If the eigenvalue \(\lambda\) of a matrix \(M\) has algebraic multiplicity 3 and geometric multiplicity 2, then which one of the following options is true?
Show Hint
- \(M\) is diagonalizable
- The characteristic polynomial of \(M\) has degree 3
- The eigenspace associated with \(\lambda\) has dimension 3
- \(M\) is not diagonalizable
The Correct Option is D
Solution and Explanation
Step 1: For any eigenvalue \(\lambda\) of a matrix \(M\), the geometric multiplicity (the number of linearly independent eigenvectors for \(\lambda\), i.e. the dimension of its eigenspace) is always less than or equal to the algebraic multiplicity (the multiplicity of \(\lambda\) as a root of the characteristic polynomial).
Step 2: Here the algebraic multiplicity is 3 and the geometric multiplicity is 2. Since \(2 < 3\), the geometric multiplicity is strictly less than the algebraic multiplicity for this eigenvalue.
Step 3: A matrix is diagonalizable if and only if, for every eigenvalue, the geometric multiplicity equals the algebraic multiplicity. Since this equality fails for \(\lambda\), \(M\) cannot be diagonalized.
Step 4: The characteristic polynomial's degree equals the size of \(M\), which must be at least 3 but need not be exactly 3, so option (B) is not forced. Also the eigenspace for \(\lambda\) has dimension equal to the geometric multiplicity, which is 2, not 3, so option (C) is false.
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