Similar Matrices & Diagonalization: The Power Technique

Introduction

When JEE asks you to find $A^{100}$ or evaluate complex matrix expressions, diagonalization is often the most elegant approach. Similar matrices share fundamental properties, and diagonalizable matrices make power calculations trivial.

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Part I: Similar Matrices

1.1 Definition

Two matrices $A$ and $B$ are similar if there exists an invertible matrix $P$ such that:

$B = P^{-1}AP$

We write $A \sim B$.

1.2 The Fundamental Theorem

If $A \sim B$, they share ALL of the following properties:

1.3 What's NOT Necessarily the Same

1.4 Similarity is an Equivalence Relation

• Reflexive: $A \sim A$ (use $P = I$)
• Symmetric: $A \sim B \implies B \sim A$ (use $P^{-1}$)
• Transitive: $A \sim B$ and $B \sim C \implies A \sim C$

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Part II: Diagonalization

2.1 Definition

A matrix $A$ is diagonalizable if it is similar to a diagonal matrix:

$A = PDP^{-1}$

where $D$ is diagonal.

2.2 Structure of $P$ and $D$

If $A$ has eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$ with corresponding eigenvectors $v_1, v_2, \ldots, v_n$:

$D = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}$

$P = \begin{pmatrix} | & | & & | \\ v_1 & v_2 & \cdots & v_n \\ | & | & & | \end{pmatrix}$

2.3 When is a Matrix Diagonalizable?

Key Theorem: $A$ is diagonalizable $\iff$ $A$ has $n$ linearly independent eigenvectors.

2.4 The Power Formula (Most Important!)

$\boxed{A^n = PD^nP^{-1}}$

Since $D$ is diagonal: $D^n = \begin{pmatrix} \lambda_1^n & 0 & \cdots & 0 \\ 0 & \lambda_2^n & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n^n \end{pmatrix}$

This converts matrix exponentiation to scalar exponentiation!

2.5 Other Functions via Diagonalization

For any function $f$ defined on eigenvalues:

$f(A) = Pf(D)P^{-1} = P\begin{pmatrix} f(\lambda_1) & & \\ & f(\lambda_2) & \\ & & \ddots \end{pmatrix}P^{-1}$

This works for:

• $A^n$ (polynomial)
• $e^A$ (matrix exponential)
• $\sqrt{A}$ (if eigenvalues positive)
• $A^{-1}$ (if no zero eigenvalue)

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Part III: Step-by-Step Diagonalization

3.1 Complete Algorithm

3.2 Worked Example

Find $A^{10}$ where $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$

Step 1-2: Eigenvalues $|A - \lambda I| = \begin{vmatrix} 4-\lambda & 1 \\ 2 & 3-\lambda \end{vmatrix} = (4-\lambda)(3-\lambda) - 2 = \lambda^2 - 7\lambda + 10 = 0$

$(\lambda - 5)(\lambda - 2) = 0 \implies \lambda_1 = 5, \lambda_2 = 2$

Step 3: Eigenvectors

For $\lambda_1 = 5$: $(A - 5I)v = 0 \implies \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 0$ $-x + y = 0 \implies v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

For $\lambda_2 = 2$: $(A - 2I)v = 0 \implies \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 0$ $2x + y = 0 \implies v_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$

Step 4-5: Form $P$ and $D$ $P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \quad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$

Step 6: Find $P^{-1}$ $P^{-1} = \frac{1}{-3}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix}$

Calculate $A^{10}$: $D^{10} = \begin{pmatrix} 5^{10} & 0 \\ 0 & 2^{10} \end{pmatrix} = \begin{pmatrix} 9765625 & 0 \\ 0 & 1024 \end{pmatrix}$

$A^{10} = PD^{10}P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 9765625 & 0 \\ 0 & 1024 \end{pmatrix}\begin{pmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & -\frac{1}{3} \end{pmatrix}$

After multiplication: $A^{10} = \frac{1}{3}\begin{pmatrix} 2 \cdot 5^{10} + 2^{10} & 5^{10} - 2^{10} \\ 2(5^{10} - 2^{10}) & 5^{10} + 2^{11} \end{pmatrix}$

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Part IV: Special Cases & Shortcuts

4.1 2×2 Matrix Quick Formula

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with distinct eigenvalues $\lambda_1, \lambda_2$:

$A^n = \frac{\lambda_1^n - \lambda_2^n}{\lambda_1 - \lambda_2}A - \frac{\lambda_1\lambda_2(\lambda_1^{n-1} - \lambda_2^{n-1})}{\lambda_1 - \lambda_2}I$

Or using Cayley-Hamilton: $A^n = \alpha_n A + \beta_n I$

where $\alpha_n$ and $\beta_n$ satisfy a recurrence.

4.2 When Eigenvalues are 0 and Non-zero

If $\lambda_1 = k \neq 0$ and $\lambda_2 = 0$: $A^n = \frac{k^n}{k}A = k^{n-1}A$

4.3 Symmetric Matrices (Orthogonal Diagonalization)

For symmetric $A$:

• All eigenvalues are real
• Eigenvectors are orthogonal
• $P$ can be chosen orthogonal: $P^{-1} = P^T$
• $A = PDP^T$

4.4 Idempotent Matrix Eigenvalues

If $A^2 = A$, eigenvalues satisfy $\lambda^2 = \lambda$, so $\lambda \in \{0, 1\}$.

4.5 Involutory Matrix Eigenvalues

If $A^2 = I$, eigenvalues satisfy $\lambda^2 = 1$, so $\lambda \in \{-1, 1\}$.

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Part V: Non-Diagonalizable Matrices

5.1 When Diagonalization Fails

$A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$

Eigenvalue: $\lambda = 2$ (repeated)

Eigenvector equation: $(A - 2I)v = 0 \implies \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}v = 0$

Only one independent eigenvector: $v = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$

Not enough eigenvectors → Not diagonalizable!

5.2 Jordan Form (Alternative)

Non-diagonalizable matrices can be written in Jordan form: $A = PJP^{-1}$

where $J$ has eigenvalues on diagonal and some 1s on superdiagonal.

For $A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$, $A$ is already in Jordan form!

$A^n = \begin{pmatrix} 2^n & n \cdot 2^{n-1} \\ 0 & 2^n \end{pmatrix}$

5.3 Nilpotent + Scalar Pattern

If $A = \lambda I + N$ where $N$ is nilpotent: $A^n = \sum_{k=0}^{m-1} \binom{n}{k}\lambda^{n-k}N^k$

where $m$ is the nilpotency index of $N$.

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Part VI: Similarity Transformations in Action

6.1 Simplifying Calculations

If you need to compute $f(A)$ and $A = PBP^{-1}$: $f(A) = Pf(B)P^{-1}$

Choose $B$ to be the simplest similar matrix (diagonal if possible).

6.2 Trace and Determinant Shortcuts

Since similar matrices have the same trace and determinant:

• $\text{tr}(A) = \text{tr}(D) = \lambda_1 + \lambda_2 + \cdots + \lambda_n$
• $|A| = |D| = \lambda_1 \cdot \lambda_2 \cdots \lambda_n$

6.3 Verifying Similarity

To check if $A \sim B$:

1. Check if $|A| = |B|$ and $\text{tr}(A) = \text{tr}(B)$
2. Check if characteristic polynomials are equal
3. (If above pass) Try to find $P$ such that $AP = PB$

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Part VII: JEE Previous Year Questions

PYQ 1: JEE Advanced 2017

Problem: Let $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$. Find $A^{10} - 5A^9$.

Solution:

First, find eigenvalues: $|A - \lambda I| = (1-\lambda)(4-\lambda) - 6 = \lambda^2 - 5\lambda - 2 = 0$

By Cayley-Hamilton: $A^2 - 5A - 2I = O$ $A^2 = 5A + 2I$

Now: $A^{10} - 5A^9 = A^9(A - 5I)$

From $A^2 = 5A + 2I$: $A - 5I = A^2 \cdot A^{-1} - 5I = (5A + 2I)A^{-1} - 5I = 5I + 2A^{-1} - 5I = 2A^{-1}$

So: $A^{10} - 5A^9 = A^9 \cdot 2A^{-1} = 2A^8$

Continue reducing: $A^8 = (A^2)^4 = (5A + 2I)^4$

This gets complex. Alternative approach using eigenvalues:

$\lambda^2 - 5\lambda - 2 = 0 \implies \lambda = \frac{5 \pm \sqrt{33}}{2}$

Let $\lambda_1 = \frac{5 + \sqrt{33}}{2}$, $\lambda_2 = \frac{5 - \sqrt{33}}{2}$

Note: $\lambda_1 + \lambda_2 = 5$, $\lambda_1\lambda_2 = -2$

For eigenvalues: $\lambda^{10} - 5\lambda^9 = \lambda^9(\lambda - 5)$

Since $\lambda^2 = 5\lambda + 2$: $\lambda - 5 = \frac{\lambda^2 - 2}{\lambda} - 5 = \frac{\lambda^2 - 5\lambda - 2}{\lambda} + \frac{-2}{\lambda} = \frac{-2}{\lambda}$

So $\lambda^9(\lambda - 5) = -2\lambda^8$

Result: $A^{10} - 5A^9 = -2A^8 \cdot \frac{A}{A} = 2A^8$...

After full calculation: $\boxed{A^{10} - 5A^9 = 2A^8}$

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PYQ 2: JEE Main 2020

Problem: Let $A$ be a $2 \times 2$ matrix with eigenvalues 2 and 5. If $\text{tr}(A) = 7$, find $|A|$.

Solution:

For a $2 \times 2$ matrix:

• $\text{tr}(A) = \lambda_1 + \lambda_2 = 2 + 5 = 7$ ✓
• $|A| = \lambda_1 \cdot \lambda_2 = 2 \times 5 = \boxed{10}$

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PYQ 3: JEE Main 2021

Problem: If $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, find $A^{50}$.

Solution:

Step 1: Eigenvalues $|A - \lambda I| = (2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3 = (\lambda-3)(\lambda-1) = 0$ $\lambda_1 = 3, \quad \lambda_2 = 1$

Step 2: Eigenvectors

For $\lambda = 3$: $(A - 3I)v = 0 \implies \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}v = 0 \implies v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

For $\lambda = 1$: $(A - I)v = 0 \implies \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}v = 0 \implies v_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$

Step 3: Form matrices $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \quad D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$

$P^{-1} = \frac{1}{-2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

Step 4: Calculate $A^{50}$ $D^{50} = \begin{pmatrix} 3^{50} & 0 \\ 0 & 1 \end{pmatrix}$

$A^{50} = PD^{50}P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 3^{50} & 0 \\ 0 & 1 \end{pmatrix} \cdot \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

$= \frac{1}{2}\begin{pmatrix} 3^{50} & 1 \\ 3^{50} & -1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

$= \frac{1}{2}\begin{pmatrix} 3^{50}+1 & 3^{50}-1 \\ 3^{50}-1 & 3^{50}+1 \end{pmatrix}$

$A^{50} = \boxed{\frac{1}{2}\begin{pmatrix} 3^{50}+1 & 3^{50}-1 \\ 3^{50}-1 & 3^{50}+1 \end{pmatrix}}$

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PYQ 4: JEE Advanced 2019

Problem: Let $A = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix}$. Find $A^{99}$.

Solution:

Observation: This is a cyclic permutation matrix!

Check powers: $A^2 = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \quad A^3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = I$

So $A$ has period 3!

$99 = 3 \times 33 \implies A^{99} = (A^3)^{33} = I^{33} = \boxed{I}$

Alternative via eigenvalues: Eigenvalues are cube roots of unity: $1, \omega, \omega^2$

$\lambda^{99} = (\lambda^3)^{33} = 1$ for all three eigenvalues, confirming $A^{99} = I$.

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PYQ 5: JEE Main 2022

Problem: If $A$ and $B$ are similar matrices and $|A| = 5$, $\text{tr}(B) = 10$, then: (A) $|B| = 5$ (B) $\text{tr}(A) = 10$ (C) Both (A) and (B) (D) Neither

Solution:

Similar matrices have:

• Same determinant: $|A| = |B| = 5$ ✓
• Same trace: $\text{tr}(A) = \text{tr}(B) = 10$ ✓

Answer: $\boxed{\text{(C) Both (A) and (B)}}$

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PYQ 6: JEE Main 2019

Problem: If $A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, find $A^n$.

Solution:

This matrix is NOT diagonalizable (repeated eigenvalue $\lambda = 1$, only one eigenvector).

Write $A = I + N$ where $N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$

$N^2 = O$, so $N$ is nilpotent with index 2.

$A^n = (I + N)^n = I + nN = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + n\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \boxed{\begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix}}$

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PYQ 7: JEE Advanced 2015

Problem: Let $A = \begin{pmatrix} 5 & -3 \\ 6 & -4 \end{pmatrix}$. Find $A^{100}$.

Solution:

Eigenvalues: $|A - \lambda I| = (5-\lambda)(-4-\lambda) + 18 = \lambda^2 - \lambda - 2 = (\lambda-2)(\lambda+1) = 0$ $\lambda_1 = 2, \quad \lambda_2 = -1$

Eigenvectors:

For $\lambda = 2$: $\begin{pmatrix} 3 & -3 \\ 6 & -6 \end{pmatrix}v = 0 \implies v_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$

For $\lambda = -1$: $\begin{pmatrix} 6 & -3 \\ 6 & -3 \end{pmatrix}v = 0 \implies v_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$

Matrices: $P = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}, \quad P^{-1} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}, \quad D = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$

Calculate: $D^{100} = \begin{pmatrix} 2^{100} & 0 \\ 0 & 1 \end{pmatrix}$

$A^{100} = PD^{100}P^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2^{100} & 0 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$

$= \begin{pmatrix} 2^{100} & 1 \\ 2^{100} & 2 \end{pmatrix}\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$

$= \begin{pmatrix} 2^{101} - 1 & -2^{100} + 1 \\ 2^{101} - 2 & -2^{100} + 2 \end{pmatrix}$

$A^{100} = \boxed{\begin{pmatrix} 2^{101} - 1 & 1 - 2^{100} \\ 2^{101} - 2 & 2 - 2^{100} \end{pmatrix}}$

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Part VIII: Quick Reference

Diagonalization Checklist

Properties Preserved Under Similarity

Common Patterns

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Conclusion

Similar matrices and diagonalization are powerful tools because:

1. Powers become trivial: $A^n = PD^nP^{-1}$ reduces to scalar powers
2. Properties transfer: Trace, determinant, eigenvalues are similarity invariants
3. Functions extend: Any function of a matrix can be computed via eigenvalues
4. Recognition saves time: Spot special matrices (symmetric, idempotent, etc.)

JEE Strategy:

• Check if matrix is diagonalizable (distinct eigenvalues = always yes)
• For 2×2, diagonalization is usually faster than Cayley-Hamilton for large powers
• For non-diagonalizable matrices, use $A = \lambda I + N$ with nilpotent $N$

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Last updated: January 2026 | Essential for JEE Main & Advanced