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Eigenvectors
We know that the vectors change its magnitude and direction when some linear transformation is applied to it. But some vectors do not change much (or in other words they change at most by its scale factor) even after the application of transformations on them. Such vectors are called eigenvectors. We have to find eigenvalues always before finding the eigenvectors.
Let us learn how to find the eigenvalues and eigenvectors for 2 × 2 and 3 × 3 matrices along with examples.
What are Eigenvectors?
In linear algebra, the eigenvectors of a square matrix are nonzero vectors which when multiplied by the square matrix would result in just the scalar multiple of the vectors. i.e., a vector v is said to be an eigenvector of a square matrix A if and only if Av = λv, for some scalar λ. Here, v is an eigenvector as when it multiplied by A resulted in λv, which is a scalar multiple of v. Here, the scalar λ is called the eigenvalue that corresponds to the eigenvector v.
The word "eigen" means "own", "characteristic", "proper", "typical", etc and it is originated from the German language. Thus, eigenvectors of a matrix are also known as characteristic vectors of the matrix.
In the above formula, if A is a square matrix of order n × n then v is a column vector of order nx1. Here, v is the right eigenvector of A as it is multiplied on the right side of A. Mostly, eigenvector means that it is right eigenvector. In the contrary, a left eigenvector of a matrix of order n × n is a row vector of order 1xn such that vA = λv.
Eigenvalues and Eigenvectors
In the previous section, we have seen that a column vector v is an eigenvector of a square matrix A (of order n × n) if and only if Av = λv. We can write this equation as Av = λIv, where I is the identity matrix of the same order n × n. The equation can be written as
Av  λIv = O
Here, O is a zero matrix.
(A  λI) v = O ... (1)
For this homogeneous system to have a nontrivial solution, the determinant should be equal to 0.
A  λI = 0 ... (2)
We can finalize the steps to find the eigenvalues and eigenvectors as follows for any square matrix A:
 To find the eigenvalues of A, solve the characteristic equation A  λI = 0 (equation (2)) for λ and all such values of λ would give the eigenvalues.
 To find the eigenvectors of A, substitute each eigenvalue (i.e., the value of λ) in equation (1) (A  λI) v = O and solve for v using the method of your choice.
(This would result in a system of homogeneous linear equations. To know how to solve such systems, click here.)
Let us see how to find the eigenvectors of a 2 × 2 matrix and 3 × 3 matrix using these steps.
How to Find Eigenvectors of a 2 × 2 matrix?
To find the eigenvectors, we first have to compute the eigenvalues using the abovementioned steps. Let us understand the process by an example.
Example: Find the eigenvalues and eigenvectors of matrix A = \(\left[\begin{array}{ll}
5 & 4 \\ \\
1 & 2
\end{array}\right]\).
Solution:
Let λ represent the eigenvalue(s) and v = \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) represent the eigenvector(s).
Then the characteristic equation is:
A  λI = 0
\(\left\begin{array}{ll}
5 λ & 4 \\ \\
1 & 2 λ
\end{array}\right\) = 0
(5  λ) (2  λ)  (4)(1) = 0
10  5λ  2λ + λ^{2}  4 = 0
λ^{2}  7λ + 6 = 0
(λ  6)(λ  1) = 0
λ = 6, λ = 1.
Thus, the eigenvalues are 1 and 6. Let us find the corresponding eigenvector to each eigenvalue in each case.
When λ = 1:
Substitute λ = 1 in the equation:
(A  λI) v = O
\(\left[\begin{array}{ll}
5 1 & 4 \\ \\
1 & 2 1
\end{array}\right]\) \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\ \\
0
\end{array}\right]\)
\(\left[\begin{array}{ll}
4& 4 \\ \\
1 & 1
\end{array}\right]\) \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\ \\
0
\end{array}\right]\)
To solve this system, we apply the elementary row transformations (Alternatively, we can use Cramer's rule as well) on the coefficient matrix:
Apply R_{2} → 4R_{2}  R_{1},
\(\left[\begin{array}{ll}
4& 4 \\ \\
0 & 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\ \\
0
\end{array}\right]\)
Expanding as equations,
4x + 4y = 0
We have one equation in two variables. So assume y = t. Then
4x + 4t = 0 ⇒ 4x = 4t ⇒ x = t
So the solution is, \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
t \\ \\
t
\end{array}\right]\) = t \(\left[\begin{array}{l}
1 \\ \\
1
\end{array}\right]\)
Thus, the eigenvector is \(\left[\begin{array}{l}
1 \\ \\
1
\end{array}\right]\).
When λ = 6:
Substitute λ = 1 in the equation:
(A  λI) v = O
\(\left[\begin{array}{ll}
5 6 & 4 \\ \\
1 & 2 6
\end{array}\right]\) \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\ \\
0
\end{array}\right]\)
\(\left[\begin{array}{ll}
1& 4 \\ \\
1 & 4
\end{array}\right]\) \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\ \\
0
\end{array}\right]\)
To solve this system, we apply the elementary row transformations (Alternatively, we can find Cramer's rule as well) on the coefficient matrix:
Apply R_{2} → R_{2} + R_{1},
\(\left[\begin{array}{ll}
1& 4 \\ \\
0 & 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\ \\
0
\end{array}\right]\)
Expanding as equations,
x + 4y = 0
Let y = t. Then x + 4t = 0 ⇒ x = 4t.
So the solution is, \(\left[\begin{array}{l}
x \\ \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
4t \\ \\
t
\end{array}\right]\) = t \(\left[\begin{array}{l}
4 \\ \\
1
\end{array}\right]\)
Thus, the eigenvector is \(\left[\begin{array}{l}
4 \\ \\
1
\end{array}\right]\).
Thus, the eigenvectors of the given 2 × 2 matrix are \(\left[\begin{array}{l}
1 \\ \\
1
\end{array}\right]\) and \(\left[\begin{array}{l}
4 \\ \\
1
\end{array}\right]\).
How to Find Eigenvectors of a 3 × 3 matrix?
We follow the same steps as above for the 3 × 3 matrix as well. Here is an example.
Example: Find the eigenvectors of 3 × 3 matrix A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\).
Solution:
Let λ be the eigenvalue and v = \(\left[\begin{array}{l}
x \\
y \\z
\end{array}\right]\) be the eigenvector of A.
Finding eigenvalues:
The characteristic equation is:
A  λI = 0
\(\left\begin{array}{lll}
1 λ & 1 & 1 \\
1 & 1 λ & 1 \\
1 & 1 & 1 λ
\end{array}\right\) = 0
Calculating the determinant, we get
λ^{3 }+ 3λ^{2} = 0
λ^{2} (λ+3) = 0
λ = 0, λ = 3.
Thus, the eigenvalues are 0 and 3. Let us find the corresponding eigenvectors.
When λ = 0:
(A  λI) v = 0
\(\left[\begin{array}{lll}
1 0 & 1 & 1 \\
1 & 1 0 & 1 \\
1 & 1 & 1 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y\\z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0\\0
\end{array}\right]\)
\(\left[\begin{array}{lll}
1& 1 & 1 \\
1 & 1& 1 \\
1 & 1 & 1
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y\\z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0\\0
\end{array}\right]\)
Apply R_{2} → R_{2}  R_{1} and R_{3} → R_{3}  R_{1},
\(\left[\begin{array}{lll}
1& 1 & 1 \\
0 & 0& 0\\
0 & 0 & 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y\\z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0\\0
\end{array}\right]\)
Expanding the above matrix, we get
x + y + z = 0
We have only one equation with two unknowns. So let us assume two of the variables to be y = t_{1} and z = t_{2}. Then the above equation becomes:
x + t_{1} + t_{2} = 0 ⇒ x =  t_{1}  t_{2}.
Thus, the eigenvector is:
\(\left[\begin{array}{l}
x \\
y \\z
\end{array}\right]\) = \(\left[\begin{array}{l}
 t_{1}  t_{2} \\
t_{1} \\t_{2}
\end{array}\right]\) = t_{1} \(\left[\begin{array}{l}
1 \\
1 \\ 0
\end{array}\right]\) + t_{2} \(\left[\begin{array}{l}
1 \\
0 \\ 1
\end{array}\right]\).
Thus, the eigenvectors that correspond to λ = 0 are \(\left[\begin{array}{l}
1 \\
1 \\ 0
\end{array}\right]\) and \(\left[\begin{array}{l}
1 \\
0 \\ 1
\end{array}\right]\).
When λ = 3:
(A  λI) v = 0
\(\left[\begin{array}{lll}
1 3 & 1 & 1 \\
1 & 1 3 & 1 \\
1 & 1 & 1 3
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y\\z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0\\0
\end{array}\right]\)
\(\left[\begin{array}{lll}
2& 1 & 1 \\
1 & 2& 1 \\
1 & 1 & 2
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y\\z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0\\0
\end{array}\right]\)
Apply R_{2} → 2R_{2} + R_{1} and R_{3} → 2R_{3} + R_{1},
\(\left[\begin{array}{lll}
2& 1 & 1 \\
0 & 3& 3\\
0 & 3 & 3
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y\\z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0\\0
\end{array}\right]\)
Now, apply R_{3} → R_{3} + R_{2},
\(\left[\begin{array}{lll}
2& 1 & 1 \\
0 & 3& 3\\
0 & 0 & 0
\end{array}\right]\) \(\left[\begin{array}{l}
x \\
y\\z
\end{array}\right]\) = \(\left[\begin{array}{l}
0 \\
0\\0
\end{array}\right]\)
Expanding the above matrix, we get
2x + y + z = 0
3y + 3z = 0
Let z = t. Then
3y + 3z = 0 ⇒ z = y = t.
2x + y + z = 0 ⇒ 2x + t + t = 0 ⇒ x = t.
Thus, the eigenvector is:
\(\left[\begin{array}{l}
x \\
y \\z
\end{array}\right]\) = \(\left[\begin{array}{l}
t \\
t \\t
\end{array}\right]\) = t \(\left[\begin{array}{l}
1 \\
1 \\ 1
\end{array}\right]\).
Thus, the eigenvector that correspond to λ = 3 is \(\left[\begin{array}{l}
1 \\
1 \\ 1
\end{array}\right]\).
Thus, the eigenvectors of the given 3 × 3 matrix are \(\left[\begin{array}{l}
1 \\
1 \\ 0
\end{array}\right]\), \(\left[\begin{array}{l}
1 \\
0 \\ 1
\end{array}\right]\), and \(\left[\begin{array}{l}
1 \\
1 \\ 1
\end{array}\right]\).
Finding Eigenspace
The eigenspace of a matrix (linear transformation) is the set of all of its eigenvectors. i.e., to find the eigenspace:
 Find eigenvalues first.
 Then find the corresponding eigenvectors.
 Just enclose all the eigenvectors in a set (Order doesn't matter).
From the above example, the eigenspace of A is, \(\left\{\left[\begin{array}{l}
1 \\
1 \\ 0
\end{array}\right], \left[\begin{array}{l}
1 \\
0 \\ 1
\end{array}\right], \left[\begin{array}{l}
1 \\
1 \\ 1
\end{array}\right]\right\}\).
Note that the vectors in the eigenspace are linearly independent.
Diagonalize Matrix Using Eigenvalues and Eigenvectors
Diagonalizing a matrix A is the process of writing it as the product of three matrices such that the middle one is a diagonal matrix, i.e. A = XDX^{1}, where D is the matrix of eigenvalues (to find D, take the identity matrix of the same order as A, replace 1s in it by eigenvalues) and X is the matrix of eigenvectors that are written in the same order as eigenvalues in D. Here is an example.
Example: Diagonalize the matrix \(=\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\).
Solution:
We have already found that the eigenvalues are 0, 0, and 3. The corresponding eigenvectors are, \(\left[\begin{array}{l}
1 \\
1 \\ 0
\end{array}\right]\), \(\left[\begin{array}{l}
1 \\
0 \\ 1
\end{array}\right]\), and \(\left[\begin{array}{l}
1 \\
1 \\ 1
\end{array}\right]\). Thus,
D = \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 3
\end{array}\right]\) and X = \(\left[\begin{array}{rcr}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{array}\right]\).
Then the inverse of the matrix X is, X^{1 }_{=}^{ }\(\left[\begin{array}{rcr}
1/3 & 2/3 & 1/3 \\
1/3 & 1/3 & 2/3 \\
1/3 & 1/3 & 1/3
\end{array}\right]\). To know how to find inverse, click here.
Thus, the diagonalization of A is:
A = \(\left[\begin{array}{rcr}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{array}\right] \left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 3
\end{array}\right] \left[\begin{array}{rcr}
1/3 & 2/3 & 1/3 \\
1/3 & 1/3 & 2/3 \\
1/3 & 1/3 & 1/3
\end{array}\right]\).
We can diagonalize a matrix A only when the determinant of the corresponding matrix X is NOT zero. Because if det (X) = 0, we cannot find X^{1}.
Eigenvectors Examples

Example 1: Find the eigenvectors of \(\left[\begin{array}{lll}
2 & 1 & 0 \\
0 & 2 & 1 \\
0 & 0 & 2
\end{array}\right]\).Solution:
To find the eigenvectors (v = \(\left[\begin{array}{lll}
x \\
y \\
z
\end{array}\right]\)), we should first find the eigenvalues(λ). For this,A  λI = 0
\(\left\begin{array}{lll}
2 λ & 1 & 0 \\
0 & 2 λ & 1 \\
0 & 0 & 2 λ
\end{array}\right\) = 0(2  λ)^{3} = 0
λ = 2, 2, 2
Each eigenvalue is equal to 2. So A has 3 identical eigenvectors. Let us find the eigenvector by substituting λ = 2 in the equation (A  λI) v = O.
\(\left[\begin{array}{lll}
2 2 & 1 & 0 \\
0 & 2 2 & 1 \\
0 & 0 & 2 2
\end{array}\right]\) \(\left[\begin{array}{lll}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 \\
0 \\
0
\end{array}\right]\)\(\left[\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0
\end{array}\right]\) \(\left[\begin{array}{lll}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{lll}
0 \\
0 \\
0
\end{array}\right]\)Expanding the matrix equation as equations, we get y = 0, and z = 0. Let x = t. Then the eigenvector is :
\(\left[\begin{array}{lll}
x \\
y \\
z
\end{array}\right]\) = \(\left[\begin{array}{lll}
t \\
0 \\
0
\end{array}\right]\) = t \(\left[\begin{array}{lll}
1 \\
0 \\
0
\end{array}\right]\)Thus, the eigenvector is \(\left[\begin{array}{lll}
1 \\
0 \\
0
\end{array}\right]\).Answer: The given 3 × 3 matrix has only one eigenvector which is \(\left[\begin{array}{lll}
1 \\
0 \\
0
\end{array}\right]\). 
Example 2: Find the eigenspace that corresponds to the matrix given in Example 1.
Solution:
We know that eigenspace is the set of eigenvectors.
The matrix A given in Example 1 has only one eigenvector and hence its eigenspace is, \(\left\{\left[\begin{array}{lll}
1 \\
0 \\
0
\end{array}\right]\right\}\).Answer: \(\left\{\left[\begin{array}{lll}
1 \\
0 \\
0
\end{array}\right]\right\}\). 
Example 3: Is it possible to diagonalize the matrix A given in Example 1? Justify your answer.
Solution:
Let us assume that the diagonalization of A = XDX^{1}. Here,
D = diagonal matrix formed by the eigenvalues = \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right]\).The corresponding matrix X is formed by eigenvectors. So X = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\).Since one of its rows is formed by all zeros, det (X) = 0 (verify the properties of determinants). Hence X^{1} is NOT defined. Hence A is nondiagonalizable.
Answer: A is NOT diagonalizable.
FAQs on Eigenvectors
How to Find Eigenvectors?
The eigenvector v of a square matrix A is a vector that satisfies Av = λv. Here, λ is a scalar and is called the eigenvalue that corresponds to the eigenvector v. To find the eigenvectors of a matrix A:
 First find its eigenvalues by solving the equation (with determinant) A  λI = 0 for λ.
 Then substitute each eigenvalue in Av = λv and solve it for v.
What is the Difference Between Eigen Value and Eigen Vector?
For any square matrix A, if Av = λv, then v is called the eigenvector and λ is called the eigenvalue. In other words, a vector that transforms at most by a scale factor when a transformation is applied is known as an eigenvector and the corresponding scale factor is known as the eigenvalue.
How to Calculate Eigenvalues and Eigenvectors?
For any square matrix A,
 To find eigenvalues: Solve the characteristic equation A  λI = 0 for λ.
 To find the eigenvectors: Solve the equation (A  λI) v = O for v.
What is the Diagonalizable Matrix?
A diagonalizable matrix is a matrix that can be expressed as the product of 3 matrices XDX^{1}, where D is a diagonal matrix. We diagonalize a matrix using eigenvalues and eigenvectors.
How to Diagonalize Matrix?
To diagonalize matrix A:
 Find its eigenvalues and replace them in the place of 1 in the identity matrix of the same order as A and denote the resultant matrix as D.
 Find the corresponding eigenvectors and write the matrix X with eigenvectors in the same order as eigenvalues of D are written.
 Find the inverse of the matrix. i.e., X^{1}.
 Write A = XDX^{1}, which is the diagonalization of matrix A.
Are Eigenvalues Same as Eigenvectors?
No, they are not same. Here are the differences between eigenvalues and eigenvectors.
Eigenvalues  Eigenvectors 

Eigenvalues are scalars.  Eigenvectors are vectors. 
Eigenvalues of a square matrix A are obtained by solving the equation A  λI = 0 for λ.  Eigenvectors of a square matrix A are obtained by solving the equation (A  λI) v = O for v. 
The concept of determinants is involved in finding eigenvalues.  Elementary transformations might be useful in finding eigenvectors. 
Can Eigenvector be a Zero Vector?
No, by the definition of eigenvectors, they are never zero vectors. If the zero vector is an eigenvector then A (O) = λ (O) is true for all values of λ and hence the corresponding eigenvalue is NOT defined.
What is Eigenvectors Formula?
To find the eigenvectors of a square matrix A, it is necessary to find its eigenvectors first by solving the characteristic equation A  λI = 0. Here, the values of λ represent the eigenvalues. Then by substituting each of them in the equation (A  λI) v = O and solving it for v would give the eigenvectors.
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