How to use Eigenvalue and Eigenvector calculator?
Here are simple steps to use the Eigenvalue and Eigenvector Calculator:
Step 1: Define the Matrix Size.
If the size of the matrix is 2\times 2, use the first calculator.
If the size of the matrix is 3\times 3, use the second calculator.
If the size of the matrix is 4\times 4, use the third calculator.
Step 2: Enter Matrix Entries.
Fill in the Matrix Entries: Input the values of the matrix into the given fields. For instance, if you have a 2 \times 2 matrix:
Enter the value for the 1st row, and 1st column in the first input box.
Enter the value for the 1st row, and 2nd column in the second input box.
Enter the value for the 2nd row, and 1st column in the third input box.
Enter the value for the 2nd row, and 2nd column in the fourth input box.
Step 3: Calculate Eigenvalues and Eigenvectors.
- Submit the Form: Click the “Calculate” button to compute the eigenvalues and eigenvectors.
- View Results: The results will be displayed below the input fields, showing the eigenvalues and corresponding eigenvectors.
What Are Eigenvalues and Eigenvectors?
Eigenvalues: Eigenvalues are scalar values that, when a linear transformation is applied to an eigenvector, only scale the eigenvector and do not change its direction.
In other words, they are the factors by which the eigenvectors are stretched or compressed.
Eigenvectors: Eigenvectors are non-zero vectors that, when a linear transformation is applied to them, change in magnitude but not in direction.
These vectors remain parallel to their original direction after the transformation.
Mathematical Explanation:
If ๐ is anย eigenvalue and ๐ฃ is an eigenvector of a matrix of the square matrix A, then we can write,
Av=\lambda v
Where,
A is a square matrix.
๐ฃ is an eigenvector of matrix A.
๐ is an eigenvalue corresponding to the eigenvector ๐ฃ.
How to find Eigenvalues and Eigenvectors?
To find the eigenvalues, solve the characteristic equation:
\det(A-\lambda I)=0
Here, \det denotes the determinant of the matrix, ๐ is the eigenvalue, and I is the identity matrix of the same dimension as A. This equation results in a polynomial known as the characteristic polynomial, and its roots are the eigenvalues of A.
After determining the eigenvalues, we can find the eigenvectors by solving this equation:
(A-\lambda I)v=0
This involves substituting each eigenvalue ๐ back into the equation and solving for the vector v.
Example of Eigenvalues and Eigenvectors:
Find the eigenvalue and eigenvector of 2\times 2 matrix A=\begin{pmatrix}4&1\\ 2&3\end{pmatrix}
Step-1: Find eigenvalues.
To find the eigenvalues, we solve the characteristic equation:
\det(A-\lambda I)=0\\
\det \left(\begin{pmatrix}4&1\\ 2&3\end{pmatrix}- \lambda \begin{pmatrix}1&0\\ 0&1\end{pmatrix}\right)=0\\
\det \begin{pmatrix}4- \lambda &1\\ 2&3- \lambda \end{pmatrix}=0\\
\left(4- \lambda \right)\left(3- \lambda \right)-1\cdot 2=0\\
12-7 \lambda + \lambda ^2-2=0\\
\lambda ^2-7 \lambda +10=0\\
\lambda ^2-5 \lambda -2 \lambda +10=0\\
\lambda ( \lambda -5)-2( \lambda -5)=0\\
( \lambda -2)( \lambda -5)=0\\
\lambda =2, \lambda =5\\
The eigenvalues of the matrix A are \lambda =2 and \lambda =5.
Step-2: Find eigenvector for the eigenvalue \lambda =2.
To find the eigenvector for \lambda =2, we solve the equation:
(Aโ2I)v=0\\
\left(\begin{pmatrix}4&1\\ 2&3\end{pmatrix}-2\begin{pmatrix}1&0\\ 0&1\end{pmatrix}\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\\
\begin{pmatrix}2&1\\ 2&1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\\
2x+y=0 \:\:\:\:\: \rightarrow \:\:\:\:\: y=-2x\\
x=x \:\:\:\:\: \rightarrow \:\:\:\:\: \text{free\:variable}\\
The solution is,
\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1\\ -2\end{pmatrix}x
The eigenvector for the eigenvalue \lambda =2 is,
v_1=\begin{pmatrix}1\\ -2\end{pmatrix}
Or we can write,
v_1=\begin{pmatrix}-1\\ 2\end{pmatrix}
Step-3: Find eigenvector for the eigenvalue \lambda =5.
To find the eigenvector for \lambda =5, we solve the equation:
(Aโ5I)v=0\\
\left(\begin{pmatrix}4&1\\ 2&3\end{pmatrix}-5\begin{pmatrix}1&0\\ 0&1\end{pmatrix}\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\\
\begin{pmatrix}-1&1\\ 2&-2\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\\
x-y=0 \:\:\:\:\: \rightarrow \:\:\:\:\: y=x\\
x=x \:\:\:\:\: \rightarrow \:\:\:\:\: \text{free\:variable}\\
The solution is,
\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1\\ 1\end{pmatrix}x
The eigenvector for the eigenvalue \lambda =5 is,
v_2=\begin{pmatrix}1\\ 1\end{pmatrix}
Eigenvectors of the matrix A are v_1=\begin{pmatrix}-1\\ 2\end{pmatrix} and v_2=\begin{pmatrix}1\\ 1\end{pmatrix}.
Find the eigenvalue and eigenvector of 3\times 3 matrix A=\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}
Step-1: Find eigenvalues.
To find the eigenvalues, we solve the characteristic equation:
\det(A-\lambda I)=0\\
\det \left(\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}- \lambda \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)=0\\
\det \begin{pmatrix}1- \lambda &2&1\\ 6&-1- \lambda &0\\ -1&-2&-1- \lambda \end{pmatrix}=0\\
\left(1- \lambda \right)\det \begin{pmatrix}-1- \lambda &0\\ -2&-1- \lambda \end{pmatrix}-2\cdot \det \begin{pmatrix}6&0\\ -1&-1- \lambda \end{pmatrix}+1\cdot \det \begin{pmatrix}6&-1- \lambda \\ -1&-2\end{pmatrix}=0\\
(1- \lambda )[(-1- \lambda )^2-0]-2[6(-1- \lambda )-0]+1[-12+1(-1- \lambda )]=0\\
(1- \lambda )(-1- \lambda )^2-2(-6-6 \lambda )+1(-12-1- \lambda )=0\\
(1- \lambda )( \lambda +1)^2+12( \lambda +1)-13- \lambda =0\\
(1- \lambda )( \lambda ^2+2 \lambda +1)+12 \lambda +12-13- \lambda =0\\
\lambda ^2+2 \lambda +1- \lambda ^3-2 \lambda ^2- \lambda +11 \lambda -1=0\\
- \lambda ^3- \lambda ^2+12 \lambda =0\\
- \lambda ( \lambda ^2+ \lambda -12)=0\\
\lambda ( \lambda ^2+4 \lambda -3 \lambda -12)=0\\
\lambda ( \lambda ( \lambda +4)-3( \lambda +4))=0\\
\lambda ( \lambda -3)( \lambda +4)=0\\
\lambda =0, \lambda =3, \lambda =-4\\
The eigenvalues of the matrix A are \lambda =0, \lambda =3 and, \lambda =-4.
Step-2: Find eigenvector for the eigenvalue \lambda =0.
To find the eigenvector for \lambda =0, we solve the equation:
(Aโ0I)v=0\\
\left(\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-0\cdot \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_2\leftarrow R_2-6R_1\\
R_3\leftarrow R_3+1R_1\\
\begin{pmatrix}1&2&1\\ 0&-13&-6\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_2\leftarrow -\frac{1}{13}R_2\\
\begin{pmatrix}1&2&1\\ \:0&1&\frac{6}{13}\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_1\leftarrow R_1-2R_2\\
\begin{pmatrix}1&0&\frac{1}{13}\\ 0&1&\frac{6}{13}\\0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
x+\frac{1}{13}z=0 \:\:\:\:\: \rightarrow \:\:\:\:\: x=-\frac{1}{13}z\\
y+\frac{6}{13}z=0 \:\:\:\:\: \rightarrow \:\:\:\:\: y=-\frac{6}{13}z\\
z=z \:\:\:\:\: \rightarrow \:\:\:\:\: \text{free\:variable}\\
The solution is,
\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-\frac{1}{13}z\\ -\frac{6}{13}z\\ z\end{pmatrix}
To avoid fractions, choose z=13.
\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-1\\ -6\\ 13\end{pmatrix}
The eigenvector for the eigenvalue \lambda =0 is,
v_1=\begin{pmatrix}-1\\ -6\\ 13\end{pmatrix}
Step-3: Find eigenvector for the eigenvalue \lambda =3.
To find the eigenvector for \lambda =3, we solve the equation:
(Aโ3I)v=0\\
\left(\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-3\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
\begin{pmatrix}-2&2&1\\ 6&-4&0\\ -1&-2&-4\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_1\leftarrow -\frac{1}{2}R_1\\
\begin{pmatrix}1&-1&-\frac{1}{2}\\ 6&-4&0\\ -1&-2&-4\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_2\leftarrow R_2-6R_1\\
R_3\leftarrow R_3+1R_1\\
\begin{pmatrix}1&-1&-\frac{1}{2}\\ 0&2&3\\ 0&-3&-\frac{9}{2}\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_2\leftarrow \frac{1}{2}R_2\\
\begin{pmatrix}1&-1&-\frac{1}{2}\\ 0&1&\frac{3}{2}\\ 0&-3&-\frac{9}{2}\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_1\leftarrow R_1+1R_2\\
R_3\leftarrow R_3+3R_2\\
\begin{pmatrix}1&0&1\\ 0&1&\frac{3}{2}\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
x+z=0 \:\:\:\:\: \rightarrow \:\:\:\:\: x=-z\\
y+\frac{3}{2}z=0 \:\:\:\:\: \rightarrow \:\:\:\:\: y=-\frac{3}{2}z\\
z=z \:\:\:\:\: \rightarrow \:\:\:\:\: \text{free\:variable}\\
The solution is,
\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-z\\ -\frac{3}{2}z\\ z\end{pmatrix}
To avoid fractions, choose z=2.
\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-2\\ -3\\ 2\end{pmatrix}
The eigenvector for the eigenvalue \lambda =3 is,
v_2=\begin{pmatrix}-2\\ -3\\ 2\end{pmatrix}
Step-4: Find eigenvector for the eigenvalue \lambda =-4.
To find the eigenvector for \lambda =-4, we solve the equation:
(Aโ(-4)I)v=0\\
\left(\begin{pmatrix}1&2&1\\ 6&-1&0\\ -1&-2&-1\end{pmatrix}-\left(-4\right)\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\right)\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
\begin{pmatrix}5&2&1\\ 6&3&0\\ -1&-2&3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_1\leftarrow \frac{1}{5}R_1\\
\begin{pmatrix}1&\frac{2}{5}&\frac{1}{5}\\ 6&3&0\\ -1&-2&3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_2\leftarrow R_2-6R_1\\
R_3\leftarrow R_3+1R_1\\
\begin{pmatrix}1&\frac{2}{5}&\frac{1}{5}\\ 0&\frac{3}{5}&-\frac{6}{5}\\ 0&-\frac{8}{5}&\frac{16}{5}\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_2\leftarrow \frac{5}{3}R_2\\
\begin{pmatrix}1&\frac{2}{5}&\frac{1}{5}\\ 0&1&-2\\ 0&-\frac{8}{5}&\frac{16}{5}\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
R_1\leftarrow R_1-\frac{2}{5}R_2\\
R_3\leftarrow R_3+\frac{8}{5}R_2\\
\begin{pmatrix}1&0&1\\ 0&1&-2\\ 0&0&0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\end{pmatrix}=0\\
x+z=0 \:\:\:\:\: \rightarrow \:\:\:\:\: x=-z\\
y-2z=0 \:\:\:\:\: \rightarrow \:\:\:\:\: y=2z\\
z=z \:\:\:\:\: \rightarrow \:\:\:\:\: \text{free\:variable}\\
The solution is,
\begin{pmatrix}x\\ y\\ z\end{pmatrix}=z\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}
The eigenvector for the eigenvalue \lambda =-4 is,
v_3=\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}
Eigenvectors of the matrix A are v_1=\begin{pmatrix}-1\\ -6\\ 13\end{pmatrix}, v_2=\begin{pmatrix}-2\\ -3\\ 2\end{pmatrix}, v_3=\begin{pmatrix}-1\\ 2\\ 1\end{pmatrix}
FAQs on Eigenvalues and Eigenvectors:
1) What is the geometric interpretation of eigenvalues and eigenvectors?
- Eigenvectors represent directions in which the transformation stretches or compresses the space.
- Eigenvalues indicate the factor by which the transformation stretches or compresses along the eigenvector direction.
2) Can eigenvalues be zero?
Yes, an eigenvalue can be zero. This indicates that the transformation compresses the eigenvector to a point.
3) Are eigenvectors unique?
Eigenvectors corresponding to a particular eigenvalue are not unique.
If ๐ฃ is an eigenvector, then any scalar multiple of ๐ฃ is also an eigenvector corresponding to the same eigenvalue.
4) What is the difference between left and right eigenvectors?
- Right Eigenvectors: Satisfy Av=\lambda v.
- Left Eigenvectors: Satisfy v^T A=\lambda v^T.
5) Can non-square matrices have eigenvalues and eigenvectors?
No, only square matrices have eigenvalues and eigenvectors. Non-square matrices do not have well-defined eigenvalues or eigenvectors.
6) How do eigenvalues relate to the trace and determinant of a matrix?
- The eigenvalues of a matrix, when added together, give the same result as adding the elements on the matrix’s main diagonal.
- The product of the eigenvalues of a matrix gives you its determinant.
