Question

Let \( M \) be a \( 7 \times 7 \) matrix with entries in \( \mathbb{R} \) and having the characteristic polynomial \[ c_M(x) = (x - 1)^\alpha (x - 2)^\beta (x - 3)^2, \] where \( \alpha>\beta \). Let \( {rank}(M - I_7) = {rank}(M - 2I_7) = {rank}(M - 3I_7) = 5 \), where \( I_7 \) is the \( 7 \times 7 \) identity matrix. 
If \( m_M(x) \) is the minimal polynomial of \( M \), then \( m_M(5) \) is equal to __________ (in integer).

Hint:

For the minimal polynomial, consider the eigenvalues and their multiplicities. The minimal polynomial has each eigenvalue appearing only once. Evaluate it at the desired value to find the answer.

Answer 1

We are given that the characteristic polynomial of a 7×7 matrix M is:

c_M(x) = (x − 1)^α(x − 2)^β(x − 3)²

along with the conditions:

rank(M − I₇) = rank(M − 2I₇) = rank(M − 3I₇) = 5

We are asked to find m_M(5), where m_M(x) is the minimal polynomial.

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Step 1: Use rank–nullity theorem

For any matrix A of order 7:
nullity(A) = 7 − rank(A)

Hence,
nullity(M − λI) = 7 − 5 = 2

for λ = 1, 2, and 3. This means that the eigenspace corresponding to each eigenvalue 1, 2, and 3 has dimension 2.

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Step 2: Deduce algebraic multiplicities

The geometric multiplicity of an eigenvalue is always less than or equal to its algebraic multiplicity.

Since the eigenspace dimension for each eigenvalue is 2, the algebraic multiplicity of each must be at least 2.

But the matrix is of order 7, and the characteristic polynomial already contains (x − 3)². Therefore, the only possibility is:

c_M(x) = (x − 1)²(x − 2)²(x − 3)²

(the remaining degree accounts for no higher powers).

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Step 3: Determine the minimal polynomial

When the geometric multiplicity equals the algebraic multiplicity for an eigenvalue, no Jordan block larger than size 1 exists.

Thus, no squared factors are required in the minimal polynomial.

Hence,

m_M(x) = (x − 1)(x − 2)(x − 3)

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Step 4: Evaluate m_M(5)

m_M(5) = (5 − 1)(5 − 2)(5 − 3)

m_M(5) = 4 × 3 × 2

m_M(5) = 24

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Final Answer:

m_M(5) = 24