1. Theoretical Concept
Diagonalization is the process of finding a coordinate system (a basis of eigenvectors) in which a linear transformation acts simply as scaling along coordinate axes. In this basis, the matrix representing the transformation is **diagonal**.
If \(P\) is the matrix whose columns are the linearly independent eigenvectors of \(A\), then: \[P^{-1} A P = D = \begin{pmatrix} \lambda_1 & 0 & \dots & 0 \\ 0 & \lambda_2 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & \lambda_n \end{pmatrix}\] Where \(\lambda_i\) are the eigenvalues corresponding to the eigenvectors in \(P\).
2. Algebraic vs Geometric Multiplicity
To audit whether a matrix can be diagonalized, we calculate two forms of multiplicity for each eigenvalue \(\lambda\):
- Algebraic Multiplicity (AM): The number of times the root \(\lambda\) appears in the characteristic equation \(\det(A - \lambda I) = 0\).
- Geometric Multiplicity (GM): The number of linearly independent eigenvectors associated with \(\lambda\). This is equal to the dimension of the eigenspace \(E_\lambda = \text{nullspace}(A - \lambda I)\): \[\text{GM} = n - \text{rank}(A - \lambda I)\]
Theorem: A matrix is diagonalizable if and only if \(\text{AM} = \text{GM}\) for every eigenvalue.
3. Quick Tests for Diagonalizability
- Distinct Eigenvalues: If an \(n \times n\) matrix has \(n\) distinct eigenvalues, it is guaranteed to be diagonalizable (since each eigenvalue yields at least one eigenvector, summing to \(n\) independent eigenvectors).
- Symmetric Matrices: Any real symmetric matrix (\(A = A^T\)) is orthogonally diagonalizable (spectral theorem).
4. Solved GATE & CSIR-NET Questions
Question 1 (GATE Mathematics)
Determine if the following matrix \(A\) is diagonalizable: \[A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}\]
Detailed Solution:
- Find eigenvalues:
Since \(A\) is upper-triangular, its eigenvalues are the diagonal entries: \(\lambda = 2, 2\).
Thus, the eigenvalue \(\lambda = 2\) has **Algebraic Multiplicity (AM) = 2**. - Compute the Geometric Multiplicity (GM) for \(\lambda = 2\):
We find the nullspace of \((A - 2I)\): \[A - 2I = \begin{pmatrix} 2-2 & 1 \\ 0 & 2-2 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\] - Calculate Rank and Nullity:
• The rank of \((A - 2I)\) is 1 (one non-zero row).
• Nullity (GM) = \(n - \text{rank} = 2 - 1 = 1\). - Compare AM and GM:
\(\text{AM} = 2\), but \(\text{GM} = 1\).
Since \(\text{AM} \ne \text{GM}\), the matrix is **not diagonalizable**.
Answer: Non-diagonalizable (it is a Jordan block of size 2).
Question 2 (GATE Mathematics)
Let \(A\) be a \(3 \times 3\) matrix with eigenvalues \(1, -1, 0\). Is \(A\) diagonalizable? Compute \(A^3\).
Detailed Solution:
- Test diagonalizability:
The matrix size is \(3 \times 3\). The eigenvalues are \(1, -1, 0\) (all 3 are distinct).
Since it has 3 distinct eigenvalues, it is **guaranteed to be diagonalizable**. - Calculate \(A^3\) using eigenvalues:
By diagonalization, \(A = PDP^{-1}\), which implies: \[A^3 = P D^3 P^{-1}\] Where the diagonal matrix \(D = \text{diag}(1, -1, 0)\). - Compute \(D^3\): \[D^3 = \begin{pmatrix} 1^3 & 0 & 0 \\ 0 & (-1)^3 & 0 \\ 0 & 0 & 0^3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} = D\]
- Substitute back:
Since \(D^3 = D\), we have \(A^3 = P D^3 P^{-1} = P D P^{-1} = A\).
Answer: Yes, A is diagonalizable, and \(A^3 = A\).
Abdul Raheem K
Mathematics Head | MSc (IIT Madras) | BS (IIT Bombay) | GATE AIR-27Abdul Raheem K is an IIT Bombay and IIT Madras alumnus. He qualified CSIR JRF and secured an exceptional **GATE AIR 27** in Mathematics. He directs the Pure and Applied Mathematics curriculum at Benzil.