Compute eigenvalues and for each eigenvalue compare algebraic multiplicity (from characteristic polynomial) with geometric multiplicity (dimension of nullspace of A-λI). Diagonalizable iff for every eigenvalue these multiplicities match and the total number of independent eigenvectors equals n.
For real symmetric (or Hermitian) matrices, eigenvectors corresponding to distinct eigenvalues are orthogonal. You can orthonormalize within eigenspaces to get an orthonormal eigenbasis.
Use symmetry/patterns: constant rows, circulant structure, triangular or block forms often give shortcuts. For exam problems look for invariant subspaces or obvious eigenvectors like (1,1,...1)^T
Algebraic multiplicity = multiplicity of eigenvalue as root of characteristic polynomial. Geometric multiplicity = dimension of eigenspace (number of independent eigenvectors for that eigenvalue). Geometric ≤ algebraic.
Sum of eigenvalues (with algebraic multiplicity) = trace of matrix. Product of eigenvalues = determinant of matrix.
They reveal invariant directions of linear maps: modal analysis, diagonalization for powers/exponentials, PCA in data science, stability in dynamical systems, etc.
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Eigenvectors of a Matrix
In linear algebra, eigenvalues and eigenvectors of a matrix are fundamental concepts used in mathematics, physics, computer science, and data analysis. They help in understanding how linear transformations act on vectors and are applied in fields like facial recognition, vibration analysis, Google’s PageRank algorithm, and machine learning.
1.0What Are Eigenvalues and Eigenvectors of a Matrix?
Eigenvector of a matrix: A non-zero vector v that changes only in scale (not in direction) when a linear transformation represented by matrix A is applied.
Eigenvalue: The scalar λ that represents how much the eigenvector is scaled.
Mathematical definition:v=λv
Where:
A is a square matrix
λ is the eigenvalue
v is the eigenvector
2.0Finding Eigenvalues and Eigenvectors of a Matrix
Step 1: Find Eigenvalues
Start from: v=λv
Rewrite as: (A−λI)v=0
For non-trivial solutions (), we require: det(A−λI)=0
This is called the characteristic equation. Solve it to find the eigenvalues.
Step 2: Find Eigenvectors
Substitute each eigenvalue λ back into: (A−λI)v=0
Solve the resulting system of linear equations for the eigenvector(s).
(If m < 0 the eigenvalues are complex; if m = 0 there is a repeated eigenvalue λ=2.)
(b) Over R, A is diagonalizable when it has two distinct real eigenvalues, i.e. when m > 0. If m = 0 we get λ=2 of algebraic multiplicity 2. Check geometric multiplicity: for m = 0,
A=(2012),(A−2I)=(0010).
Nullspace is {(x,0)T} → geometric multiplicity 1 → not diagonalizable.
If m < 0 eigenvalues are complex conjugates, so not diagonalizable over R (but are diagonalizable over C if they are distinct).
Answer: diagonalizable over R iff m > 0.
Example 2: Find eigenvalues and a basis of eigenvectors of A=411141114.
Solution.
Observe the vector : u=(1,1,1)T
Au=(4+1+1,1+4+1,1+1+4)T=(6,6,6)T=6u,
so λ1=6 with eigenvector u.
For eigenvectors orthogonal to u (i.e. sum of components = 0), pick v1=(1,−1,0)T. Then
Similarly, v2=(1,0,−1)T also yields eigenvalue 3. So λ2=λ3=3 with eigenspace x:x1+x2+x3=0. Because A is symmetric, eigenvectors corresponding to different eigenvalues are orthogonal. A convenient orthogonal basis:
λ=6:(1,1,1)T;λ=3:(1,−1,0)T,(1,0,−1)T
Answer: eigenvalues 6,3,3 with the eigenvectors above.
Example 3 : Let A=200120012. Find eigenvalues and the dimension of the eigenspace. Is A diagonalizable?
Solution.
Characteristic polynomial: (2−λ)3. So eigenvalue λ=2 with algebraic multiplicity 3. Solve :
(A−2I)x=0:A−2I=000100010.
Equations give x2=0,x3=0. So eigenvectors are (x1,0,0)T → eigenspace dimension = 1.
Because geometric multiplicity (1) < algebraic multiplicity (3), A is not diagonalizable (it is a single Jordan block of size 3).
Answer:eigenvalue 2 only; eigenspace dimension 1; not diagonalizable.
Example 4: If Av=λv for nonzero v, show that for any polynomial p(t)=a0+a1t+⋯+aktk, p(A)v=p(λ)v
Solution.
Apply powers: A2v=A(Av)=A(λv)=λ(Av)=λ2v,, and by induction Akv=λkv
Then, p(A)v=∑j=0kajAjv=∑j=0kajλjv=p(λ)v.
So v is an eigenvector of the matrix p(A) with eigenvalue p(λ).
Answer: proved.
Example 5: Let P be a square matrix with P2=P. Show the only possible eigenvalues are 0 and 1, and describe the eigenspaces.
Solution. If Pv=λv for nonzero v, apply P2v=Pv:
P(Pv)=P(λv)=λPv=λ2v.
But P2v=Pv=λv. So λ2v=λv
Since
v=0 , λ2=λ⇒λ(λ−1)=0. Thus λ=0 or λ=0 .
Eigenspaces:
λ=1 : vectors fixed by P (range of P).
λ=0 : vectors sent to 0 by P (nullspace of P).
If P is an orthogonal projection, these spaces are orthogonal complements.
Answer: eigenvalues 0,1; eigenspaces = nullspace and range.
Example 6: Let J be the n x n matrix of all ones. Find the eigenvalues and eigenvectors of J.
Solution. Let u=(1,1,....1)T. Then Ju= nu. So nn is an eigenvalue with an eigenvector uu. For any vector vv with entries summing to zero (1Tv=0), we have Jv = 0 because every row of J sums the entries of v. So eigenvalues are n (multiplicity 1) and 0 (multiplicity n-1).
Eigenspaces: span{u} and the hyperplane {v:∑v1=0}
Answer: eigenvalues n and 0 with described eigenspaces.
4.0Practice questions on Eigenvectors of a Matrix
For A=(32−10)., find eigenvalues and eigenvectors.
If A is 2×2 with trace 5 and determinant 6, find its eigenvalues.
Show that if A is symmetric, eigenvectors corresponding to distinct eigenvalues are orthogonal.
Find eigenvalues of Ak in terms of eigenvalues of A.
Let A=001100010. Find eigenvalues and eigenvectors.
If A has eigenvalues 1,2,3, what is tr(A) and det(A)?
For A=(1011)., compute eigenvalues and geometric multiplicities.
Answers (brief)
Solve char poly λ2−3λ+2=0⇒λ=1,2. Find eigenvectors by substitution.
Eigenvalues are roots of λ2−5λ+6=0⇒λ=2,3.
Use (λ1−λ2)uTv=0 argument from Au=λ1andAv=λ2v
Eigenvalues of Ak are λk.
This permutation matrix represents a cycle (1 2 3). Eigenvalues are 1,ω,ω2 where ω=e2πi/3. Eigenvectors are corresponding discrete Fourier vectors.