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| Boilerpipe Text | Eigenvectors
are concepts used in linear algebra to analyze the properties of square matrices. They are crucial for understanding how a matrix transformation affects the direction of vectors along them. Eigenvalues provide scaling information for these transformations, finding use in physics, engineering, computer graphics, and other fields involving matrices to represent systems and transformations.
Formal definition
An eigenvector of a square matrix is a non-zero vector that becomes equal to a scaled version of the same vector when multiplied by the matrix.
Consider a square matrix
A
A
of size
n
×
n
n \times n
, and if x is a non-zero vector, the product of
A
A
and
x
x
can be represented as
λ
x
\lambda x
, where
x
x
is the eigenvector, and
λ
\lambda
is the scalar value known as the eigenvalue associated with matrix
A
A |
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# How to calculate eigenvectors
**Eigenvectors** are concepts used in linear algebra to analyze the properties of square matrices. They are crucial for understanding how a matrix transformation affects the direction of vectors along them. Eigenvalues provide scaling information for these transformations, finding use in physics, engineering, computer graphics, and other fields involving matrices to represent systems and transformations.
##
### Formal definition
An eigenvector of a square matrix is a non-zero vector that becomes equal to a scaled version of the same vector when multiplied by the matrix.
Consider a square matrix A A A of size n × n n \\times n n×n, and if x is a non-zero vector, the product of A A A and x x x can be represented as λ x \\lambda x λx, where x x x is the eigenvector, and λ \\lambda λ is the scalar value known as the eigenvalue associated with matrix A A A
A
x
\=
λ
x
Ax=\\lambda x
Ax\=λx
As an example, we can see below that we have an eigenvector x \= \[ 1 2 \] x=\\begin{bmatrix} 1 \\\\ 2 \\end{bmatrix} x\=\[12\]and a square matrix A \= \[ 3 0 8 − 1 \] A=\\begin{bmatrix} 3 & 0 \\\\ 8 & -1 \\end{bmatrix} A\=\[380−1\]where the result of A x Ax Ax gives us the vector x x x back but with a scaling of λ \= 3 \\lambda =3 λ\=3.
Example of an eigenvector with its eigen value
### Calculating eigenvectors
To find the eigenvectors of a square matrix, follow these steps:
1. Given a square matrix A A A of size n × n n \\times n n×n, start by finding the eigenvalues of the matrix. To do this, solve the characteristic equation:
d
e
t
(
A
−
λ
I
)
\=
0
det(A-\\lambda I) = 0
det(A−λI)\=0
> **Note:** Here, I I I represents the identity matrix of the same size as A A A, and λ \\lambda λ is the unknown eigenvalue.
1. After obtaining the eigenvalues ( λ 1 , λ 2 , . . . , λ n ) (\\lambda\_{1},\\lambda\_{2},...,\\lambda\_{n}) (λ1,λ2,...,λn), each eigenvalue corresponds to a specific eigenvector.
2. For each eigenvalue λ i \\lambda\_{i} λi, solve the equation ( A − λ i I ) x i \= 0 (A-\\lambda\_{i} I)x\_{i} = 0 (A−λiI)xi\=0, where x i x\_{i} xiis the eigenvector corresponding to λ i \\lambda\_{i} λi. This equation can be rewritten as:
(
A
−
λ
i
I
)
x
i
\=
0
⟹
A
x
i
\=
λ
i
x
i
(A-\\lambda\_{i}I)x\_{i} = 0 \\implies Ax\_{i} =\\lambda\_{i}x\_{i}
(A−λiI)xi\=0⟹Axi\=λixi
1. Solve the system of linear equations ( A − λ i I ) x i \= 0 (A-\\lambda\_{i} I)x\_{i} = 0 (A−λiI)xi\=0 to find the non-zero vector x i x\_{i} xi. Keep in mind that the eigenvector is unique only up to a scalar multiple, so the resulting eigenvector may be normalized.
2. Repeat steps 3 and 4 for each eigenvalue to find all the corresponding eigenvectors.
> **Note:** It's important to remember that not all square matrices have distinct eigenvalues, and in some cases, the process of finding eigenvalues and eigenvectors might involve more complex calculations or numerical methods, especially for larger matrices. However, for
>
> (
>
> 2
>
> ×
>
> 2
>
> )
>
> (2 \\times 2)
>
> (2×2)
>
> and
>
> (
>
> 3
>
> ×
>
> 3
>
> )
>
> (3 \\times 3)
>
> (3×3)
>
> matrices, the above steps are generally sufficient to find eigenvalues and eigenvectors.
### Example
We will see a basic example of finding eigenvectors and eigenvalues given a square matrix A A A.
- **Question:** Find the eigenvectors of the matrix A \= \[ 3 0 8 − 1 \] A=\\begin{bmatrix} 3 & 0 \\\\ 8 & -1 \\end{bmatrix} A\=\[380−1\]. Note that this matrix is similar to the one we saw above. Let's calculate its eigenvectors using the above steps.
- **Answer:** Following the first step, we must set up the characteristic equation and solve for eigenvalues.
d
e
t
(
A
−
λ
I
)
\=
0
⟹
d
e
t
(
\[
3
0
8
−
1
\]
−
λ
\[
1
0
0
1
\]
)
\=
0
det(A-\\lambda I) = 0 \\implies det (\\begin{bmatrix} 3 & 0 \\\\ 8 & -1 \\end{bmatrix} -\\lambda \\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix} )=0
det(A−λI)\=0⟹det(\[380−1\]−λ\[1001\])\=0
d
e
t
(
\[
3
−
λ
0
8
−
1
−
λ
\]
)
\=
0
det (\\begin{bmatrix} 3-\\lambda & 0 \\\\ 8 & -1-\\lambda \\end{bmatrix} )= 0
det(\[3−λ80−1−λ\])\=0
(
3
−
λ
)
(
−
1
−
λ
)
\=
0
(3-\\lambda)(-1-\\lambda) = 0 \\text{ }
(3−λ)(−1−λ)\=0
λ
1
\=
3
,
λ
2
\=
−
1
\\lambda\_{1} =3,\\lambda\_{2} = -1
λ1\=3,λ2\=−1
Next, we need to solve the equation ( A − λ i I ) x i \= 0 (A-\\lambda\_{i} I)x\_{i} = 0 (A−λiI)xi\=0 for each λ i \\lambda\_{i} λi.
We will start solving with λ 1 \= 3 \\lambda\_{1} =3 λ1\=3 first, and then we'll solve for λ 2 \= − 1 \\lambda\_{2}=-1 λ2\=−1
### Solving for eigenvector x 1 x\_{1} x1
Suppose x 1 \= \[ x y \] x\_{1}= \\begin{bmatrix} x \\\\ y \\end{bmatrix} x1\=\[xy\]then we can find it using the ( A − λ 1 I ) x 1 \= 0 (A-\\lambda\_{1} I)x\_{1} = 0 (A−λ1I)x1\=0 formula.
(
A
−
λ
1
I
)
x
1
\=
0
⟹
(
\[
3
0
8
−
1
\]
−
3
\[
1
0
0
1
\]
)
x
1
\=
\[
0
0
\]
(A-\\lambda\_{1} I)x\_{1} = 0 \\implies (\\begin{bmatrix} 3 & 0 \\\\ 8 & -1 \\end{bmatrix}-3\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix})x\_{1} = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}
(A−λ1I)x1\=0⟹(\[380−1\]−3\[1001\])x1\=\[00\]
\[
0
0
8
−
4
\]
x
1
\=
\[
0
0
\]
\\begin{bmatrix} 0 & 0 \\\\ 8 & -4 \\end{bmatrix}x\_{1} = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}
\[080−4\]x1\=\[00\]
8
x
−
4
y
\=
0
⟹
y
\=
2
x
8x-4y=0 \\implies y = 2x
8x−4y\=0⟹y\=2x
We can introduce a free variable s s s and let s \= x s=x s\=x then y \= 2 s y=2s y\=2s. Now we can get our eigenvector in matrix form x 1 \= \[ 1 2 \] s x\_{1} = \\begin{bmatrix} 1 \\\\ 2 \\end{bmatrix}s x1\=\[12\]s .
### Solving for eigenvector x 2 x\_{2} x2
Suppose x 2 \= \[ x y \] x\_{2}= \\begin{bmatrix} x \\\\ y \\end{bmatrix} x2\=\[xy\]then we can find it using the ( A − λ 2 I ) x 2 \= 0 (A-\\lambda\_{2} I)x\_{2} = 0 (A−λ2I)x2\=0 formula.
(
A
−
λ
2
I
)
x
2
\=
0
⟹
(
\[
3
0
8
−
1
\]
−
(
−
1
)
\[
1
0
0
1
\]
)
x
2
\=
\[
0
0
\]
(A-\\lambda\_{2} I)x\_{2} = 0 \\implies (\\begin{bmatrix} 3 & 0 \\\\ 8 & -1 \\end{bmatrix}-(-1)\\begin{bmatrix} 1 & 0 \\\\ 0 & 1 \\end{bmatrix})x\_{2} = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}
(A−λ2I)x2\=0⟹(\[380−1\]−(−1)\[1001\])x2\=\[00\]
\[
4
0
8
0
\]
x
2
\=
\[
0
0
\]
\\begin{bmatrix} 4 & 0 \\\\ 8 & 0 \\end{bmatrix}x\_{2} = \\begin{bmatrix} 0 \\\\ 0 \\end{bmatrix}
\[4800\]x2\=\[00\]
4
x
\+
0
\=
0
8
x
\+
0
\=
0
x
\=
0
4x+0 =0\\\\8x+0=0 \\\\x=0
4x\+0\=08x\+0\=0x\=0
We can introduce a free variable s s s and let s \= y s=y s\=y. Now we can get our second eigenvector in matrix form x 2 \= \[ 0 1 \] s x\_{2} = \\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix}s x2\=\[01\]s .
We can summarize the results as follows:
- λ 1 \= 3 , x 1 \= \[ 1 2 \] \\lambda\_{1} = 3 , x\_{1}= \\begin{bmatrix} 1 \\\\ 2 \\end{bmatrix} λ1\=3,x1\=\[12\]
- λ 2 \= − 1 , x 2 \= \[ 0 1 \] \\lambda\_{2} = -1 , x\_{2}= \\begin{bmatrix} 0 \\\\ 1 \\end{bmatrix} λ2\=−1,x2\=\[01\]
### Quiz
Now that you know how to find eigenvectors and eigenvalues, try this quiz to test your understanding.
Eigenvectors and eigenvalues
Q
Find eigenvectors and corresponding eigen values of the matrix A \= \[ − 1 3 2 0 \] A=\\begin{bmatrix} -1 & 3 \\\\ 2 & 0 \\end{bmatrix} A\=\[−1230\]
A)
x 1 \= \[ 1 1 \] , λ 1 \= 2 x\_{1}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix},\\lambda\_{1}=2 x1\=\[11\],λ1\=2
x 2 \= \[ − 3 2 1 \] , λ 2 \= − 3 x\_{2}=\\begin{bmatrix} \\frac{-3}{2} \\\\ 1 \\end{bmatrix},\\lambda\_{2}=-3 x2\=\[2−31\],λ2\=−3
B)
x 1 \= \[ 0 5 \] , λ 1 \= 2 x\_{1}=\\begin{bmatrix} 0 \\\\ 5 \\end{bmatrix},\\lambda\_{1}=2 x1\=\[05\],λ1\=2
x 2 \= \[ 5 2 1 \] , λ 2 \= 10 x\_{2}=\\begin{bmatrix} \\frac{5}{2} \\\\ 1 \\end{bmatrix},\\lambda\_{2}=10 x2\=\[251\],λ2\=10
C)
x 1 \= \[ 1 1 \] , λ 1 \= 6 x\_{1}=\\begin{bmatrix} 1 \\\\ 1 \\end{bmatrix},\\lambda\_{1}=6 x1\=\[11\],λ1\=6
x 2 \= \[ − 1 2 1 \] , λ 2 \= − 3 x\_{2}=\\begin{bmatrix} \\frac{-1}{2} \\\\ 1 \\end{bmatrix},\\lambda\_{2}=-3 x2\=\[2−11\],λ2\=−3
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### Conclusion
Understanding eigenvectors and eigenvalues is crucial in various fields, including linear algebra, physics, and data analysis. These concepts help us analyze the behavior of matrices and systems, enabling us to find important patterns and solutions.
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| Readable Markdown | **Eigenvectors** are concepts used in linear algebra to analyze the properties of square matrices. They are crucial for understanding how a matrix transformation affects the direction of vectors along them. Eigenvalues provide scaling information for these transformations, finding use in physics, engineering, computer graphics, and other fields involving matrices to represent systems and transformations.
### Formal definition
An eigenvector of a square matrix is a non-zero vector that becomes equal to a scaled version of the same vector when multiplied by the matrix.
Consider a square matrix A A of size n × n n \\times n, and if x is a non-zero vector, the product of A A and x x can be represented as λ x \\lambda x, where x x is the eigenvector, and λ \\lambda is the scalar value known as the eigenvalue associated with matrix A A |
| Shard | 28 (laksa) |
| Root Hash | 12990463358539855228 |
| Unparsed URL | io,educative!www,/answers/how-to-calculate-eigenvectors s443 |