Proof Of Matrix Identity (A⁻¹ + B⁻¹)⁻¹ = A(A + B)⁻¹B
This article delves into the fascinating realm of matrix algebra, providing a comprehensive proof for a notable identity involving matrix inverses. Specifically, we will demonstrate that given two matrices, A and B, of the same size, and assuming that the necessary inverses exist, the following equation holds true:
(A⁻¹ + B⁻¹)⁻¹ = A(A + B)⁻¹B
This identity is a valuable tool in various mathematical and engineering applications, and understanding its derivation is crucial for anyone working with matrices. We will walk through a step-by-step proof, highlighting the key properties of matrix inverses and matrix multiplication used in each stage. Before diving into the proof, let's briefly recap some fundamental concepts that will be instrumental in our discussion.
Prerequisites: Matrix Inverses and Properties
Before we embark on the proof, it's essential to have a firm grasp of matrix inverses and their fundamental properties. A matrix A is said to be invertible (or non-singular) if there exists a matrix A⁻¹, called the inverse of A, such that their product results in the identity matrix I:
A A⁻¹ = A⁻¹ A = I
Where I is the identity matrix of the same size as A. The identity matrix is a square matrix with ones on the main diagonal and zeros elsewhere. It acts as the multiplicative identity in matrix algebra, meaning that for any matrix A, we have AI = IA = A. Matrix inverses possess several crucial properties that we will leverage in our proof:
- (A⁻¹)⁻¹ = A: The inverse of the inverse of a matrix is the original matrix itself.
- (AB)⁻¹ = B⁻¹A⁻¹: The inverse of a product of matrices is the product of their inverses in reverse order. This property is crucial when dealing with products of invertible matrices.
- If A is invertible, then the equation AX = B has a unique solution given by X = A⁻¹B.
- The inverse of a matrix, if it exists, is unique. This means that there is only one matrix that satisfies the definition of the inverse.
These properties will serve as the building blocks for our proof. With these concepts in mind, we are now well-equipped to tackle the main identity.
Proof of (A⁻¹ + B⁻¹)⁻¹ = A(A + B)⁻¹B
Our goal is to demonstrate the equivalence of the two expressions (A⁻¹ + B⁻¹)⁻¹ and A(A + B)⁻¹B. To achieve this, we will start with one side of the equation and manipulate it using valid matrix operations and properties until we arrive at the other side. A common strategy in proving matrix identities is to work with one side and try to transform it into the other side. In this case, we will start with the right-hand side, A(A + B)⁻¹B, and show that it is equal to the left-hand side, (A⁻¹ + B⁻¹)⁻¹. This approach often involves strategic multiplication, distribution, and the application of inverse properties.
Let's consider the expression A(A + B)⁻¹B. We will try to manipulate this expression to get to (A⁻¹ + B⁻¹)⁻¹. To do this, we'll try to bring (A⁻¹ + B⁻¹) into the picture. We can consider multiplying by (A⁻¹ + B⁻¹) and its inverse strategically.
Let's multiply A(A + B)⁻¹B by (A⁻¹ + B⁻¹) on the left:
(A⁻¹ + B⁻¹)A(A + B)⁻¹B
Now, let's focus on simplifying the first part of this expression, specifically (A⁻¹ + B⁻¹)A. By using the distributive property of matrix multiplication, we can expand this as follows:
(A⁻¹ + B⁻¹)A = A⁻¹A + B⁻¹A
Since A⁻¹A = I (by the definition of the inverse), we have:
A⁻¹A + B⁻¹A = I + B⁻¹A
Substituting this back into our original expression, we get:
(A⁻¹ + B⁻¹)A(A + B)⁻¹B = (I + B⁻¹A)(A + B)⁻¹B
This form looks closer to our target, but we need to manipulate it further. Let's try to rewrite the expression (I + B⁻¹A) to make it more compatible with (A + B)⁻¹. To do this, we can try to find a common 'denominator' of sorts. Let's multiply the identity matrix I by B⁻¹B (which is equal to I) on the left:
I = B⁻¹B
So, we can rewrite (I + B⁻¹A) as:
I + B⁻¹A = B⁻¹B + B⁻¹A
Now, we can factor out B⁻¹ from both terms:
B⁻¹B + B⁻¹A = B⁻¹(B + A)
Substituting this back into our expression, we get:
(I + B⁻¹A)(A + B)⁻¹B = B⁻¹(B + A)(A + B)⁻¹B
Here, we encounter the term (B + A)(A + B)⁻¹. Because matrix addition is commutative (A + B = B + A), we can rewrite this as:
(B + A)(A + B)⁻¹ = (A + B)(A + B)⁻¹
By the definition of the inverse, (A + B)(A + B)⁻¹ = I (assuming (A + B) is invertible). Therefore, our expression simplifies to:
B⁻¹(B + A)(A + B)⁻¹B = B⁻¹IB = B⁻¹B
And B⁻¹B = I, so we have:
(A⁻¹ + B⁻¹)A(A + B)⁻¹B = I
Now, we know that when (A⁻¹ + B⁻¹) is multiplied on the left of A(A + B)⁻¹B, the result is the identity matrix. To complete the proof, we need to show that when A(A + B)⁻¹B is multiplied on the right of (A⁻¹ + B⁻¹), the result is also the identity matrix. This is because the definition of the inverse requires multiplication in both orders to yield the identity matrix. However, the algebra is neater if, instead of right-multiplying, we left-multiply by the inverse, which we can do because we already suspect what the inverse is:
(A⁻¹ + B⁻¹) A(A + B)⁻¹B = I.
Multiply by the inverse of (A⁻¹ + B⁻¹) on the left gives
(A(A + B)⁻¹B = (A⁻¹ + B⁻¹)^(-1)
Which is what we wanted to prove. Therefore, we have successfully demonstrated that:
(A⁻¹ + B⁻¹)⁻¹ = A(A + B)⁻¹B
Alternative Proof using Inverse Definition
Another way to approach this proof is to use the definition of the inverse directly. We want to show that (A⁻¹ + B⁻¹)⁻¹ = A(A + B)⁻¹B. This means we need to show that when we multiply (A⁻¹ + B⁻¹) by A(A + B)⁻¹B (in either order), we get the identity matrix I. Let's multiply (A⁻¹ + B⁻¹) on the left:
(A⁻¹ + B⁻¹) [A(A + B)⁻¹B]
Distribute the (A⁻¹ + B⁻¹) term:
= A⁻¹A(A + B)⁻¹B + B⁻¹A(A + B)⁻¹B
Since A⁻¹A = I, this simplifies to:
= I(A + B)⁻¹B + B⁻¹A(A + B)⁻¹B
= (A + B)⁻¹B + B⁻¹A(A + B)⁻¹B
Now, we need to manipulate this expression to see if it simplifies to I. Let's factor out the (A + B)⁻¹ term from the right:
= [(A + B)⁻¹ + B⁻¹A(A + B)⁻¹]B
This doesn't seem to lead directly to the identity matrix. Let's try a slightly different approach. Instead of distributing (A⁻¹ + B⁻¹) right away, let's try to find a common denominator within the parentheses:
(A⁻¹ + B⁻¹) = (B⁻¹B)A⁻¹ + (A⁻¹A)B⁻¹ = B⁻¹BA⁻¹ + A⁻¹AB⁻¹
This step is designed to allow us to work with a single fraction, which might simplify the multiplication process.
However, this also doesn't seem to lead to a direct simplification. Let's revert to our initial approach of multiplying (A⁻¹ + B⁻¹) [A(A + B)⁻¹B] and re-examine the expression we derived earlier:
(A + B)⁻¹B + B⁻¹A(A + B)⁻¹B
To further simplify, let's find a common denominator for the two terms by multiplying the first term by B⁻¹B on the left (which is equivalent to multiplying by the identity matrix):
= B⁻¹B(A + B)⁻¹B + B⁻¹A(A + B)⁻¹B
Now, we can factor out B⁻¹ on the left and (A + B)⁻¹B on the right:
= B⁻¹[B(A + B)⁻¹B + A(A + B)⁻¹B]
Now, factor out (A + B)⁻¹B from the right:
= B⁻¹[B + A](A + B)⁻¹B
Since (B + A) = (A + B), we have:
= B⁻¹(A + B)(A + B)⁻¹B
By the definition of the inverse, (A + B)(A + B)⁻¹ = I, so:
= B⁻¹IB
= B⁻¹B
= I
Thus, we have shown that (A⁻¹ + B⁻¹) [A(A + B)⁻¹B] = I. This confirms that A(A + B)⁻¹B is indeed the inverse of (A⁻¹ + B⁻¹). Therefore, we have successfully proven the identity.
Conclusion: Significance of the Identity
In conclusion, we have rigorously proven the matrix identity (A⁻¹ + B⁻¹)⁻¹ = A(A + B)⁻¹B using two different approaches. The first proof involved manipulating the expression A(A + B)⁻¹B to arrive at (A⁻¹ + B⁻¹)⁻¹, while the second proof directly utilized the definition of the inverse to show that the product of (A⁻¹ + B⁻¹) and A(A + B)⁻¹B yields the identity matrix. This identity is a valuable result in linear algebra, offering a way to express the inverse of a sum of inverses in terms of the original matrices and their sum. This identity finds applications in various fields, including circuit analysis, where it can simplify calculations involving parallel combinations of impedances, and in statistics, where it can be used in the context of covariance matrices. Understanding and being able to apply this identity is a testament to a strong foundation in matrix algebra and its applications.
