Diagonalization via Eigenvectors

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2017)
• Priority tier: T4
• Marks (count): 10 (1)
• Average solve time: ~15 min
• Difficulty mix: medium 1
• Section: A | Dominant type: computation

Why This Chapter Matters

Diagonalization translates a matrix into its simplest equivalent form — diagonal — making powers, exponentials, and differential equations trivial to compute. UPSC 2017 posed a 10-mark Section A question asking candidates to diagonalize a specific matrix or to show it cannot be diagonalized, testing both the eigenvector computation and the invertibility check on $P$. The topic also connects directly to the spectral theorem for symmetric matrices (a higher-frequency atom), so fluency here pays compound dividends.

Minimum Theory

Eigenvalues and eigenvectors: For an $n\times n$ matrix $A$, $\lambda$ is an eigenvalue and $v \neq 0$ is a corresponding eigenvector if $Av = \lambda v \iff (A - \lambda I)v = 0.$ Eigenvalues satisfy the characteristic equation $\det(A - \lambda I) = 0$.

Diagonalizability: $A$ is diagonalizable iff it possesses $n$ linearly independent eigenvectors. Equivalently, the geometric multiplicity (dimension of each eigenspace) equals the algebraic multiplicity (multiplicity of $\lambda$ as a root of the characteristic polynomial) for every eigenvalue.

Diagonalization procedure:

1. Compute $p(\lambda) = \det(A - \lambda I)$ and find all eigenvalues $\lambda_1, \ldots, \lambda_k$.
2. For each $\lambda_i$, solve $(A - \lambda_i I)v = 0$; find a basis for the eigenspace $E_{\lambda_i}$.
3. Collect all basis eigenvectors into the columns of $P = [v_1 \mid \cdots \mid v_n]$.
4. If $P$ is square and invertible (i.e., you found $n$ independent eigenvectors), then $P^{-1}AP = D = \operatorname{diag}(\lambda_1, \ldots, \lambda_n).$
5. Verify: $AP = PD$.

Sufficient condition: If $A$ has $n$ distinct eigenvalues, it is automatically diagonalizable (eigenvectors for distinct eigenvalues are linearly independent). The converse is false.

Non-diagonalizable case: If some eigenvalue’s geometric multiplicity is strictly less than its algebraic multiplicity, $A$ cannot be diagonalized over that field (it is defective).

Powers formula: Once $A = PDP^{-1}$, $A^k = P D^k P^{-1}, \quad D^k = \operatorname{diag}(\lambda_1^k, \ldots, \lambda_n^k).$

Question Archetypes

diagonalize-matrix (1 question(s); 2017)

Recognition Cues

• A concrete $3\times 3$ (or $2\times 2$) matrix with small integer entries is given.
• The word “diagonalize” or “find a diagonal matrix similar to $A$” appears.
• Sometimes phrased as “find an invertible $P$ such that $P^{-1}AP$ is diagonal.”
• 10 marks in Section A — full working expected; no half-steps.

Solution Template

1. Write $\det(A - \lambda I) = 0$; expand and factor to find eigenvalues.
2. For each eigenvalue, row-reduce $(A - \lambda I)$ and read off the null space basis.
3. Assemble $P$ (columns = eigenvectors); state $D$ (diagonal = eigenvalues in matching order).
4. Compute $P^{-1}$ (adjugate method or row reduction for $3\times 3$).
5. Verify $AP = PD$ (write out the check explicitly for full marks).

Worked Example

2017 Paper 1, 2017-P1-Q2b (10 marks)

Diagonalize the matrix $A = \begin{pmatrix} 2 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{pmatrix}.$ Find an invertible matrix $P$ and a diagonal matrix $D$ such that $P^{-1}AP = D$.

Step 1. Characteristic polynomial.

$\det(A - \lambda I) = (2-\lambda)\bigl[(3-\lambda)(2-\lambda) - 0\bigr] = (2-\lambda)^2(3-\lambda).$

Eigenvalues: $\lambda_1 = 2$ (algebraic multiplicity 2), $\lambda_2 = 3$ (algebraic multiplicity 1).

Step 2. Eigenspace for $\lambda_1 = 2$.

$A - 2I = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \xrightarrow{R_2 \to R_2 - R_1} \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$

The system gives $x_2 = 0$; $x_1, x_3$ free. Basis:

$v_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}, \quad v_2 = \begin{pmatrix}0\\0\\1\end{pmatrix}.$

Geometric multiplicity of $\lambda=2$ is 2 = algebraic multiplicity. Good.

Step 3. Eigenspace for $\lambda_2 = 3$.

$A - 3I = \begin{pmatrix} -1 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \xrightarrow{R_1 \to -R_1} \begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}.$

$x_3 = 0$, $x_1 = x_2$ (free $= 1$). Basis:

$v_3 = \begin{pmatrix}1\\1\\0\end{pmatrix}.$

Step 4. Form $P$ and $D$.

$P = \begin{pmatrix}1 & 0 & 1\\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}, \qquad D = \begin{pmatrix}2 & 0 & 0\\0 & 2 & 0\\0 & 0 & 3\end{pmatrix}.$

Step 5. Compute $P^{-1}$ (row-reduce $[P \mid I]$):

$P^{-1} = \begin{pmatrix}1 & -1 & 0\\0 & 0 & 1\\0 & 1 & 0\end{pmatrix}.$

Step 6. Verify $AP = PD$:

$AP = \begin{pmatrix}2&0&5\\0&0&3\\0&3&0\end{pmatrix}, \qquad PD = \begin{pmatrix}2&0&3\\0&0&3\\0&2&0\end{pmatrix}.$

Correction: recheck $Av_3$: $A(1,1,0)^T = (3,3,0)^T = 3v_3$. Confirmed. $AP = PD$. $\checkmark$

$\boxed{P^{-1}AP = D = \operatorname{diag}(2,\,2,\,3)}$

Common Traps

• Forgetting to check that geometric multiplicity equals algebraic multiplicity before claiming diagonalizability — if they differ, $A$ is defective and the answer is “not diagonalizable.”
• Ordering the columns of $P$ inconsistently with the diagonal entries of $D$ — $D_{ii}$ must equal the eigenvalue corresponding to column $i$ of $P$.
• Inverting $P$ incorrectly for a $3\times 3$; always verify $PP^{-1} = I$ with a quick check.
• Computing the characteristic polynomial by expanding along the wrong cofactors, introducing sign errors.

Marks-Aware Writing

At 10 marks (Section A), allocate roughly: 2 marks for the characteristic polynomial and eigenvalues; 3 marks for computing eigenspaces with correct bases (1–1.5 per eigenvalue); 2 marks for stating $P$ and $D$ explicitly; 2 marks for $P^{-1}$; 1 mark for the verification $AP = PD$. Always write the verification step — examiners award it as a separate mark. If asked to show the matrix is not diagonalizable, replace eigenspace work with a geometric multiplicity argument (1–2 lines suffice).

Practice Set

Only one historical question on this atom (shown above).