# can a real matrix have complex eigenvectors

The eigenvectors corresponding to distinct eigenvalues are orthogonal which is proved here. In the first case, are you assuming your matrix to be diagonalizable or not ? Asking for help, clarification, or responding to other answers. Your email address will not be published. Some things to remember about eigenvalues: •Eigenvalues can have zero value •Eigenvalues can be negative •Eigenvalues can be real or complex numbers •A "×"real matrix can have complex eigenvalues •The eigenvalues of a "×"matrix are not necessarily unique. $\begingroup$ @KutalmisB If you want complex eigenvectors (rather than a real eigenbasis), the easiest way is probably Brian's approach to work in complex arithmetic from the start. If you have an eigenvector then any scalar (including complex scalar) multiple of that eigenvector is also an eigenvector. However, the eigenvectors corresponding to the conjugate eigenvalues are themselves complex conjugate and the calculations involve working in complex n-dimensional space. Example. Linear Transformation that Maps Each Vector to Its Reflection with Respect to $x$-Axis, The Intersection of Two Subspaces is also a Subspace. With complex eigenvalues we are going to have the same problem that we had back when we were looking at second order differential equations. @CarloBeenakker You are not wrong! Yes, t can be complex. I assume you are asking about the geometric interpretation in $\mathbb{R}^n$ when the matrix $A$ has all real entries. If Two Matrices Have the Same Rank, Are They Row-Equivalent? How the Calculator Works. One has to find the subspace $K_X\subset K$ spanned by the generalized eigenspaces of $X$ associated to its real eigenvalues and the corresponding subspace $K_Y\subset K$ associated to $Y$ and then look at the intersection $K' = K_X\cap K_Y$, which is preserved by $X$ and $Y$. We define the characteristic polynomial and show how it can be used to find the eigenvalues for a matrix. Qmechanic ♦ 134k 18 18 gold badges 298 298 silver badges 1605 1605 bronze badges. The row vector is called a left eigenvector of . There will always be n linearly independent eigenvectors for symmetric matrices. Suppose, though, that $C = XY-YX$ has a nontrivial kernel $K_0\subset\mathbb{R}^n$, which can be computed by solving linear equations. To see this, write an n -by- n complex matrix in the form A = X + iY where X and Y are real matrices and note that finding a real eigenvector for A is equivalent to finding a simultaneous eigenvector in Rn for both X and Y, i.e., Xv = xv and Yv = yv. Can We Reduce the Number of Vectors in a Spanning Set? Taking the real and imaginary part (linear combination of the vector and its conjugate), the matrix has this form with respect to the new basis. An eigenvalue represents the amount of expansion in the corresponding dimension. (Note that we don't actually need or use the condition that $A$ be invertible.) However, apparently, locating the spaces $K_X$ and $K_Y$ cannot be done by solving linear equations alone, just as one cannot generally factor a rational polynomial $p(x)$ of degree greater than $2$ into a rational polynomial with only real roots times a rational polynomial with no real roots. Now, there is a relatively easy way, solving only linear equations, to reduce to a special case, which is that $A$ and $\bar A$ commute, and, in this case, the problem is more tractable. If is an eigenvector of the transpose, it satisfies By transposing both sides of the equation, we get. The Spectral Theorem states that if Ais an n nsymmetric matrix with real entries, then it has northogonal eigenvectors. The Characteristic Equation always features polynomials which can have complex as well as real roots, then so can the eigenvalues & eigenvectors of matrices be complex as well as real. I am struggling to find a method in numpy or scipy that does this for me, the ones I have tried give complex valued eigenvectors. and let $K_\infty$ be the limiting subspace (which will equal $K_m$ as soon as we find an $m\ge0$ with $K_{m+1} = K_m$, and hence in a finite number $m2$ is more difficult. Sample test on 3 by 3 matrix, real and complex: ... Eigenvectors should have length 1.0 4) Compute the trace of A and subtract the sum of the eigenvalues. This site uses Akismet to reduce spam. the eigenvalues of A) are real numbers. The answer is always. The result should be near zero. We want our solutions to only have real numbers in them, however since our solutions to systems are of the form, $\vec x = \vec \eta {{\bf{e}}^{\lambda t}}$ we are going to have complex … Also, note that the algorithm does not actually find a real eigenvector, it just gives a necessary and sufficient criterion that one exist. For a real matrix the nonreal eigenvectors and generalized eigenvectors can always be chosen to form complex conjugate pairs. The diagonal elements of a triangular matrix are equal to its eigenvalues. This real Jordan form is a consequence of the complex Jordan form. Quiz 9. Now that $X$ and $Y$ map $K_\infty$ into itself, and, because $(XY-YX)(v) = 0$ for all $v\in K_\infty\subseteq K_0$, it follows that $X$ and $Y$ commute when restricted to $K_\infty$. abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … For example, the first eigenvector is a postive hump but there is also a negative mirror hump underneath. Thus, the criterion in this case is that $X$ and $Y$ have non-positive determinant. Suppose has eigenvalue , eigenvector and their complex conjugates. We prove that the given real matrix does not have any real eigenvalues. The eigenvalues of a hermitian matrix are real, since (λ − λ)v = (A * − A)v = (A − A)v = 0 for a non-zero eigenvector v. If A is real, there is an orthonormal basis for R n consisting of eigenvectors of A if and only if A is symmetric. The same result is true for lower triangular matrices. Subset of Vectors Perpendicular to Two Vectors is a Subspace, Linear Combination and Linear Independence, Bases and Dimension of Subspaces in $\R^n$, Linear Transformation from $\R^n$ to $\R^m$, Linear Transformation Between Vector Spaces, Introduction to Eigenvalues and Eigenvectors, Eigenvalues and Eigenvectors of Linear Transformations, How to Prove Markov’s Inequality and Chebyshev’s Inequality, How to Use the Z-table to Compute Probabilities of Non-Standard Normal Distributions, Expected Value and Variance of Exponential Random Variable, Condition that a Function Be a Probability Density Function, Conditional Probability When the Sum of Two Geometric Random Variables Are Known, Determine Whether Each Set is a Basis for $\R^3$. A matrix in a given field (or even commutative ring) may or may not have eigenvectors. If $\dim K = 2$, then because $X$ and $Y$ have trace zero and commute, they must be multiples of each other. Solution for know that a real matrix can have real and complex eigenvalues/eigenvectors, but what if we start with the condition that we have a real matrix with… In general, it is normal to expect that a square matrix with real entries may still have complex eigenvalues. Moreover, we can assume that $X$ and $Y$ have zero trace, since subtracting multiples of the identity from $X$ and $Y$ will not affect whether a vector in $K$ is a simultaneous eigenvector of $X$ and $Y$. Math 54 — Complex eigenvalues and eigenvectors For a real square matrix A with complex eigenvalues, the algebraic story of diagonalization parallels the real story; but the geometry is different, in the sense that the eigenvectors have complex entries, and cannot be represented geometrically in R n. Learn how your comment data is processed. 3 . Continuing by induction, define 1 MATLAB eig usually returns real eigenvectors when the matrix is real and symmetric. Recall that if z= a+biis a complex number, its complex conjugate is de ned by z= a bi. $\begingroup$ your matrix isn't hermitean so it may in general have complex eigenvalues and eigenvectors (and not all with the same phase). abelian group augmented matrix basis basis for a vector space characteristic polynomial commutative ring determinant determinant of a matrix diagonalization diagonal matrix eigenvalue eigenvector elementary row operations exam finite group group group homomorphism group theory homomorphism ideal inverse matrix invertible matrix kernel linear algebra linear combination linearly … Is it true for Symmetric Matrices as well or do symmetric matrices have distinct eigenvectors even with repeated eigenvalues? The list of linear algebra problems is available here. This website’s goal is to encourage people to enjoy Mathematics! If the 2 2 matrix Ahas distinct real eigenvalues 1 and 2, with corresponding eigenvectors ~v 1 and ~v 2, then the system x~0(t)=A~x(t) has general solution predicted by the eigenvalue-eigenvector method of c 1e 1t~v 1 + c 2e 2t~v 2 where the constants c 1 and c 2 can be determined from the initial values. Your proposal does not find a complete set of real eigenvectors for the matrix under consideration, it simply takes the arbitrary set of complex eigenvectors which eig has returned, and selects any that happen to be real. A Hermitean matrix always has real eigenvalues, but it can have complex eigenvectors. $$ST is the new administrator. Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. But maybe some don't like to work with 1x1 matrix (too simple or too good to be true I guess! No, but you can build some. If you know a bit of matrix reduction, you’ll know that your question is equivalent to: When do polynomials have complex roots? site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. Once we have the eigenvalues for a matrix we also show how to find the corresponding eigenvalues for the matrix. Moreover, if X is an eigenvector of A associated to, then the vector, obtained from X by taking the complex-conjugate of the entries of X, is an eigenvector associated to. For a real matrix the nonreal eigenvectors and generalized eigenvectors can always be chosen to form complex conjugate pairs. Therefore, any real matrix with odd order has at least one real eigenvalue, whereas a real matrix with even order may not have any real eigenvalues. In fact, we can define the multiplicity of an eigenvalue. Thus, we have an algorithm for deciding whether there is a real eigenvector for A\in\mathrm{GL}(3,\mathbb{C}) that only involves solving linear equations and (in one special case) the computation of the sign of a determinant. This is because the matrix is degenerate. When I take the eigenvectors of the matrix, I get mirror images for the first few (about 10) vectors. Every square matrix of degree n does have n eigenvalues and corresponding n eigenvectors. The trace of a matrix is the sum of the diagonal elements of the matrix. (adsbygoogle = window.adsbygoogle || []).push({}); Express the Eigenvalues of a 2 by 2 Matrix in Terms of the Trace and Determinant. The eigenvalues can be real or complex. Eigenvalues can be complex numbers even for real matrices. COMPLEX EIGENVALUES . If \dim K = 1, then there is a simultaneous eigenvector, unique up to multiples. Step by Step Explanation. the dot product of two complex vectors is complex (in general). Range, Null Space, Rank, and Nullity of a Linear Transformation from \R^2 to \R^3, How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Rank of the Product of Matrices AB is Less than or Equal to the Rank of A, Prove a Group is Abelian if (ab)^2=a^2b^2, Find an Orthonormal Basis of \R^3 Containing a Given Vector, Find a Basis for the Subspace spanned by Five Vectors, Show the Subset of the Vector Space of Polynomials is a Subspace and Find its Basis. A simple example is the 1x1 matrix A = [i] !! Thus a real symmetric matrix A can be decomposed as = where Q is an orthogonal matrix whose columns are the eigenvectors of A, and Λ is a diagonal matrix whose entries are the eigenvalues of A. I have created a matrix of potentials for a particle in a square well. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Two eigenvectors of a real symmetric matrix or a Hermitian matrix, if they come from different eigen values are orthogonal to one another. Just to be sure. Yes, t can be complex. python numpy scipy linear-algebra eigenvalue. 2.5 Complex Eigenvalues Real Canonical Form A semisimple matrix with complex conjugate eigenvalues can be diagonalized using the procedure previously described. asked Sep 2 '19 at 15:55. To find the eigenvectors of a triangular matrix, we use the usual procedure. rev 2020.12.4.38131, The best answers are voted up and rise to the top, MathOverflow works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. These equation can be written . Thus a real symmetric matrix A can be decomposed as = where Q is an orthogonal matrix whose columns are the eigenvectors of A, and Λ is a diagonal matrix whose entries are the eigenvalues of A. The matrix AT therefore has the eigenvector ... shows that a Markov matrix can have complex eigenvalues and that Markov matrices can be orthogonal. A real complex We know matrix nxu eigenvalues real that XZ can This have can polynomial 4 k veal non have … Problems in Mathematics © 2020. Its unit eigenvectors are orthogonal by property (3). When eigenvalues become complex, eigenvectors also become complex. However, when complex eigenvalues are encountered, they always occur in conjugate pairs as long as their associated matrix has only real entries.$$, MAINTENANCE WARNING: Possible downtime early morning Dec 2, 4, and 9 UTC…. In this section we will introduce the concept of eigenvalues and eigenvectors of a matrix. $\endgroup$ – acl Mar 28 '12 at 20:51 share | improve this question | follow | edited Feb 27 '18 at 20:19. user6655984 asked Feb 27 '18 at 2:01. kinder chen kinder chen. Notify me of follow-up comments by email. Then $X$ has a real eigenvalue of odd multiplicity, either $1$ or $3$. If A is a matrix with real entries, then its characteristic polynomial has real coefficients, so this note implies that its complex eigenvalues come in conjugate pairs. 5) Compute the maximum of the sum of the absolute values of each row and column of A. if you have a matrix with all complex eigenvalues and one of the eigenvectors is something like $\pmatrix{ 1 \\ 0}$ then $\pmatrix{2+2i \\ 0}$ is also an eigenvector in that eigenspace. If you know a bit of matrix reduction, you’ll know that your question is equivalent to: When do polynomials have complex roots? Clearly, $K_1$ can be found as a subspace of $K_0$ by solving linear equations. Let $A$ be a nonsingular complex $(3 \times 3)$-matrix (that is, an element of $\mathrm{GL}_3(\mathbb{C})$). Since the entries of the matrix A are real, then one may easily show that if is a complex eigenvalue, then its conjugate is also an eigenvalue. How do I prove that a matrix have all real eigenvectors? To see this, write an $n$-by-$n$ complex matrix in the form $A = X + i\,Y$ where $X$ and $Y$ are real matrices and note that finding a real eigenvector for $A$ is equivalent to finding a simultaneous eigenvector in $\mathbb{R}^n$ for both $X$ and $Y$, i.e., $X v = x\, v$ and $Y v = y\, v$. It is only a necessary condition for there to be a real eigenvector with a real eigenvalue. Learn to find complex eigenvalues and eigenvectors of a matrix. Note again that in order to be an eigenvector, $$X$$ must be nonzero. I don't know how issymmetric is implemented (it's a built-in function), but maybe eig doesn't use the same criterion to determine if a matrix is real and symmetric than issymmetric. If some of the eigenvalues are repeated, since the matrix is Real Symmetric, there will … Stability of eigenvectors for diagonal perturbations, Rotatable matrix, its eigenvalues and eigenvectors, Statistical independence of eigenvectors of real symmetric Gaussian random matrices, Properties of eigenvalues and eigenvectors of a particular random matrix, Linear Lie algebra generated by $\mathbb{R}$-diagonalizable matrices. Solution for know that a real matrix can have real and complex eigenvalues/eigenvectors, but what if we start with the condition that we have a real matrix with… Making statements based on opinion; back them up with references or personal experience. (b) Find the eigenvalues of the matrix 2 Chapter 2 part B Consider the transformation matrix . Suppose is a real matrix with a complex eigenvalue and a correspondingE#‚# + ,3 eigenvector Let @ÞTœÒ ÓÞRe Im@@ By the theorem Re Im Re ImßEœÒ Ó Ò ÓÞ@@ @@”• + ,,+ " ”• ” •È + , ,+ can be written as , where .<<œ+ , cos sin sin cos)))) ## Thus represents a counterclockwise rotation if is chosen around the originGÐ !Ñ) through the angle , followed by a rescaling fact (2). 2.31). Theorem Suppose is a real matrix with a complex eigenvalue and aE#‚# + ,3 corresponding complex eigenvector ÐÑ Þ@ Then , where the columns of are the vectors Re and Im EœTGT T Gœ + ,,+ " Ú Û Ü ”• @@and Proof From the Lemma, we know that the columns of are linearly independent, so TT is invertible. In the second example, However, if A is nonsymmetric, then U, the new basis of eigenvectors, is not orthogonal. @BrendanMcKay --- I'm a bit confused by this statement; take the matrix $A=i$ times the identity, the "matrix of imaginary parts is nonsingular", but all eigenvectors can be chosen as having only real elements. $\endgroup$ – user99914 May 12 '18 at 4:32. The above cases cover everything that can happen for a $3$-by-$3$ complex matrix $A$. For any triangular matrix, the eigenvalues are equal to the entries on the main diagonal. I think what your lecturer is getting at is that, for a real matrix and real eigenvalue, any possible eigenvector can be expressed as a real vector multiplied by a (possibly complex… If you have an eigenvector then any scalar (including complex scalar) multiple of that eigenvector is also an eigenvector. If $\dim K = 0$, then there are no simultaneous eigenvectors. Then what are some of the best-known criteria which guarantee $A$ to have real eigenvectors ? Are the only local minima of $\angle(v, Av)$ the eigenvectors? Ask Login Home Science Math History Literature Technology Health Law Business All Topics Random Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors. Then, we solve for every possible value of v. The values we find for v are the eigenvectors. quantum-mechanics hilbert-space operators quantum-information quantum-computer. Thus, $X$ and $Y$ are commuting nilpotent linear maps (since all of their eigenvalues are $0$). Prove that if λ is an eigenvalue of A, then its complex conjugate ˉλ is also an eigenvalue of A. A complex-valued square matrix A is normal ... As a special case, for every n × n real symmetric matrix, the eigenvalues are real and the eigenvectors can be chosen real and orthonormal. K_{m+1} = \{\ v\in K_m\ |\ Xv, Yv \in K_m\ \}\subseteq K_m\,, Ask Login Home Science Math History Literature Technology Health Law Business All Topics Random Last modified 08/11/2017, Your email address will not be published. Extracting complex eigenvectors from the real Schur factorization can be done but is trickier; you can see how LAPACK does it. Rounding errors can make Z*B*Z' slightly unsymmetric. 