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What is significance of eigenvectors of a matrix?

Updated: 8/19/2019
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Q: What is significance of eigenvectors of a matrix?
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Can a Hermitian Matrix possess Complex Eigenvectors?

Yes. Simple example: a=(1 i) (-i 1) The eigenvalues of the Hermitean matrix a are 0 and 2 and the corresponding eigenvectors are (i -1) and (i 1). A Hermitean matrix always has real eigenvalues, but it can have complex eigenvectors.


What is the eigen value?

This is the definition of eigenvectors and eigenvalues according to Wikipedia:Specifically, a non-zero column vector v is a (right) eigenvector of a matrix A if (and only if) there exists a number λ such that Av = λv. The number λ is called the eigenvalue corresponding to that vector. The set of all eigenvectors of a matrix, each paired with its corresponding eigenvalue, is called the eigensystemof that matrix


Do similar matrices have the same eigenvectors?

No, in general they do not. They have the same eigenvalues but not the same eigenvectors.


What are eigenvalues and eigenvectors?

An eigenvector is a vector which, when transformed by a given matrix, is merely multiplied by a scalar constant; its direction isn't changed. An eigenvalue, in this context, is the factor by which the eigenvector is multiplied when transformed.


What are eigen values and eigen vectors?

This is a complicated subject, which can't be explained in a few words. Read the Wikipedia article on "eigenvalue"; or better yet, read a book on linear algebra. Briefly, and quoting from the Wikipedia, "The eigenvectors of a square matrix are the non-zero vectors that, after being multiplied by the matrix, remain parallel to the original vector. For each eigenvector, the corresponding eigenvalue is the factor by which the eigenvector is scaled when multiplied by the matrix."


Is eigenvalue applicable only to n x n matrices?

The answer is yes, and here's why: Remember that for the eigenvalues (k) and eigenvectors (v) of a matrix (M) the following holds: M.v = k*v, where "." denotes matrix multiplication. This operation is only defined if the number of columns in the first matrix is equal to the number of rows in the second, and the resulting matrix/vector will have as many rows as the first matrix, and as many columns as the second matrix. For example, if you have a 3 x 2 matrix and multiply with a 2 x 4 matrix, the result will be a 3 x 4 matrix. Applying this to the eigenvalue problem, where the second matrix is a vector, we see that if the matrix M is m x n and the vector is n x 1, the result will be an m x 1 vector. Clearly, this can never be a scalar multiple of the original vector.


Do commutative matrices have the same eigenvectors?

It is true that diagonalizable matrices A and B commute if and only if they are simultaneously diagonalizable. This result can be found in standard texts (e.g. Horn and Johnson, Matrix Analysis, 1999, Theorem 1.3.12.) One direction of the if and only if proof is straightforward, but the other direction is more technical: If A and B are diagonalizable matrices of the same order, and have the same eigenvectors, then, without loss of generality, we can write their diagonalizations as A = VDV-1 and B = VLV-1, where V is the matrix composed of the basis eigenvectors of A and B, and D and L are diagonal matrices with the corresponding eigenvalues of A and B as their diagonal elements. Since diagonal matrices commute, DL = LD. So, AB = VDV-1VLV-1 = VDLV-1 = VLDV-1 = VLV-1VDV-1 = BA. The reverse is harder to prove, but one online proof is given below as a related link. The proof in Horn and Johnson is clear and concise. Consider the particular case that B is the identity, I. If A = VDV-1 is a diagonalization of A, then I = VIV-1 is a diagonalization of I; i.e., A and I have the same eigenvectors.


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Yes. Neo is an anagram of One, a reference to his destiny of being The One who would bring peace.


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What type of matrix is a vector?

Vector matrix has both size and direction. There are different types of matrix namely the scalar matrix, the symmetric matrix, the square matrix and the column matrix.