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 ____.
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 ____.
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For diagonalizability, each repeated eigenvalue must have enough linearly independent eigenvectors. Here \(\lambda=1\) is repeated, so its eigenspace must have dimension \(2\).
Updated On: Jul 4, 2026
- \(a=bc\)
- \(b=ac\)
- \(c=ab\)
- \(a=b=c\)
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The Correct Option is B
Solution and Explanation
Concept:
A matrix is diagonalizable if the total number of linearly independent eigenvectors is equal to the order of the matrix. Here \(P\) is a \(3\times 3\) upper triangular matrix. For an upper triangular matrix, eigenvalues are the diagonal entries.
Step 1: Find eigenvalues of \(P\).
The diagonal entries of \(P\) are: \[ 1,\ 2,\ 1 \] So the eigenvalues are: \[ \lambda=1,\ 1,\ 2 \] Thus: \[ \lambda=1 \] has algebraic multiplicity \(2\), and \[ \lambda=2 \] has algebraic multiplicity \(1\).
Step 2: Condition for diagonalizability.
Since \(\lambda=2\) has multiplicity \(1\), it will give one eigenvector. For diagonalizability, eigenvalue \(\lambda=1\) must give two linearly independent eigenvectors. So we need: \[ \dim E_1=2 \] where \(E_1\) is the eigenspace corresponding to \(\lambda=1\).
Step 3: Find eigenspace for \(\lambda=1\).
We solve: \[ (P-I)X=0 \] Now: \[ P-I= \begin{pmatrix} 1 & a & b\\ 0 & 2 & c\\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \] \[ P-I= \begin{pmatrix} 0 & a & b\\ 0 & 1 & c\\ 0 & 0 & 0 \end{pmatrix} \] Let: \[ X= \begin{pmatrix} x y z \end{pmatrix} \] Then: \[ (P-I)X=0 \] gives: \[ ay+bz=0 \] and: \[ y+cz=0 \]
Step 4: Solve the equations.
From: \[ y+cz=0 \] we get: \[ y=-cz \] Substitute this in: \[ ay+bz=0 \] \[ a(-cz)+bz=0 \] \[ -acz+bz=0 \] \[ (b-ac)z=0 \]
Step 5: Need two independent eigenvectors.
The variable \(x\) is already free. For the eigenspace corresponding to \(\lambda=1\) to have dimension \(2\), \(z\) must also be free. So we need: \[ b-ac=0 \] \[ b=ac \]
Step 6: Final conclusion.
Therefore, \(P\) is diagonalizable iff: \[ b=ac \] \[ \therefore \text{Correct Answer is (B)} \]
A matrix is diagonalizable if the total number of linearly independent eigenvectors is equal to the order of the matrix. Here \(P\) is a \(3\times 3\) upper triangular matrix. For an upper triangular matrix, eigenvalues are the diagonal entries.
Step 1: Find eigenvalues of \(P\).
The diagonal entries of \(P\) are: \[ 1,\ 2,\ 1 \] So the eigenvalues are: \[ \lambda=1,\ 1,\ 2 \] Thus: \[ \lambda=1 \] has algebraic multiplicity \(2\), and \[ \lambda=2 \] has algebraic multiplicity \(1\).
Step 2: Condition for diagonalizability.
Since \(\lambda=2\) has multiplicity \(1\), it will give one eigenvector. For diagonalizability, eigenvalue \(\lambda=1\) must give two linearly independent eigenvectors. So we need: \[ \dim E_1=2 \] where \(E_1\) is the eigenspace corresponding to \(\lambda=1\).
Step 3: Find eigenspace for \(\lambda=1\).
We solve: \[ (P-I)X=0 \] Now: \[ P-I= \begin{pmatrix} 1 & a & b\\ 0 & 2 & c\\ 0 & 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix} \] \[ P-I= \begin{pmatrix} 0 & a & b\\ 0 & 1 & c\\ 0 & 0 & 0 \end{pmatrix} \] Let: \[ X= \begin{pmatrix} x y z \end{pmatrix} \] Then: \[ (P-I)X=0 \] gives: \[ ay+bz=0 \] and: \[ y+cz=0 \]
Step 4: Solve the equations.
From: \[ y+cz=0 \] we get: \[ y=-cz \] Substitute this in: \[ ay+bz=0 \] \[ a(-cz)+bz=0 \] \[ -acz+bz=0 \] \[ (b-ac)z=0 \]
Step 5: Need two independent eigenvectors.
The variable \(x\) is already free. For the eigenspace corresponding to \(\lambda=1\) to have dimension \(2\), \(z\) must also be free. So we need: \[ b-ac=0 \] \[ b=ac \]
Step 6: Final conclusion.
Therefore, \(P\) is diagonalizable iff: \[ b=ac \] \[ \therefore \text{Correct Answer is (B)} \]
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