Question

If \[ P= \begin{pmatrix} 1 & a & b\\ 0 & 2 & c\\ 0 & 0 & 1 \end{pmatrix}, \quad a,b,c\in\mathbb{Z}, \] then \(P\) is diagonalizable iff ____.

• A. \(a=bc\)
• B. \(b=ac\)
• C. \(c=ab\)
• D. \(a=b=c\)

Hint:

For diagonalizability, each repeated eigenvalue must have enough linearly independent eigenvectors. Here \(\lambda=1\) is repeated, so its eigenspace must have dimension \(2\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
An \(n \times n\) matrix \(P\) is diagonalizable if and only if, for every eigenvalue \(\lambda\), its geometric multiplicity equals its algebraic multiplicity. 

Step 2: Key Formula or Approach
1. Find the eigenvalues using the characteristic equation \[ \det(P-\lambda I)=0. \] 2. Since \(P\) is upper triangular, the eigenvalues are its diagonal entries.
3. For the repeated eigenvalue, compute \[ \dim\bigl(\ker(P-\lambda I)\bigr) = n-\operatorname{rank}(P-\lambda I). \] 4. The matrix is diagonalizable only if the geometric multiplicity equals the algebraic multiplicity. 

Step 3: Detailed Explanation
Given \[ P= \begin{pmatrix} 1 & amp; a & amp; b\\ 0 & amp; 2 & amp; c\\ 0 & amp; 0 & amp; 1 \end{pmatrix}. \] 1. Find the Eigenvalues
Since \(P\) is upper triangular, its eigenvalues are the diagonal entries: \[ \lambda=1,\;2,\;1. \] Thus, - Eigenvalue \(1\) has algebraic multiplicity \(2\). - Eigenvalue \(2\) has algebraic multiplicity \(1\). The eigenvalue \(2\) automatically has geometric multiplicity \(1\). Hence, we only need to check the repeated eigenvalue \(\lambda=1\). 
2. Compute \(P-I\)
\[ P-I= \begin{pmatrix} 0 & amp; a & amp; b\\ 0 & amp; 1 & amp; c\\ 0 & amp; 0 & amp; 0 \end{pmatrix}. \] Since the algebraic multiplicity of \(\lambda=1\) is \(2\), its geometric multiplicity must also be \(2\). Using \[ \dim\bigl(\ker(P-I)\bigr) = 3-\operatorname{rank}(P-I), \] we require \[ 3-\operatorname{rank}(P-I)=2, \] which gives \[ \operatorname{rank}(P-I)=1. \] 
3. Rank Condition
The first two rows of \(P-I\) are \[ (0,a,b) \quad\text{and}\quad (0,1,c). \] For the rank to be \(1\), these two rows must be linearly dependent. Hence, \[ (0,a,b) = a(0,1,c) = (0,a,ac). \] Comparing the third components, \[ b=ac. \] Therefore, the necessary and sufficient condition for diagonalizability is \[ b=ac. \] 

Step 4: Conclusion
The matrix \(P\) is diagonalizable if and only if \[ b=ac. \] 

Final Answer: (B)