How To Find The Eigenvalues Of A Matrix

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The matrix equation $A\mathbf {ten} =\mathbf {b}$ involves a matrix acting on a vector to produce another vector. In general, the manner $A$ acts on $\mathbf {x}$ is complicated, but in that location are certain cases where the action maps to the same vector, multiplied past a scalar factor.

Eigenvalues and eigenvectors accept immense applications in the physical sciences, specially breakthrough mechanics, among other fields.

Steps

1

Understand determinants. The determinant of a matrix $\det A=0$ when $A$ is non-invertible. When this occurs, the null space of $A$ becomes non-trivial - in other words, at that place are non-cypher vectors that satisfy the homogeneous equation $A\mathbf {x} =0.$ ^[one]
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iv
5
Solve the characteristic polynomial for the eigenvalues. This is, in general, a difficult step for finding eigenvalues, as there exists no general solution for quintic functions or higher polynomials. Nevertheless, nosotros are dealing with a matrix of dimension 2, so the quadratic is easily solved.
- ${\begin{aligned}&(\lambda -v)(\lambda +2)=0\\&\lambda =5,-2\stop{aligned}}$
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Substitute the eigenvalues into the eigenvalue equation, one past one. Allow's substitute $\lambda _{1}=5$ beginning.^[iii]
- $(A-5I)\mathbf {x} ={\begin{pmatrix}-iv&four\\3&-3\end{pmatrix}}$
- The resulting matrix is obviously linearly dependent. We are on the right track here.
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Row-reduce the resulting matrix. With larger matrices, information technology may not be so obvious that the matrix is linearly dependent, and so we must row-reduce. Here, however, nosotros can immediately perform the row operation $R_{2}\to 4R_{2}+3R_{one}$ to obtain a row of 0's.^[4]
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Obtain the basis for the eigenspace. The previous step has led us to the basis of the null space of $A-5I$ - in other words, the eigenspace of $A$ with eigenvalue 5.

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Question

Why do we supplant y with one and not whatsoever other number while finding eigenvectors?

For simplicity. Eigenvectors are only defined up to a multiplicative constant, and then the choice to set the constant equal to 1 is often the simplest.
Question

How do y'all find the eigenvectors of a 3x3 matrix?

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Community Answer

Kickoff, find the solutions 10 for det(A - 11) = 0, where I is the identity matrix and x is a variable. The solutions x are your eigenvalues. Let'southward say that a, b, c are your eignevalues. Now solve the systems [A - aI | 0], [A - bI | 0], [A - cI | 0]. The footing of the solution sets of these systems are the eigenvectors.

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The determinant of a triangular matrix is easy to observe - it is merely the production of the diagonal elements. The eigenvalues are immediately found, and finding eigenvectors for these matrices and then becomes much easier.^[five]
- Beware, however, that row-reducing to row-echelon form and obtaining a triangular matrix does not give you lot the eigenvalues, every bit row-reduction changes the eigenvalues of the matrix in general.

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