Three matrices can look equally harmless and still lead to three different verdicts. One has a repeated eigenvalue but too few eigenvectors, one turns every nonzero real vector off its own line, and one has perpendicular eigenvectors that make its powers and its quadratic form easy to read. This workshop practices deciding which scalar field and which hypotheses are in play before calculating, then running the method that applies and checking the result against the original matrix. Tuples such as stand for column vectors.
Key terms
Choice of scalar field
Definition Let be a real matrix and let be or . An eigenvalue of over is a number in with for some nonzero in , and these are exactly the roots of that lie in . Over the characteristic polynomial always factors into linear factors, counted with multiplicity, while over it may not. The matrix is diagonalizable over if with invertible and diagonal, both with entries in , which happens exactly when has a basis of eigenvectors of . Diagonalizable over implies diagonalizable over , but not conversely. Changing the field changes which scalars and vectors are allowed, never the entries of .
In plain words The same real matrix can be diagonalizable over but not over , so a claim that a matrix is diagonalizable should name the field. Fix the field first, then look for eigenvalues and eigenvectors in it.
Example The matrix
turns the -plane through a quarter turn and stretches the -axis by . Its characteristic polynomial is . Over the only eigenvalue is , and has rank , so its eigenspace is the line spanned by . One line cannot contain a basis of , so is not diagonalizable over . Over it has the three distinct eigenvalues , and , with eigenvectors , and , so it is diagonalizable over .
Test for a defective matrix
Definition Let be a real matrix and let be an eigenvalue of , real or complex. The algebraic multiplicity is the multiplicity of as a root of , and the geometric multiplicity is
For a nonreal the null space and the rank are taken over . These always satisfy . The matrix is defective if for at least one of its eigenvalues, real or complex. To test a matrix, compute for each repeated eigenvalue. A simple root has both multiplicities equal to and needs no check. A defective matrix has fewer than independent eigenvectors, so it is diagonalizable neither over nor over . If every root of lies in the field , then is diagonalizable over exactly when it is not defective. Without that condition a matrix can fail to be defective and still not be diagonalizable over , as the quarter turn over shows.
In plain words Count independent eigenvectors, not roots. A repeated root is a signal to check the eigenspace, not a verdict.
Example The matrices
both have the characteristic polynomial , so for each. The matrix has rank , its only equation is , and . So is defective, and all its eigenvectors lie on the line spanned by . Complex scalars change nothing, because the equation stays the same. For the matrix is zero, so and is already diagonal. Replacing by would predict , but in fact
Complex eigenpairs of a rotation
Definition Let and be real numbers with , and let
With , the matrix rotates the plane through the angle with and and scales by . Its characteristic polynomial has no real root, so has no real eigenvalue and no real eigenvector. Over it has the conjugate eigenvalues , with conjugate eigenvectors,
The eigenvector matrix with columns and has determinant , so is diagonalizable over . Both eigenvalues have modulus , the scale factor.
In plain words A rotation that is not a multiple of a half turn moves every real line off itself, so there is no real eigenvector. Complex coordinates still provide two eigenvectors, and they come as a conjugate pair. They are columns of complex numbers, not arrows in the real plane.
Example The matrix
has and , so it turns every vector through a quarter turn and doubles its length. Its characteristic polynomial is , with roots . For the check is , using . Real consequences can be checked with real arithmetic. Since , the diagonalization gives , and indeed two quarter turns make a half turn while lengths are multiplied by , so .
Real orthogonal diagonalization
Definition A real matrix has a real orthogonal diagonalization
with real orthogonal and real diagonal, if and only if is symmetric. This is the real spectral theorem together with the fact that is always symmetric. The method is to check , find the eigenvalues, which are real, find an orthonormal basis of each eigenspace, with Gram-Schmidt inside any eigenspace of dimension or more, and place these vectors in with their eigenvalues in matching order in . Because the columns of are orthonormal, , so no inverse has to be computed.
In plain words Symmetry guarantees a basis of perpendicular eigenvectors, and perpendicular unit eigenvectors make the inverse free. Without symmetry a real diagonalization may still exist, but it uses a slanted eigenbasis and a genuine inverse.
Example For
the characteristic polynomial is . The matrix gives the equation and the eigenvector , and gives and the eigenvector . Their dot product is , and normalizing gives and . With and , the spectral decomposition checks the result,
and the sum of these two matrices is .
Matrix power from eigenvalues
Definition Suppose has an eigenbasis with , so that with and . For every integer , with . To apply to a vector, write , where , and then
With a real orthonormal eigenbasis the coefficients are dot products, . The method needs an eigenbasis, so it does not apply to a defective matrix, and every coefficient must be kept, not only the one with the largest eigenvalue.
In plain words Split the vector along eigenvectors, multiply each piece by its eigenvalue raised to the th power, and add the pieces back up.
Example For above and , the coefficients are , so . Then
For this returns , and for it gives . Direct multiplication agrees, since and .
Hypothesis of a method
Definition The hypothesis of a method is the condition under which its conclusion is guaranteed, and it is checked before the method is used. The methods of this unit need the following.
- Diagonalization over a field needs linearly independent eigenvectors in .
- Real orthogonal diagonalization needs a real symmetric matrix.
- The eigenvalue test for definiteness needs the symmetric matrix of the form, which for is .
- The power formula needs an eigenbasis.
When a hypothesis fails, the conclusion can fail too, and the failure points to what to do instead, such as working over , computing powers directly or replacing a matrix by its symmetric part.
In plain words Before calculating, name the theorem you plan to use and check its conditions. Being square does not make a matrix eligible for every method.
Example The matrix
has the single eigenvalue , which is positive, but its form takes the value at . So the eigenvalues of do not decide the sign of its form, because the eigenvalue test needs a symmetric matrix and is not symmetric. The symmetric matrix of this form has rows and , with eigenvalues and , so the form is indefinite. Similarly, the matrix with rows and is diagonalizable, with eigenvectors and for the eigenvalues and , but it has no orthonormal eigenbasis. Each of its eigenspaces is a line, and the two lines are not perpendicular.
Check yourself
- Is the matrix with rows and diagonalizable over ? Over ?
Show the answer
The characteristic polynomial is
so is the only eigenvalue, with . The matrix has rows and and rank , so , and the eigenspace is spanned by . Check that . Since , the matrix is defective and is diagonalizable neither over nor over . Complex scalars do not help, because the only eigenvalue is already real and has rank over either field. The matrix is not symmetric either, so the spectral theorem gives no guarantee here.
- Find the complex eigenvalues and eigenvectors of the matrix with rows and . Explain why has no real eigenvector, and use the eigenvalues to find .
Show the answer
The matrix has the rotation form with and . Its characteristic polynomial is , whose discriminant is negative, so there is no real eigenvalue and therefore no real eigenvector. Over the eigenvalues are , with eigenvector for and for . Check the first, , and . Both eigenvalues have modulus , and rotates through the angle with and and scales by .
Let have the columns and and let , so that . The matrix has the same eigenvectors with the eigenvalues . The rotation form with and has exactly these eigenpairs, and a matrix is determined by its action on a basis, so
Direct multiplication agrees. Both diagonal entries are , the entry above the diagonal is , and the entry below it is .
- Let be the matrix with rows and . Orthogonally diagonalize , use the result to find , and classify the quadratic form .
Show the answer
The matrix is symmetric, so the real spectral theorem applies. Its characteristic polynomial is . For , the system reduces to , giving . For , it reduces to , giving . These are orthogonal with length , so
The coefficients of are and , so . Then
Direct multiplication agrees, since , and .
The form is . In the coordinates and it becomes . One eigenvalue is positive and one negative, so the form is indefinite, with witnesses and .