Problem 7
Question
If \(a x=1, x\) is a right inverse of \(a\); if \(y a=1, y\) is a left inverse of \(a\). Prove that if \(a\) has a right inverse \(x\) and a left inverse \(y\), then \(a\) is invertible, and its inverse is equal to \(x\) and to \(y\). (First show that \(y a x a=1\).)
Step-by-Step Solution
Verified Answer
The conditions \(ax = 1\) and \(ya = 1\) imply \(a\) is invertible with inverse \(x = y = a^{-1}\).
1Step 1: Understanding Given Conditions
We are given that for a matrix or element \(a\), there exists a right inverse \(x\) such that \(ax = 1\), and a left inverse \(y\) such that \(ya = 1\). We need to prove that these conditions imply \(a\) is invertible and both \(x\) and \(y\) serve as the inverse of \(a\).
2Step 2: Express Compound Identity
Start by considering the expression \(yaxa\). We want to show \(yaxa = 1\). Begin by substituting the given inverse conditions."
3Step 3: Simplify Using Right and Left Inverse
Substitute \(ya = 1\), and replace \(ax\) with \(1\), thus the expression becomes \((ya)x(ax) = (1)x(1) = x\). Simplify further, \(x\).
4Step 4: Prove Compound Identity Equals 1
Next, consider \(yax = (ya)x = 1 \, \text{using } \,(ya = 1) \). Ensure \((yax)\) is equal to \(1\), hence \( yax = 1.\)
5Step 5: Show Equivalence of Inverses
Since \(yax = 1\) and \(xa = 1\), it demonstrates that right inverse \(x\) and left inverse \(y\) satisfy complete inverse properties for both sides. Thus, \(x = y = a^{-1}\).
6Step 6: Conclusion that \(a\) is Invertible
We have shown \(a\) is invertible, with its inverse being \(x=y\), because both achieve identity when multiplied by \(a\) from left or right.
Key Concepts
Right InverseLeft InverseMatrix InversesInverse Properties
Right Inverse
A right inverse in linear algebra refers to an element or matrix that, when multiplied by another matrix on the right, results in the identity matrix. Specifically, if you have a matrix or element \( a \) and a matrix \( x \) such that \( ax = 1 \), then \( x \) is called the right inverse of \( a \). This means multiplying \( a \) by \( x \) results in the identity element, which is 1 for numbers or the identity matrix for matrices.
Finding a right inverse is particularly useful when you want to verify if an operation or transformation can be "undone" from a single side. However, a right inverse does not guarantee a left inverse, meaning you cannot assume \( xa = 1 \) unless explicitly stated.
Finding a right inverse is particularly useful when you want to verify if an operation or transformation can be "undone" from a single side. However, a right inverse does not guarantee a left inverse, meaning you cannot assume \( xa = 1 \) unless explicitly stated.
- Right inverses are unique only if \( a \) is a square matrix with full row rank.
- Just having a right inverse does not confirm a matrix's invertibility unless supported by a left inverse too.
Left Inverse
A left inverse is similar to a right inverse but operates from the left side. For a matrix \( a \), if there exists a matrix or element \( y \) such that \( ya = 1 \), \( y \) is the left inverse of \( a \). This condition implies that when \( y \) is multiplied before \( a \), it results in the identity matrix.
The left inverse is important when considering transformations that need to be reversed from the beginning. For instance, you might have a function or transformation represented by the matrix \( a \). To "undo" or retrieve the original input, you need \( y \), ensuring that applying \( y \) before \( a \) cancels out the operation.
The left inverse is important when considering transformations that need to be reversed from the beginning. For instance, you might have a function or transformation represented by the matrix \( a \). To "undo" or retrieve the original input, you need \( y \), ensuring that applying \( y \) before \( a \) cancels out the operation.
- Left inverses exist when \( a \) is a square matrix with full column rank.
- Having a left inverse does not automatically mean \( a \) can also have a right inverse without additional conditions.
Matrix Inverses
In the realm of linear algebra, a matrix inverse is a matrix that, when multiplied with the original matrix, results in the identity matrix. A matrix \( a \) is said to have an inverse if there exists another matrix \( a^{-1} \) such that \( aa^{-1} = a^{-1}a = I \), where \( I \) is the identity matrix.
For a matrix to be invertible, it must be square, meaning it has the same number of rows and columns, and full rank, indicating there is no loss in dimension from input to output.
For a matrix to be invertible, it must be square, meaning it has the same number of rows and columns, and full rank, indicating there is no loss in dimension from input to output.
- The inverse is unique; if it exists, it is the same from both the left and the right side.
- Matrix inverses are a cornerstone in solving systems of linear equations, allowing for straightforward solutions using methods like matrix factorizations.
Inverse Properties
Inverse properties are fundamental mathematical rules that govern the behavior of matrices and numbers when considering their inverses. These properties ensure consistency and predictability in mathematical operations.
For any element or matrix, provided it is invertible, we can consider the following:
For any element or matrix, provided it is invertible, we can consider the following:
- Uniqueness: Each invertible matrix has a unique inverse.
- Associativity: When dealing with multiple matrices, the order of multiplication can be reconsidered. However, inverses themselves directly follow the standard associative property.
- Identity: The product of a matrix and its inverse results in the identity matrix: \( aa^{-1} = a^{-1}a = I \).
- Double Inversion: The inverse of the inverse returns the original matrix: \((a^{-1})^{-1} = a \).
Other exercises in this chapter
Problem 6
By part 5, the set of all the nonzero elements in a field is a multiplicative group. Now use Lagrange's theorem to prove that in a finite field with \(m\) eleme
View solution Problem 6
Let \(A\) be any ring. Prove that if the additive group of \(A\) is cyclic, then \(A\) is a commutative ring
View solution Problem 7
In any integral domain, if \(a^{n}=0\) for some integer \(n\), then \(a=0\).
View solution Problem 6
Prove that each of the following is true in a nontrivial ring with unity. \(A \times B\) always has divisors of zero.
View solution