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: $v=\lambda v$

Where:

• A is a square matrix
• $\lambda$ is the eigenvalue
• $v$ is the eigenvector

2.0Finding Eigenvalues and Eigenvectors of a Matrix

Step 1: Find Eigenvalues

1. Start from: $v=\lambda v$
2. Rewrite as: $(A-\lambda I)v=0$
3. For non-trivial solutions (), we require: $\det (A-\lambda I)=0$

This is called the characteristic equation. Solve it to find the eigenvalues.

Step 2: Find Eigenvectors

1. Substitute each eigenvalue λ back into: $(A-\lambda I)v=0$ 
2. Solve the resulting system of linear equations for the eigenvector(s).

3.0Solved Example on Eigenvectors of a Matrix 

Example 1: Let $A=\left[\begin{array}{cc}2& 1\\ m& 2\end{array}\right].$

1. Find the eigenvalues of A.
2. For which values of m is A diagonalizable over $\mathbb{R}$?

Solution.

(a) Characteristic polynomial: $\det (A-\lambda I)=(2-\lambda {)}^2-m=0.$

So . $(2-\lambda {)}^2=m$

Hence, eigenvalues $\lambda =2\pm \sqrt{m}.$

(If m < 0 the eigenvalues are complex; if m = 0 there is a repeated eigenvalue $\lambda =2$.)

(b) Over $\mathbb{R}$, A is diagonalizable when it has two distinct real eigenvalues, i.e. when m > 0. If m = 0 we get $\lambda =2$ of algebraic multiplicity 2. Check geometric multiplicity: for m = 0,

$A=\left(\begin{array}{cc}2& 1\\ 0& 2\end{array}\right),(A-2I)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right).$

Nullspace is $\{(x,0{)}^T\}$ → geometric multiplicity 1 → not diagonalizable.

If m < 0 eigenvalues are complex conjugates, so not diagonalizable over $\mathbb{R}$ (but are diagonalizable over $\mathbb{C}$ if they are distinct).

Answer: diagonalizable over $\mathbb{R}$ iff m > 0.

Example 2: Find eigenvalues and a basis of eigenvectors of  $A=\left(\begin{array}{ccc}4& 1& 1\\ 1& 4& 1\\ 1& 1& 4\end{array}\right).$

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 ${\lambda }_1=6$ with eigenvector u.

For eigenvectors orthogonal to u (i.e. sum of components = 0), pick $v_1=(1,-1,0{)}^T$. Then

$A{v}_1=\left(\begin{array}{c}4(1)+1(-1)+1(0)\\ 1(1)+4(-1)+1(0)\\ 1(1)+1(-1)+4(0)\end{array}\right)=\left(\begin{array}{c}3\\ -3\\ 0\end{array}\right)=3v_1.$

Similarly, $v_2=(1,0,-1{)}^T$ also yields eigenvalue 3. So ${\lambda }_2={\lambda }_3=3$ with eigenspace $x:x_1+x_2+x_3=0$. Because A is symmetric, eigenvectors corresponding to different eigenvalues are orthogonal. A convenient orthogonal basis:

$\lambda =6:(1,1,1{)}^T;\lambda =3:(1,-1,0{)}^T,(1,0,-1{)}^T$

Answer: eigenvalues 6,3,3 with the eigenvectors above.

Example 3 : Let $A=\left(\begin{array}{ccc}2& 1& 0\\ 0& 2& 1\\ 0& 0& 2\end{array}\right).$ Find eigenvalues and the dimension of the eigenspace. Is A diagonalizable?

Solution.

Characteristic polynomial: $(2-\lambda {)}^3.$ So eigenvalue $\lambda =2$ with algebraic multiplicity 3. Solve :

$(A-2I)x=0:A-2I=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right).$

Equations give $x_2=0,x_3=0$. So eigenvectors are $(x_1,0,0{)}^T$ → 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 $Av=\lambda v$ for nonzero v, show that for any polynomial $p(t)=a_0+a_{1}t+\cdots +a_k{t}^k$, $p(A)v=p(\lambda )v$

Solution.

Apply powers: $A^{2}v=A(Av)=A(\lambda v)=\lambda (Av)={\lambda }^{2}v,$, and by induction $A^{k}v={\lambda }^{k}v$

Then, $p(A)v={\sum }_{j=0}^k{a}_j{A}^{j}v={\sum }_{j=0}^k{a}_j{\lambda }^{j}v=p(\lambda )v.$

So v is an eigenvector of the matrix p(A) with eigenvalue $p(\lambda )$.

Answer: proved.

Example 5: Let P be a square matrix with $P^2=P$. Show the only possible eigenvalues are 0 and 1, and describe the eigenspaces.

Solution.
If $Pv=\lambda v$ for nonzero v, apply $P^{2}v=Pv$:

$P(Pv)=P(\lambda v)=\lambda Pv={\lambda }^{2}v.$

But $P^{2}v=Pv=\lambda v$. So ${\lambda }^{2}v=\lambda v$

Since

$v\ne 0$ , ${\lambda }^2=\lambda \Rightarrow \lambda (\lambda -1)=0.$ Thus $\lambda =0$ or $\lambda =0$ .

Eigenspaces:

• $\lambda =1$ : vectors fixed by P (range of P).
• $\lambda =0$ : 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 ($1^{T}v=0$), 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 $\{v:\sum v_1=0\}$

Answer: eigenvalues n and 0 with described eigenspaces.

4.0Practice questions on Eigenvectors of a Matrix

1. For $A=\left(\begin{array}{cc}3& -1\\ 2& 0\end{array}\right).$, find eigenvalues and eigenvectors.
2. If A is $2\times 2$ with trace 5 and determinant 6, find its eigenvalues.
3. Show that if A is symmetric, eigenvectors corresponding to distinct eigenvalues are orthogonal.
4. Find eigenvalues of $A^k$ in terms of eigenvalues of A.
5. Let $A=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 1\\ 1& 0& 0\end{array}\right).$ Find eigenvalues and eigenvectors.
6. If A has eigenvalues 1,2,3, what is $\mathrm{tr}(A)$ and $\det (A)$?
7. For $A=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right).$, compute eigenvalues and geometric multiplicities.

Answers (brief)

1. Solve char poly ${\lambda }^2-3\lambda +2=0\Rightarrow \lambda =1,2.$ Find eigenvectors by substitution.
2. Eigenvalues are roots of ${\lambda }^2-5\lambda +6=0\Rightarrow \lambda =2,3.$
3. Use $({\lambda }_1-{\lambda }_2)u^{T}v=0$ argument from $Au={\lambda }_{1}andAv={\lambda }_{2}v$
4. Eigenvalues of $A^k$ are ${\lambda }^k$.
5. This permutation matrix represents a cycle (1 2 3). Eigenvalues are $1,\omega ,{\omega }^2$ where $\omega =e^{2\pi i/3}$. Eigenvectors are corresponding discrete Fourier vectors.
6. $\mathrm{tr}(A)=1+2+3=6,\det (A)=1\cdot 2\cdot 3=6$
7. Char poly $(1-\lambda {)}^2$ → eigenvalue 1 with algebraic multiplicity 2; geometric multiplicity 1 (not diagonalizable).

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