Characteristic Polynomial: Complete Guide to Finding det(A - λI)

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• What is the Characteristic Polynomial?
• How to Compute the Characteristic Polynomial
• Properties of the Characteristic Polynomial
• Connection to Eigenvalues
• Special Cases
• Step-by-Step Examples
• Common Questions
• Practice Problems
• Related Topics

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What is the Characteristic Polynomial?

Definition

The characteristic polynomial of a square matrix $A$ is a polynomial in $\lambda$ (lambda) defined as:

$$ \boxed{p(\lambda) = \det(A - \lambda I)} $$

Where:

• $A$ is your $n \times n$ square matrix
• $\lambda$ is a scalar variable
• $I$ is the identity matrix
• $\det()$ denotes the determinant

Why Is It Called "Characteristic"?

The characteristic polynomial completely characterizes the matrix's eigenvalues. The roots of this polynomial are exactly the eigenvalues of the matrix, which capture the essential "characteristics" of the linear transformation represented by $A$.

Simple Example (2×2)

For a generic 2×2 matrix: $$ A = \left[\begin{array}{cc} a & b \\ c & d \end{array}\right] $$

Subtract $\lambda$ from the diagonal: $$ A - \lambda I = \left[\begin{array}{cc} a - \lambda & b \\ c & d - \lambda \end{array}\right] $$

Compute the determinant: $$ p(\lambda) = (a - \lambda)(d - \lambda) - bc $$

Expanding this gives the standard form: $$ p(\lambda) = \lambda^2 - (a + d)\lambda + (ad - bc) $$

Notice the pattern:

• Coefficient of $\lambda^2$ is $1$.
• Coefficient of $\lambda$ is negative trace: $-\text{tr}(A) = -(a + d)$.
• Constant term is determinant: $\det(A) = ad - bc$.

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How to Compute the Characteristic Polynomial

Method 1: Direct Determinant Calculation (2×2)

Step-by-step process:

1. Form $A - \lambda I$: Subtract $\lambda$ from each diagonal entry.
2. Compute determinant: Use the formula $\det = (ad - bc)$.
3. Expand and simplify: Combine like terms to get a polynomial in $\lambda$.
4. Standard Form: Write as $\lambda^2 - \text{tr}(A)\lambda + \det(A)$.

Method 2: Cofactor Expansion (3×3 and larger)

For a 3×3 matrix, expand along the first row:

$$ \det(A - \lambda I) = (a_{11} - \lambda)C_{11} - a_{12}C_{12} + a_{13}C_{13} $$

Where $C_{ij}$ is the cofactor associated with element $(i,j)$. This reduces the problem to computing 2×2 determinants.

Method 3: Trace and Determinant Shortcut (2×2 Only)

For any 2×2 matrix, you can skip the expansion: $$ p(\lambda) = \lambda^2 - \text{tr}(A)\lambda + \det(A) $$

Method 4: Computer Algebra (3×3 and larger)

For matrices larger than 3×3, hand calculation is prone to error. The number of terms grows factorially ($n!$). Use our calculator above for fast, accurate symbolic results!

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Properties of the Characteristic Polynomial

1. Degree Equals Matrix Size

For an $n \times n$ matrix, the characteristic polynomial has degree $n$: $$ \deg(p(\lambda)) = n $$ This means there are exactly $n$ eigenvalues (counting multiplicities and complex roots).

2. Leading Coefficient

The coefficient of the highest power $\lambda^n$ is either $1$ or $(-1)^n$, depending on the definition used. We typically normalize it to be monic (coefficient 1): $$ p(\lambda) = \lambda^n - \text{tr}(A)\lambda^{n-1} + \dots + (-1)^n \det(A) $$

3. Constant Term

The constant term is related to the determinant: $$ p(0) = \det(-A) = (-1)^n \det(A) $$

4. Trace Relationship

The coefficient of the $\lambda^{n-1}$ term is always the negative of the trace (sum of diagonal elements): $$ \text{Coeff of } \lambda^{n-1} = -\text{tr}(A) $$

5. Roots Are Eigenvalues

This is the most critical property:

• $\lambda$ is an eigenvalue of $A$ if and only if $p(\lambda) = 0$.

6. Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation. If you substitute the matrix $A$ for $\lambda$ in the polynomial, the result is the zero matrix:

$$ p(A) = \mathbf{0} $$

Example: If $p(\lambda) = \lambda^2 - 4\lambda + 3$, then $A^2 - 4A + 3I = 0$.

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Connection to Eigenvalues

How to Find Eigenvalues

1. Compute the characteristic polynomial $p(\lambda) = \det(A - \lambda I)$.
2. Solve the characteristic equation $p(\lambda) = 0$.
3. The solutions are the eigenvalues $\lambda_1, \lambda_2, \dots, \lambda_n$.

Multiplicity

• Algebraic Multiplicity: The number of times an eigenvalue appears as a root of $p(\lambda)$.
• Example: If $p(\lambda) = (\lambda - 2)^2(\lambda - 5)$, then:
  • $\lambda = 2$ has algebraic multiplicity 2.
  • $\lambda = 5$ has algebraic multiplicity 1.

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Special Cases

1. Diagonal Matrix

For a diagonal matrix $D = \text{diag}(d_1, \dots, d_n)$: $$ p(\lambda) = (d_1 - \lambda)(d_2 - \lambda)\cdots(d_n - \lambda) $$ The eigenvalues are simply the diagonal entries.

2. Triangular Matrix

For any upper or lower triangular matrix, the characteristic polynomial is the same as for a diagonal matrix: $$ p(\lambda) = \prod_{i=1}^n (a_{ii} - \lambda) $$ The eigenvalues are the diagonal entries.

3. 2×2 Matrix

$$ p(\lambda) = \lambda^2 - \text{tr}(A)\lambda + \det(A) $$

4. 3×3 Matrix

$$ p(\lambda) = -\lambda^3 + \text{tr}(A)\lambda^2 - S_2 \lambda + \det(A) $$ Where $S_2$ is the sum of the principal 2×2 minors.

5. Symmetric Matrix

• All roots of $p(\lambda)$ are real numbers.
• The polynomial factors completely over the real numbers.

6. Skew-Symmetric Matrix (Odd Dimension)

• If $n$ is odd, $\lambda = 0$ is always a root.
• The constant term of the polynomial is 0.

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Step-by-Step Examples

Example 1: 2×2 Matrix

Problem: Find the characteristic polynomial of $A = \left[\begin{array}{cc} 4 & 1 \\ 2 & 3 \end{array}\right]$.

Solution:

Step 1: Form $A - \lambda I$ $$ A - \lambda I = \left[\begin{array}{cc} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{array}\right] $$

Step 2: Compute determinant $$ \det = (4 - \lambda)(3 - \lambda) - (1)(2) $$

Step 3: Expand $$ = (12 - 4\lambda - 3\lambda + \lambda^2) - 2 $$ $$ = \lambda^2 - 7\lambda + 10 $$

Conclusion: $$ p(\lambda) = \lambda^2 - 7\lambda + 10 $$ (Check: Trace = 7, Det = 10. Matches!)

Example 2: 3×3 Matrix

Problem: Find the characteristic polynomial of $A = \left[\begin{array}{ccc} 2 & 1 & 0 \\ 1 & 2 & 1 \\ 0 & 1 & 2 \end{array}\right]$.

Solution:

Step 1: Form $A - \lambda I$ $$ A - \lambda I = \left[\begin{array}{ccc} 2 - \lambda & 1 & 0 \\ 1 & 2 - \lambda & 1 \\ 0 & 1 & 2 - \lambda \end{array}\right] $$

Step 2: Expand along the first row $$ \det = (2 - \lambda)\det\left[\begin{array}{cc} 2 - \lambda & 1 \\ 1 & 2 - \lambda \end{array}\right] - 1\det\left[\begin{array}{cc} 1 & 1 \\ 0 & 2 - \lambda \end{array}\right] + 0 $$

Step 3: Compute 2×2 minors

• First minor: $(2-\lambda)^2 - 1 = \lambda^2 - 4\lambda + 3$
• Second minor: $1(2-\lambda) - 0 = 2 - \lambda$

Step 4: Substitute and simplify $$ p(\lambda) = (2 - \lambda)(\lambda^2 - 4\lambda + 3) - (2 - \lambda) $$ Factor out $(2-\lambda)$: $$ = (2 - \lambda)[(\lambda^2 - 4\lambda + 3) - 1] $$ $$ = (2 - \lambda)(\lambda^2 - 4\lambda + 2) $$ Expand: $$ = 2\lambda^2 - 8\lambda + 4 - \lambda^3 + 4\lambda^2 - 2\lambda $$ $$ = -\lambda^3 + 6\lambda^2 - 10\lambda + 4 $$

Conclusion: $$ p(\lambda) = \lambda^3 - 6\lambda^2 + 10\lambda - 4 $$ (Normalized to make the leading coefficient positive)

Example 3: Diagonal Matrix

Problem: Find the characteristic polynomial of $A = \left[\begin{array}{ccc} 3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 7 \end{array}\right]$.

Solution:

Since $A$ is diagonal, the eigenvalues are the diagonal entries. $$ p(\lambda) = (3 - \lambda)(5 - \lambda)(7 - \lambda) $$

Expanding this yields: $$ p(\lambda) = -\lambda^3 + 15\lambda^2 - 71\lambda + 105 $$

Or in standard monic form: $$ p(\lambda) = \lambda^3 - 15\lambda^2 + 71\lambda - 105 $$

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Common Questions

Q: Why do we use $\lambda$ instead of $x$?

A: $\lambda$ is the standard symbol for eigenvalues (from the German Eigenwert). It distinguishes eigenvalue problems from general polynomial roots.

Q: Can two different matrices have the same characteristic polynomial?

A: Yes. Similar matrices ($B = P^{-1}AP$) always have the same characteristic polynomial. However, having the same characteristic polynomial does not guarantee two matrices are similar.

Q: What if all eigenvalues are zero?

A: Then $p(\lambda) = \lambda^n$. This implies the matrix is nilpotent (i.e., $A^k = 0$ for some $k$).

Q: How does the discriminant help?

A: For a 2×2 matrix, the discriminant $\Delta = \text{tr}(A)^2 - 4\det(A)$ tells us about the nature of the roots:

• $\Delta > 0$: Two distinct real eigenvalues.
• $\Delta = 0$: One repeated real eigenvalue.
• $\Delta < 0$: Two complex conjugate eigenvalues.

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Practice Problems

Beginner Level

1. Find $p(\lambda)$ for $A = \left[\begin{array}{cc} 1 & 2 \\ 3 & 4 \end{array}\right]$.
2. Find $p(\lambda)$ for $A = \left[\begin{array}{cc} 5 & 0 \\ 0 & 5 \end{array}\right]$.

Intermediate Level

3. Find $p(\lambda)$ for $A = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$.
4. Find the eigenvalues of $A = \left[\begin{array}{cc} 1 & 4 \\ 1 & 1 \end{array}\right]$ by solving $p(\lambda)=0$.

Advanced Level

5. Use the Cayley-Hamilton theorem to express $A^3$ in terms of $A$ and $I$ for $A = \left[\begin{array}{cc} 2 & 1 \\ 1 & 2 \end{array}\right]$.
6. If $p(\lambda) = \lambda^3 - 6\lambda^2 + 11\lambda - 6$, what are the eigenvalues?

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Summary

Key Takeaways

Property	Formula
Definition	$p(\lambda) = \det(A - \lambda I)$
Degree	$n$ (size of matrix)
Leading Term	$\lambda^n$
$\lambda^{n-1}$ Coeff	$-\text{tr}(A)$
Constant Term	$(-1)^n \det(A)$
Roots	Eigenvalues of $A$

When to Use This Knowledge

• ✅ Finding Eigenvalues: The primary method for exact solutions.
• ✅ Stability Analysis: In differential equations, the roots determine stability.
• ✅ Cayley-Hamilton Theorem: Simplifying matrix powers.
• ✅ Control Theory: Determining system poles.

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Related Topics

Explore these related concepts in our Linear Algebra Toolbox:

• Eigenvalue Calculator - Find eigenvalues from the characteristic polynomial
• Matrix Diagonalization - Use eigenvalues to diagonalize matrices
• Linear ODE System Solver - System stability determined by characteristic polynomial roots
• Positive Definite Checker - Check definiteness using eigenvalues