Eigenvalues and Eigenvectors:
Presume that A is a square (n × n) matrix. We utter that a nonzero vector v is an eigenvector (ev) and a number λ is its eigenvalue (ew) if:
Av = λv.
Geometrically this signifies that Av is in the same direction as v since multiplying a vector by a number changes its length however not its direction.
Mat lab has a built-in routine for finding eigenvalues as well as eigenvectors
> A = pascal(4)> [v e] = eig(A)
The results are a matrix v that holds eigenvectors as columns and a diagonal matrix e that contains eigenvalues on the diagonal. We are able to check this by:
> v1 = v(:,1)> A*v1>e(1,1)*v1
Finding Eigenvalues for 2×2 and 3×3:
If A is 2×2 or 3×3 then we are able to find its eigenvalues and eigenvectors by hand. Notice that Equation is able to be rewritten as:
Av − λv = 0.
It would be good to factor out the v from the right-hand side of this equation however we can’t for the reason that A is a matrix and λ is a number. However since Iv = v we can do the following:
Av − λv = Av − λIv= (A − λI)v= 0
If v is nonzero after that by Theorem 3 in the matrix (A − λI) should be singular. Beside the similar theorem we must have
det(A − λI) = 0.
This is called a characteristic equation.
For a 2 × 2 matrix A − λI is considered as in the following example:
The determinant of A − λI is next
det(A − λI) = (1 − λ)(5 − λ) − 4 • 3
= −7 − 6λ + λ2.
The characteristic equation det(A − λI) = 0 is merely a quadratic equation:
λ2− 6λ − 7 = 0.
The roots of this equation are λ1 = 7 and λ2 = −1. These are the ew’s of the matrix A. Now to find the corresponding ev’s we return to the equation (A − λI)v = 0. For λ1 = 7, the equation for the ev (A − λI)v = 0 is equivalent to the augmented matrix
Notice that the first also second rows of this matrix are multiples of one another. Therefore Gaussian elimination would produce all zeros on the bottom row. Therefore this equation has infinitely many solutions that are infinitely many ev’s. Since only the direction of the ev matters this is okay we only require to find one of the ev’s. Because the second row of the augmented matrix represents the equation:
3x − 2y = 0,
we can let,
This come from observe that (x, y) = (2, 3) is a solution of 3x − 2y = 0.
For λ2 = −1, (A − λI)v = 0 is equivalent to the augmented matrix:
One more time the first as well as second rows of this matrix are multiples of one another. For ease we are able to let:
One is able to always check an ev and ew by multiplying:
For a 3 × 3 matrix we could total the same process. The det(A − λI) = 0 would be a cubic polynomial as well as we would expect to usually get 3 roots, which are the ew’s.
For an × n matrix with n ≥ 4 this process is too long as well as cumbersome to complete by hand. Additional this process isn’t well suited even to implementation on a computer program since it involves determinants as well as solving an-degree polynomial. For n ≥ 4 we require more ingenious methods. These methods rely on the geometric meaning of ev’s as well as ew’s rather than solving algebraic equations.
It turns out that the eigenvalues of a few matrices are complex numbers even when the matrix only contains real numbers. When this take place the complex ew’s should occur in conjugate pairs that is:
λ1,2 = α ± iβ.
The corresponding ev’s should also come in conjugate pairs:
w = u ± iv.
In applications the imaginary part of the ew β frequently is related to the frequency of an oscillation.
This is for the reason that of Euler’s formula
eα+iβ = eα(cos β + i sin β).
Certain kinds of matrices that take place in applications can only have real ew’s and ev’s. The most common such kind of matrix is the symmetric matrix. A matrix is symmetric if it is equal to its own transpose that is it is symmetric across the diagonal. For illustration:
is symmetric and therefore we know beforehand that its ew’s will be real not complex.
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