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For a matrix to be diagonalizable, there must exist an invertible matrix P such that \(P^{-1}AP\) is a diagonal matrix. The eigenvalue \(\lambda\) of matrix A has multiplicity n, which means it has n linearly independent eigenvectors.

1. If A is diagonalizable, we can write \(A = PDP^{-1}\), where D is a diagonal matrix consisting of A's eigenvalues and P is the matrix of eigenvectors (which are linearly independent due to the multiplicity of λ being n). 2. Given the multiplicity, every entry in the diagonal matrix D should be equal to λ. 3. Now, consider the identity matrix I. 4. Multiply A by I and compare with λI: \[A = PDP^{-1}\] \[AI = PDP^{-1}I = PDP^{-1}\] 5. Since \(D = \lambda I\), the expression becomes: \[AI = P (\lambda I) P^{-1}\] 6. Comparing the left-hand side (AI) and right-hand side (λI), we can safely say that: \[A = λI\]

1. Given A = λI, we will determine if it is diagonalizable. 2. Recall that a matrix is diagonalizable if it can be written in the form \(A = PDP^{-1}\). 3. Since A = λI, D is just the diagonal matrix with λ along the diagonal entries. 4. If we choose the matrix P to be the identity matrix I, then \(P^{-1}\) is also the identity matrix I. 5. Therefore, we can write: \[A = λI = III^{-1} = PDP^{-1}\] So A is diagonalizable. In conclusion, we have proven the statement "A is diagonalizable if and only if A = λI". The analysis showed the importance of understanding the definitions of diagonalizable, eigenvalues, eigenvectors, and multiplicity. The solution demonstrated the steps needed to prove the statement in both directions (if and only if), providing a comprehensive proof.