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 ## Introduction to Eigenvalues and Eigenvectors A rectangular arrangement of numbers in the form of rows and columns is known as a matrix. In this article, we will discuss **Eigenvalues and Eigenvectors Problems and Solutions**. **Download Complete Chapter Notes of Matrices & Determinants** [Download Now]() Consider a square matrix n × n. If X is the non-trivial column vector solution of the matrix equation AX = λX, where λ is a scalar, then X is the eigenvector of matrix A, and the corresponding value of λ is the eigenvalue of matrix A. Suppose the matrix equation is written as A X – λ X = 0. Let I be the n × n identity matrix. If I X is substituted by X in the equation above, we obtain A X – λ I X = 0. The equation is rewritten as (A – λ I) X = 0. The equation above consists of non-trivial solutions if and only if the [determinant value of the matrix](https://byjus.com/jee/determinants/) is 0. The characteristic equation of A is Det (A – λ I) = 0. ‘A’ being an n × n matrix, if (A – λ I) is expanded, (A – λ I) will be the characteristic polynomial of A because its degree is n. ## Properties of Eigenvalues Let A be a matrix with eigenvalues λ1, λ2,…., λn. The following are the properties of eigenvalues. (1) The trace of A, defined as the sum of its diagonal elements, is also the sum of all eigenvalues, t r ( A ) \= ∑ i \= 1 n a i i \= ∑ i \= 1 n λ i \= λ 1 \+ λ 2 \+ ⋯ \+ λ n . (2) The determinant of A is the product of all its eigenvalues, det ( A ) \= ∏ i \= 1 n λ i \= λ 1 λ 2 ⋯ λ n . (3) The eigenvalues of the kth power of A, that is, the eigenvalues of Ak, for any positive integer k, are λ 1 k , … , λ n k . . (4) The matrix A is invertible if and only if every eigenvalue is nonzero. (5) If A is invertible, then the eigenvalues of A\-1 are 1 λ 1 , … , 1 λ n and each eigenvalue’s geometric multiplicity coincide. The characteristic polynomial of the inverse is the reciprocal polynomial of the original, the eigenvalues share the same algebraic multiplicity. (6) If A is equal to its conjugate transpose, or equivalently if A is Hermitian, then every eigenvalue is real. The same is true for any real symmetric matrix. (7) If A is not only Hermitian but also positive-definite, positive-semidefinite, negative-definite, or negative-semidefinite, then every eigenvalue is positive, non-negative, negative, or non-positive, respectively. (8) If A is unitary, every eigenvalue has absolute value \|λi\| = 1. (9) If A is a n×n matrix and {λ1, λ2,…., λk} are its eigenvalues, then the eigenvalues of the matrix I + A (where I is the identity matrix) are {λ1\+ 1, λ2\+1,…., λk\+1}. **Also, Read:** [Eigenvectors of a Matrix](https://byjus.com/jee/eigenvectors-of-a-matrix/) [Adjoint and Inverse of a Matrix](https://byjus.com/jee/adjoint-and-inverse-of-a-matrix/) [Normalized and Decomposition of Eigenvectors](https://byjus.com/jee/normalized-and-decomposition-of-eigenvectors/) ## Eigenvalues and Eigenvectors Solved Problems **Example 1:** Find the eigenvalues and eigenvectors of the following matrix. **Solution:**  **Example 2:** Find all eigenvalues and corresponding eigenvectors for the matrix A if ( 2 − 3 0 2 − 5 0 0 0 3 ) **Solution:** det ( ( 2 − 3 0 2 − 5 0 0 0 3 ) − λ ( 1 0 0 0 1 0 0 0 1 ) ) ( 2 − 3 0 2 − 5 0 0 0 3 ) − λ ( 1 0 0 0 1 0 0 0 1 ) λ ( 1 0 0 0 1 0 0 0 1 ) \= ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 − 3 0 2 − 5 0 0 0 3 ) − ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 − λ − 3 0 2 − 5 − λ 0 0 0 3 − λ ) \= det ( 2 − λ − 3 0 2 − 5 − λ 0 0 0 3 − λ ) \= ( 2 − λ ) det ( − 5 − λ 0 0 3 − λ ) − ( − 3 ) det ( 2 0 0 3 − λ ) \+ 0 ⋅ det ( 2 − 5 − λ 0 0 ) \= ( 2 − λ ) ( λ 2 \+ 2 λ − 15 ) − ( − 3 ) ⋅ 2 ( − λ \+ 3 ) \+ 0 ⋅ 0 \= − λ 3 \+ 13 λ − 12 − λ 3 \+ 13 λ − 12 \= 0 − ( λ − 1 ) ( λ − 3 ) ( λ \+ 4 ) \= 0 The eigenvalues are : λ \= 1 , λ \= 3 , λ \= − 4 Eigenvectors for λ \= 1 ( 2 − 3 0 2 − 5 0 0 0 3 ) − 1 ⋅ ( 1 0 0 0 1 0 0 0 1 ) \= ( 1 − 3 0 2 − 6 0 0 0 2 ) ( A − 1 I ) ( x y z ) \= ( 1 − 3 0 0 0 1 0 0 0 ) ( x y z ) \= ( 0 0 0 ) { x − 3 y \= 0 z \= 0 } I s o l a t e { z \= 0 x \= 3 y } Plug into ( x y z ) η \= ( 3 y y 0 ) y ≠ 0 Let y \= 1 ( 3 1 0 ) S i m i l a r l y Eigenvectors for λ \= 3 : ( 0 0 1 ) Eigenvectors for λ \= − 4 : ( 1 2 0 ) The eigenvectors for ( 2 − 3 0 2 − 5 0 0 0 3 ) \= ( 3 1 0 ) , ( 0 0 1 ) , ( 1 2 0 ) **Example 3:** Consider the matrix **** for some variable ‘a’. Find all values of ‘a’, which will prove that A has eigenvalues 0, 3, and −3. **Solution:** Let p (t) be the characteristic polynomial of A, i.e. let p (t) = det (A − tI) = 0. By expanding along the second column of A − tI, we can obtain the equation  \= (3 − t) \[(−2 −t) (−1 − t) − 4\] + 2\[(−2 − t) a + 5\] \= (3 − t) (2 + t + 2t + t2 −4) + 2 (−2a − ta + 5) \= (3 − t) (t2 \+ 3t − 2) + (−4a −2ta + 10) \= 3t2 \+ 9t − 6 − t3 − 3t2 \+ 2t − 4a − 2ta + 10 \= −t3 \+ 11t − 2ta + 4 − 4a \= −t3 \+ (11 − 2a) t + 4 − 4a For the eigenvalues of A to be 0, 3 and −3, the characteristic polynomial p (t) must have roots at t = 0, 3, −3. This implies p (t) = –t (t − 3) (t + 3) =–t(t2 − 9) = –t3 \+ 9t. Therefore, −t3 \+ (11 − 2a) t + 4 − 4a = −t3 + 9t. For this equation to hold, the constant terms on the left and right-hand sides of the above equation must be equal. This means that 4 − 4a = 0, which implies a = 1. Hence, A has eigenvalues 0, 3, and −3 precisely when a = 1. **Example 4:** Find the eigenvalues and eigenvectors of ( 2 0 0 0 3 4 0 4 9 ) **Solution:** det ( ( 2 0 0 0 3 4 0 4 9 ) − λ ( 1 0 0 0 1 0 0 0 1 ) ) ( 2 0 0 0 3 4 0 4 9 ) − λ ( 1 0 0 0 1 0 0 0 1 ) λ ( 1 0 0 0 1 0 0 0 1 ) \= ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 0 0 0 3 4 0 4 9 ) − ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 − λ 0 0 0 3 − λ 4 0 4 9 − λ ) \= det ( 2 − λ 0 0 0 3 − λ 4 0 4 9 − λ ) \= ( 2 − λ ) det ( 3 − λ 4 4 9 − λ ) − 0 ⋅ det ( 0 4 0 9 − λ ) \+ 0 ⋅ det ( 0 3 − λ 0 4 ) \= ( 2 − λ ) ( λ 2 − 12 λ \+ 11 ) − 0 ⋅ 0 \+ 0 ⋅ 0 \= − λ 3 \+ 14 λ 2 − 35 λ \+ 22 − λ 3 \+ 14 λ 2 − 35 λ \+ 22 \= 0 − ( λ − 1 ) ( λ − 2 ) ( λ − 11 ) \= 0 The eigenvalues are : λ \= 1 , λ \= 2 , λ \= 11 Eigenvectors for λ \= 1 ( 2 0 0 0 3 4 0 4 9 ) − 1 ⋅ ( 1 0 0 0 1 0 0 0 1 ) \= ( 1 0 0 0 2 4 0 4 8 ) ( A − 1 I ) ( x y z ) \= ( 1 0 0 0 1 2 0 0 0 ) ( x y z ) \= ( 0 0 0 ) { x \= 0 y \+ 2 z \= 0 } I s o l a t e { x \= 0 y \= − 2 z } Plug into ( x y z ) η \= ( 0 − 2 z z ) z ≠ 0 Let z \= 1 ( 0 − 2 1 ) S i m i l a r l y Eigenvectors for λ \= 2 : ( 1 0 0 ) Eigenvectors for λ \= 11 : ( 0 1 2 ) The eigenvectors for ( 2 0 0 0 3 4 0 4 9 ) \= ( 0 − 2 1 ) , ( 1 0 0 ) , ( 0 1 2 ) ## Frequently Asked Questions Q1 ### What do you mean by eigenvalues? Eigenvalues are the special set of scalar values associated with the set of linear equations in the matrix equations. Q2 ### Can the eigenvalue be zero? Yes, the eigenvalue can be zero. Q3 ### Can a singular matrix have eigenvalues? Every singular matrix has a 0 eigenvalue. Q4 ### How to find the eigenvalues of a square matrix A? Use the equation det(A-λI) = 0 and solve for λ. Determine all the possible values of λ, which are the required eigenvalues of matrix A. #### Comments ### Leave a Comment [Cancel reply](https://byjus.com/jee/eigenvalues-and-eigenvectors-problems-and-solutions/#respond)  #### Register with BYJU'S & Download Free PDFs #### Register with BYJU'S & Watch Live Videos

## Introduction to Eigenvalues and Eigenvectors A rectangular arrangement of numbers in the form of rows and columns is known as a matrix. In this article, we will discuss **Eigenvalues and Eigenvectors Problems and Solutions**. **Download Complete Chapter Notes of Matrices & Determinants** Download Now Consider a square matrix n × n. If X is the non-trivial column vector solution of the matrix equation AX = λX, where λ is a scalar, then X is the eigenvector of matrix A, and the corresponding value of λ is the eigenvalue of matrix A. Suppose the matrix equation is written as A X – λ X = 0. Let I be the n × n identity matrix. If I X is substituted by X in the equation above, we obtain A X – λ I X = 0. The equation is rewritten as (A – λ I) X = 0. The equation above consists of non-trivial solutions if and only if the [determinant value of the matrix](https://byjus.com/jee/determinants/) is 0. The characteristic equation of A is Det (A – λ I) = 0. ‘A’ being an n × n matrix, if (A – λ I) is expanded, (A – λ I) will be the characteristic polynomial of A because its degree is n. ## Properties of Eigenvalues Let A be a matrix with eigenvalues λ1, λ2,…., λn. The following are the properties of eigenvalues. (1) The trace of A, defined as the sum of its diagonal elements, is also the sum of all eigenvalues, t r ( A ) \= ∑ i \= 1 n a i i \= ∑ i \= 1 n λ i \= λ 1 \+ λ 2 \+ ⋯ \+ λ n . (2) The determinant of A is the product of all its eigenvalues, det ( A ) \= ∏ i \= 1 n λ i \= λ 1 λ 2 ⋯ λ n . (3) The eigenvalues of the kth power of A, that is, the eigenvalues of Ak, for any positive integer k, are λ 1 k , … , λ n k . . (4) The matrix A is invertible if and only if every eigenvalue is nonzero. (5) If A is invertible, then the eigenvalues of A\-1 are 1 λ 1 , … , 1 λ n and each eigenvalue’s geometric multiplicity coincide. The characteristic polynomial of the inverse is the reciprocal polynomial of the original, the eigenvalues share the same algebraic multiplicity. (6) If A is equal to its conjugate transpose, or equivalently if A is Hermitian, then every eigenvalue is real. The same is true for any real symmetric matrix. (7) If A is not only Hermitian but also positive-definite, positive-semidefinite, negative-definite, or negative-semidefinite, then every eigenvalue is positive, non-negative, negative, or non-positive, respectively. (8) If A is unitary, every eigenvalue has absolute value \|λi\| = 1. (9) If A is a n×n matrix and {λ1, λ2,…., λk} are its eigenvalues, then the eigenvalues of the matrix I + A (where I is the identity matrix) are {λ1\+ 1, λ2\+1,…., λk\+1}. **Also, Read:** [Eigenvectors of a Matrix](https://byjus.com/jee/eigenvectors-of-a-matrix/) [Adjoint and Inverse of a Matrix](https://byjus.com/jee/adjoint-and-inverse-of-a-matrix/) [Normalized and Decomposition of Eigenvectors](https://byjus.com/jee/normalized-and-decomposition-of-eigenvectors/) ## Eigenvalues and Eigenvectors Solved Problems **Example 1:** Find the eigenvalues and eigenvectors of the following matrix. **Solution:**  **Example 2:** Find all eigenvalues and corresponding eigenvectors for the matrix A if ( 2 − 3 0 2 − 5 0 0 0 3 ) **Solution:** det ( ( 2 − 3 0 2 − 5 0 0 0 3 ) − λ ( 1 0 0 0 1 0 0 0 1 ) ) ( 2 − 3 0 2 − 5 0 0 0 3 ) − λ ( 1 0 0 0 1 0 0 0 1 ) λ ( 1 0 0 0 1 0 0 0 1 ) \= ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 − 3 0 2 − 5 0 0 0 3 ) − ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 − λ − 3 0 2 − 5 − λ 0 0 0 3 − λ ) \= det ( 2 − λ − 3 0 2 − 5 − λ 0 0 0 3 − λ ) \= ( 2 − λ ) det ( − 5 − λ 0 0 3 − λ ) − ( − 3 ) det ( 2 0 0 3 − λ ) \+ 0 ⋅ det ( 2 − 5 − λ 0 0 ) \= ( 2 − λ ) ( λ 2 \+ 2 λ − 15 ) − ( − 3 ) ⋅ 2 ( − λ \+ 3 ) \+ 0 ⋅ 0 \= − λ 3 \+ 13 λ − 12 − λ 3 \+ 13 λ − 12 \= 0 − ( λ − 1 ) ( λ − 3 ) ( λ \+ 4 ) \= 0 The eigenvalues are : λ \= 1 , λ \= 3 , λ \= − 4 Eigenvectors for λ \= 1 ( 2 − 3 0 2 − 5 0 0 0 3 ) − 1 ⋅ ( 1 0 0 0 1 0 0 0 1 ) \= ( 1 − 3 0 2 − 6 0 0 0 2 ) ( A − 1 I ) ( x y z ) \= ( 1 − 3 0 0 0 1 0 0 0 ) ( x y z ) \= ( 0 0 0 ) { x − 3 y \= 0 z \= 0 } I s o l a t e { z \= 0 x \= 3 y } Plug into ( x y z ) η \= ( 3 y y 0 ) y ≠ 0 Let y \= 1 ( 3 1 0 ) S i m i l a r l y Eigenvectors for λ \= 3 : ( 0 0 1 ) Eigenvectors for λ \= − 4 : ( 1 2 0 ) The eigenvectors for ( 2 − 3 0 2 − 5 0 0 0 3 ) \= ( 3 1 0 ) , ( 0 0 1 ) , ( 1 2 0 ) **Example 3:** Consider the matrix **** for some variable ‘a’. Find all values of ‘a’, which will prove that A has eigenvalues 0, 3, and −3. **Solution:** Let p (t) be the characteristic polynomial of A, i.e. let p (t) = det (A − tI) = 0. By expanding along the second column of A − tI, we can obtain the equation  \= (3 − t) \[(−2 −t) (−1 − t) − 4\] + 2\[(−2 − t) a + 5\] \= (3 − t) (2 + t + 2t + t2 −4) + 2 (−2a − ta + 5) \= (3 − t) (t2 \+ 3t − 2) + (−4a −2ta + 10) \= 3t2 \+ 9t − 6 − t3 − 3t2 \+ 2t − 4a − 2ta + 10 \= −t3 \+ 11t − 2ta + 4 − 4a \= −t3 \+ (11 − 2a) t + 4 − 4a For the eigenvalues of A to be 0, 3 and −3, the characteristic polynomial p (t) must have roots at t = 0, 3, −3. This implies p (t) = –t (t − 3) (t + 3) =–t(t2 − 9) = –t3 \+ 9t. Therefore, −t3 \+ (11 − 2a) t + 4 − 4a = −t3 + 9t. For this equation to hold, the constant terms on the left and right-hand sides of the above equation must be equal. This means that 4 − 4a = 0, which implies a = 1. Hence, A has eigenvalues 0, 3, and −3 precisely when a = 1. **Example 4:** Find the eigenvalues and eigenvectors of ( 2 0 0 0 3 4 0 4 9 ) **Solution:** det ( ( 2 0 0 0 3 4 0 4 9 ) − λ ( 1 0 0 0 1 0 0 0 1 ) ) ( 2 0 0 0 3 4 0 4 9 ) − λ ( 1 0 0 0 1 0 0 0 1 ) λ ( 1 0 0 0 1 0 0 0 1 ) \= ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 0 0 0 3 4 0 4 9 ) − ( λ 0 0 0 λ 0 0 0 λ ) \= ( 2 − λ 0 0 0 3 − λ 4 0 4 9 − λ ) \= det ( 2 − λ 0 0 0 3 − λ 4 0 4 9 − λ ) \= ( 2 − λ ) det ( 3 − λ 4 4 9 − λ ) − 0 ⋅ det ( 0 4 0 9 − λ ) \+ 0 ⋅ det ( 0 3 − λ 0 4 ) \= ( 2 − λ ) ( λ 2 − 12 λ \+ 11 ) − 0 ⋅ 0 \+ 0 ⋅ 0 \= − λ 3 \+ 14 λ 2 − 35 λ \+ 22 − λ 3 \+ 14 λ 2 − 35 λ \+ 22 \= 0 − ( λ − 1 ) ( λ − 2 ) ( λ − 11 ) \= 0 The eigenvalues are : λ \= 1 , λ \= 2 , λ \= 11 Eigenvectors for λ \= 1 ( 2 0 0 0 3 4 0 4 9 ) − 1 ⋅ ( 1 0 0 0 1 0 0 0 1 ) \= ( 1 0 0 0 2 4 0 4 8 ) ( A − 1 I ) ( x y z ) \= ( 1 0 0 0 1 2 0 0 0 ) ( x y z ) \= ( 0 0 0 ) { x \= 0 y \+ 2 z \= 0 } I s o l a t e { x \= 0 y \= − 2 z } Plug into ( x y z ) η \= ( 0 − 2 z z ) z ≠ 0 Let z \= 1 ( 0 − 2 1 ) S i m i l a r l y Eigenvectors for λ \= 2 : ( 1 0 0 ) Eigenvectors for λ \= 11 : ( 0 1 2 ) The eigenvectors for ( 2 0 0 0 3 4 0 4 9 ) \= ( 0 − 2 1 ) , ( 1 0 0 ) , ( 0 1 2 ) ## Frequently Asked Questions Q1 ### What do you mean by eigenvalues? Eigenvalues are the special set of scalar values associated with the set of linear equations in the matrix equations. Q2 ### Can the eigenvalue be zero? Yes, the eigenvalue can be zero. Q3 ### Can a singular matrix have eigenvalues? Every singular matrix has a 0 eigenvalue. Q4 ### How to find the eigenvalues of a square matrix A? Use the equation det(A-λI) = 0 and solve for λ. Determine all the possible values of λ, which are the required eigenvalues of matrix A.