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what are eigenvalues and eigenvectors of a matrix

Rahul PalRahul Pal·researched on Researchly·June 19, 2026Try free
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Eigenvalues and Eigenvectors

Classification: Study Help — Linear Algebra


For a square matrix $A$, a non-zero vector $\mathbf{v}$ is an eigenvector if multiplying by $A$ only scales it — never rotates or shears it:

$$A\mathbf{v} = \lambda \mathbf{v}$$

where $\lambda$ (a scalar) is the corresponding eigenvalue.

Intuitively: $\mathbf{v}$ stays on its own line; $A$ just stretches or flips it by factor $\lambda$.


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Step 1 — Characteristic Equation

Rearrange $A\mathbf{v} = \lambda\mathbf{v}$:

$$(A - \lambda I)\mathbf{v} = \mathbf{0}$$

For a non-trivial solution ($\mathbf{v} \neq \mathbf{0}$), the matrix $(A - \lambda I)$ must be singular:

$$\det(A - \lambda I) = 0$$

This is the characteristic polynomial — solve it for $\lambda$.


Step 2 — Find Eigenvalues $\lambda$

Expand $\det(A - \lambda I)$ and solve the resulting polynomial.


Step 3 — Find Eigenvectors

For each $\lambda$, solve the linear system:

$$(A - \lambda I)\mathbf{v} = \mathbf{0}$$

using row reduction (RREF). The null space gives the eigenvectors.


Let:

$$A = \begin{pmatrix} 4 & 1 \ 2 & 3 \end{pmatrix}$$

Step 1: Characteristic polynomial

$$\det(A - \lambda I) = \det\begin{pmatrix} 4-\lambda & 1 \ 2 & 3-\lambda \end{pmatrix}$$

$$= (4-\lambda)(3-\lambda) - (1)(2)$$

$$= \lambda^2 - 7\lambda + 10 = 0$$

Step 2: Solve for $\lambda$

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

Step 3: Find eigenvectors

For $\lambda_1 = 5$:

$$(A - 5I)\mathbf{v} = \begin{pmatrix} -1 & 1 \ 2 & -2 \end{pmatrix}\mathbf{v} = \mathbf{0}$$

Row reduce → $-v_1 + v_2 = 0 \implies v_2 = v_1$

$$\mathbf{v}_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix}$$

For $\lambda_2 = 2$:

$$(A - 2I)\mathbf{v} = \begin{pmatrix} 2 & 1 \ 2 & 1 \end{pmatrix}\mathbf{v} = \mathbf{0}$$

Row reduce → $2v_1 + v_2 = 0 \implies v_2 = -2v_1$

$$\mathbf{v}_2 = \begin{pmatrix} 1 \ -2 \end{pmatrix}$$

Verification

$$A\mathbf{v}_1 = \begin{pmatrix}4&1\2&3\end{pmatrix}\begin{pmatrix}1\1\end{pmatrix} = \begin{pmatrix}5\5\end{pmatrix} = 5\begin{pmatrix}1\1\end{pmatrix} \checkmark$$


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Table
PropertyDetail
Trace$\text{tr}(A) = \sum \lambda_i$
Determinant$\det(A) = \prod \lambda_i$
Singular matrixHas eigenvalue $\lambda = 0$
Symmetric matrixAll eigenvalues are real; eigenvectors are orthogonal
Repeated eigenvaluesMay have fewer independent eigenvectors (defective matrix)

  • PCA (data science) — eigenvectors of covariance matrix give principal components
  • PageRank (Google) — dominant eigenvector of link matrix
  • Stability analysis — sign of eigenvalues determines system behavior (ODE/control theory)
  • Quantum mechanics — observable values are eigenvalues of operators

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