (correct answer)

Explanation: 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.

Question 10

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 $\mathbf{b}$ ?

Every equation $S(\mathbf{x}) = \mathbf{b}$ has exactly one solution because $\text{rank}(A) = 3$
Every equation $S(\mathbf{x}) = \mathbf{b}$ has infinitely many solutions because $\text{nullity}(A) = 2$ (correct answer)
Every equation $S(\mathbf{x}) = \mathbf{b}$ is solvable, and solutions form a 2-dimensional affine subspace
Some equations $S(\mathbf{x}) = \mathbf{b}$ have no solution because the null space is nontrivial

Explanation: 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.

Question 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?

Show that the columns of the matrix span $\mathbb{R}^3$ by computing their determinant (correct answer)
Verify that the null space contains only the zero vector by row reduction
Check if the system $S(\mathbf{x}) = \mathbf{b}$ has a solution for arbitrary $\mathbf{b} \in \mathbb{R}^3$
Confirm that all three column vectors are linearly independent using cross products

Explanation: 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.

Question 12

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?

$T$ is not one-to-one, and the dimension of $\text{ker}(T)$ is 2
$T$ is not onto, and the range of $T$ is a 2-dimensional subspace of $\mathbb{R}^3$
Both properties from choices A and B hold since $\text{rank}(T) = 2$ (correct answer)
$T$ maps $\mathbb{R}^4$ onto a 2-dimensional subspace, but we cannot determine if it's one-to-one from the given information

Explanation: 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.

Question 13

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?

The reasoning is correct, but the conclusion is wrong because linearly dependent rows actually prevent one-to-one property
The reasoning is flawed because row dependence has no relationship to the one-to-one property of transformations
The student should examine column dependence, not row dependence, to determine if $T$ is one-to-one (correct answer)
The reasoning confuses one-to-one with onto properties, since row dependence affects the range, not the kernel

Explanation: 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.

Question 14

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?

The vector $(a, b, c, d)$ is unique because $T$ is one-to-one since $4 > 3$
The vector $(a, b, c, d)$ is not unique because the null space of $T$ is nontrivial (correct answer)
The vector $(a, b, c, d)$ is unique because the coefficient matrix has rank 3
The vector $(a, b, c, d)$ is not unique because $T$ cannot be onto when domain dimension exceeds codomain dimension

Explanation: 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.

Question 15

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 $S$ ?

$S$ is one-to-one but not onto.
$S$ is onto but not one-to-one.
$S$ is neither one-to-one nor onto. (correct answer)
$S$ is both one-to-one and onto.

Explanation: 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.

Question 16

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?

$T$ is one-to-one but not onto. (correct answer)
$T$ is onto but not one-to-one.
$T$ is both one-to-one and onto.
$T$ is neither one-to-one nor onto.

Explanation: The transformation is from

\mathbb{R}^3

\mathbb{R}^4

, so its standard matrix

A

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.

Question 17

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?

$T$ is one-to-one but not onto.
$T$ is onto but not one-to-one.
$T$ is neither one-to-one nor onto. (correct answer)
$T$ is both one-to-one and onto.

Explanation: 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.

Question 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?

The dimension of the domain is less than or equal to the dimension of the codomain ( $n \le m$ ).
The dimension of the domain is greater than or equal to the dimension of the codomain ( $n \ge m$ ). (correct answer)
The columns of the standard matrix for $T$ are linearly independent.
The null space of $T$ is the zero vector.

Explanation: 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.

Question 19

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 $T$ ?

2
4 (correct answer)
6
Cannot be determined from the given information.

Explanation: 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.

Question 20

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?

$T_A$ can be onto.
$T_B$ can be one-to-one.
$T_A$ must be one-to-one.
$T_B$ cannot be one-to-one. (correct answer)

Explanation: 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.