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A matrix is said to be diagonalizable if it can be expressed as the product of three matrices. Specifically,

• An invertible matrix \( P \).
• A diagonal matrix \( D \).
• The inverse of \( P \), denoted as \( P^{-1} \).

In simpler terms, a diagonalizable matrix \( A \) allows us to rewrite it in the form \( A = PDP^{-1} \). Here, \( P \) contains the eigenvectors of \( A \) as columns, and \( D \) contains the eigenvalues of \( A \) along its diagonal.
Diagonalization is useful because it simplifies many matrix operations. For example, raising a matrix to a power becomes easier when it's diagonal. For those matrices like \( A \) in the exercise, knowing it's diagonalizable helps simplify finding its inverse too.

Eigenvalues are special numbers associated with a matrix that describe how the matrix stretches or shrinks vectors. Given a matrix \( A \), an eigenvalue \( \lambda \) is a scalar such that there exists a non-zero vector \( v \) (called an eigenvector) satisfying:
\( Av = \lambda v \).
In the given exercise, the eigenvalues of the matrix \( A \) are all either 1 or \(-1\). This specific condition plays a key role in simplifying the matrix operations. Because if eigenvalues are 1 or \(-1\), the inverse of the matrix \( D \), with these values on its diagonal, equals \( D \) itself. That's due to:

• The inverse of 1 is 1.
• The inverse of \(-1\) is \(-1\).

This direct relationship between eigenvalues and their inverses simplifies calculations significantly, proving essential in demonstrating that \( A^{-1} = A \).

Matrix decomposition involves breaking down a matrix into simpler components, making complex matrix operations easier to handle. In linear algebra, this form of decomposition in the context of diagonalizable matrices is expressed through:

• Diagonal matrix representation.
• Utilization of eigenvectors and eigenvalues.

For the matrix \( A \), it uses the decomposition \( A = PDP^{-1} \). This breaks \( A \) into the matrices \( P \), \( D \), and \( P^{-1} \).
This method is crucial, especially when dealing with powers or inverses of matrices. For \( A \), because \( D^2 = D \), it simplifies operations greatly. Each operation on \( A \) can be transferred into working with \( P \) and \( D \). It's almost like peeling away layers of complexity, making it lean towards confirming that \( A = A^{-1} \).

An inverse matrix, denoted as \( A^{-1} \), is a matrix that, when multiplied by the original matrix \( A \), yields the identity matrix \( I \). This identity matrix is like the number 1 for regular multiplication—it's the neutral element of matrix multiplication. Thus, the equation falls like this:

• \( AA^{-1} = I \).
• \( A^{-1}A = I \).

For the matrix in the exercise, we show that \( A^{-1} = A \). This means \( A \) is its own inverse. Often, only special matrices have this property, such as orthogonal matrices where \( A^T = A^{-1} \), or those with specific eigenvalues, like the ones we explore here, having all eigenvalues either 1 or \(-1\).
To find an inverse using matrix decomposition, knowing that \( D^2 = D \) helps. It means the diagonal matrix \( D \) retains its form and properties even when inversed. Therefore, with \( A = PDP^{-1} \), this special property confirms that \( A \) itself acts as its own inverse, thus proving \( A^{-1} = A \).
Understanding these properties not only solves the problem but opens doors to recognizing similar patterns in other matrix problems.