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Questions and Answers
If matrix $A$ is invertible and $B_1$ and $B_2$ are both inverses of $A$, which of the following steps demonstrates why the inverse of $A$ is unique?
If matrix $A$ is invertible and $B_1$ and $B_2$ are both inverses of $A$, which of the following steps demonstrates why the inverse of $A$ is unique?
- Using the property that $A(B_1 - B_2) = I_n$ to conclude $B_1 = B_2$.
- Demonstrating that $A(B_1 - B_2) = 0$ leads to $B_1 - B_2 = A^{-1}0 = 0$, thus $B_1 = B_2$. (correct)
- Asserting that since $A$ is invertible, any two matrices multiplied by $A$ must be equal.
- Showing that $AB_1 = AB_2$ implies $B_1 = B_2$ directly.
Given an $n \times n$ matrix $A$, which of the following conditions is sufficient to guarantee that $Ax = b$ has a unique solution for every vector $b$ in $R^n$?
Given an $n \times n$ matrix $A$, which of the following conditions is sufficient to guarantee that $Ax = b$ has a unique solution for every vector $b$ in $R^n$?
- The null space of $A$ contains more than just the zero vector.
- The matrix $A$ is invertible. (correct)
- The columns of $A$ are linearly dependent.
- The determinant of $A$ is zero.
For a matrix $A$, if the equation $Ax = 0$ has only the trivial solution, what can be concluded about the columns of $A$?
For a matrix $A$, if the equation $Ax = 0$ has only the trivial solution, what can be concluded about the columns of $A$?
- They are linearly independent. (correct)
- They span a subspace of dimension less than $n$.
- They span $R^n$ but are linearly dependent.
- They form a basis for a subspace of $R^n$ other than $R^n$ itself.
If a square matrix $A$ reduces to the identity matrix $I_n$, what does this imply about the properties of $A$?
If a square matrix $A$ reduces to the identity matrix $I_n$, what does this imply about the properties of $A$?
If the column space of a matrix $A$ is equal to $R^n$, what can be inferred about the rank of $A$?
If the column space of a matrix $A$ is equal to $R^n$, what can be inferred about the rank of $A$?
What is the relationship between the nullity of a matrix $A$ and the solutions to the equation $Ax = 0$?
What is the relationship between the nullity of a matrix $A$ and the solutions to the equation $Ax = 0$?
If the nullity of a matrix $A$ is 0, what does this imply about the solution to the equation $Ax = 0$?
If the nullity of a matrix $A$ is 0, what does this imply about the solution to the equation $Ax = 0$?
How is the image of a linear transformation $T_A$ related to the column space of the matrix $A$?
How is the image of a linear transformation $T_A$ related to the column space of the matrix $A$?
What does it mean for a linear transformation $T_A : R^n \rightarrow R^n$ to be surjective?
What does it mean for a linear transformation $T_A : R^n \rightarrow R^n$ to be surjective?
If a linear transformation $T_A$ is injective, what can be said about the solutions to the equation $Ax = b$ for any $b$ in $R^n$?
If a linear transformation $T_A$ is injective, what can be said about the solutions to the equation $Ax = b$ for any $b$ in $R^n$?
What condition must be met for a linear transformation $T_A$ to be considered bijective?
What condition must be met for a linear transformation $T_A$ to be considered bijective?
If $A$ is an $n \times n$ matrix and $\det(A) \neq 0$, how does this relate to the invertibility of $A$?
If $A$ is an $n \times n$ matrix and $\det(A) \neq 0$, how does this relate to the invertibility of $A$?
For a matrix $A$, if 0 is not an eigenvalue of $A$, what does this imply about the solutions to the equation $Ax = 0$?
For a matrix $A$, if 0 is not an eigenvalue of $A$, what does this imply about the solutions to the equation $Ax = 0$?
If the columns of matrix A form a basis for $R^n$, what does this imply about the solution to the equation $Ax = b$ for any vector $b$ in $R^n$?
If the columns of matrix A form a basis for $R^n$, what does this imply about the solution to the equation $Ax = b$ for any vector $b$ in $R^n$?
Suppose matrix $A$ is $n \times n$ and invertible. What is the dimension of the column space of $A$?
Suppose matrix $A$ is $n \times n$ and invertible. What is the dimension of the column space of $A$?
A function $T_A: R^n \rightarrow R^n$ is defined by $T_A(x) = Ax$. Which of the following is equivalent to $T_A$ being surjective?
A function $T_A: R^n \rightarrow R^n$ is defined by $T_A(x) = Ax$. Which of the following is equivalent to $T_A$ being surjective?
If a matrix $A$ is invertible, how is its null space (nul(A)) characterized?
If a matrix $A$ is invertible, how is its null space (nul(A)) characterized?
Given that $A$ is an $n \times n$ matrix, which statement is true if $Ax = b$ does not have a unique solution for every $b$ in $R^n$?
Given that $A$ is an $n \times n$ matrix, which statement is true if $Ax = b$ does not have a unique solution for every $b$ in $R^n$?
How are the concepts of injectivity and invertibility related for a linear transformation $T_A: R^n \rightarrow R^n$?
How are the concepts of injectivity and invertibility related for a linear transformation $T_A: R^n \rightarrow R^n$?
What can be concluded if $\det(A) = 0$ for a square matrix $A$?
What can be concluded if $\det(A) = 0$ for a square matrix $A$?
Flashcards
Invertible Matrix
Invertible Matrix
A matrix A is invertible if there exists a matrix A⁻¹ such that AA⁻¹ = A⁻¹A = Iₙ. The inverse is unique.
Unique Solution
Unique Solution
For every vector b in Rⁿ, the equation Ax = b has a unique solution.
Columns Span Rⁿ
Columns Span Rⁿ
The columns of matrix A span Rⁿ, meaning every vector in Rⁿ can be written as a linear combination of the columns of A.
Trivial Solution Only
Trivial Solution Only
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Linear Independence
Linear Independence
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Reduces to Iₙ
Reduces to Iₙ
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col(A) = Rⁿ
col(A) = Rⁿ
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nul(A) = {0}
nul(A) = {0}
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Basis for Rⁿ
Basis for Rⁿ
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rank(A) = n
rank(A) = n
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nullity(A) = 0
nullity(A) = 0
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im(TA) = Rⁿ
im(TA) = Rⁿ
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ker(TA) = {0}
ker(TA) = {0}
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TA is surjective
TA is surjective
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TA is injective
TA is injective
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TA is bijective
TA is bijective
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det(A) ≠ 0
det(A) ≠ 0
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0 not eigenvalue
0 not eigenvalue
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Study Notes
- For A ∈ R^{n×n}, and a linear transformation T_A: R^n → R^n with standard matrix A, the following statements are equivalent:
A is invertible
- A−1 is the unique inverse of A
- If A is invertible and B1, B2 are inverses of A, then AB1 = In = AB2
- AB1 − AB2 = A(B1 − B2) = In − In = 0, which means A−1A(B1 − B2) = B1 − B2 = A−10 = 0
- Thus B1 − B2 = 0, or B1 = B2, meaning the inverse of A is unique
Solutions to Ax = b
- Ax = b has a unique solution for all b ∈ R^n
- The solution is x = A−1b and is unique because A−1 is unique
Columns of A
- The columns of A span R^n
- The columns of A are linearly independent
- The columns of A form a basis for R^n
- A basis for subspace W is a set of linearly independent vectors that span W
Solutions to Ax = 0
- Ax = 0 has only the trivial solution
- This arises because A0 = 0, meaning it has to be a unique solution
Matrix Reduction
- A reduces to In
- A must have n pivot columns and rows for its columns to be linearly independent and a spanning set
Column Space
- col(A) = R^n
- col(A) is the span of the columns of A
Null Space
- nul(A) = 0
- nul(A) is the set of vectors x where Ax = 0
Rank and Nullity
- rank(A) = n, where rank is the dimension of the column space
- The column space is R^n , having dimension n
- nullity(A) = 0, where nullity is the dimension of the null space.
Image of Transformation
- im(TA) = R^n
- The image of a transformation is the set of possible outputs, or range
- The image of TA is col(A)
Kernel of Transformation
- ker(TA) = 0
- The kernel of a transformation is the set of inputs that map to 0
- The kernel of TA is nul(A)
Surjectivity and Injectivity
- TA is surjective if its range equals its codomain
- The codomain of TA is R^n, which is also its range
- TA is injective if each input maps to its own unique output
- For each x ∈ R^n, there is a unique b where Ax = b
Bijective Transformations
- TA is bijective if and only if it is both surjective and injective
Determinant of A
- det(A) ≠ 0
- The determinant of a matrix is a nonzero scalar multiple of its reduced form's determinant
- The determinant of a triangular matrix is the product of the entries on its diagonal
- A reduces to In, so det(A) must be a nonzero scalar multiple of 1
Eigenvalues
- 0 is not an eigenvalue of A
- If λ is an eigenvalue of A, there is a nonzero vector x such that Ax = λx
- If λ = 0, there is a nonzero vector x such that Ax = 0, which is impossible
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