Down with Determinants! Sheldon Axler 21 Decemb er 1994 1. Intro duction Ask anyone why a square matrix of complex numb ers has an eigenvalue, and you'll probably get the wrong answer, which go es something like this: The characteristic p olynomial of the matrix|which is de ned via determinants|has a ro ot (by the fundamental theorem of algebra); this ro ot is an eigenvalue of the matrix. What's wrong with that answer? It dep ends up on determinants, that's what. Determinants are dicult, non-intuitive, and often de ned without motivation. As we'll see, there is a b etter pro of|one that is simpler, clearer, provides more insight, and avoids determinants. This pap er will showhow linear algebra can b e done b etter without determinants. Without using determinants, we will de ne the multiplici ty of an eigenvalue and prove that the numb er of eigenvalues, counting multiplicities, equals the dimension of the underlying space. Without determinants, we'll de ne the characteristic and minimal p olynomials and then prove that they b ehave as exp ected. Next, we will easily prove that every matrix is similar to a nice upp er-triangular one. Turning to inner pro duct spaces, and still without mentioning determinants, we'll havea simple pro of of the nite-dimensional Sp ectral Theorem. Determinants are needed in one place in the undergraduate mathematics curricu- lum: the change of variables formula for multi-variable integrals. Thus at the end of this pap er we'll revive determinants, but not with any of the usual abstruse de ni- tions. We'll de ne the determinant of a matrix to b e the pro duct of its eigenvalues (counting multiplici ties). This easy-to-rememb er de nition leads to the usual for- mulas for computing determinants. We'll derive the change of variables formula for multi-variable integrals in a fashion that makes the app earance of the determinant there seem natural. This work was partially supp orted by the National Science Foundation. Many p eople made comments that help ed improve this pap er. I esp ecially thank Marilyn Brouwer, William Brown, Jonathan Hall, Paul Halmos, Richard Hill, Ben Lotto, and Wade Ramey. @ det @ 2 A few friends who use determinants in their researchhave expressed unease at the title of this pap er. I know that determinants play an honorable role in some areas of research, and I do not mean to b elittle their imp ortance when they are indisp ensable. But most mathematicians and most students of mathematics will have a clearer understanding of linear algebra if they use the determinant-free approach to the basic structure theorems. The theorems in this pap er are not new; they will already b e familiar to most readers. Some of the pro ofs and de nitions are new, although many parts of this approachhave b een around in bits and pieces, but without the attention they de- served. For example, at a recentannual meeting of the AMS and MAA, I lo oked through every linear algebra text on display. Out of over fty linear algebra texts o ered for sale, only one obscure b o ok gave a determinant-free pro of that eigen- values exist, and that b o ok did not manage to develop other key parts of linear algebra without determinants. The anti-determinant philosophyadvo cated in this pap er is an attempt to counter the undeserved dominance of determinant-dep endent metho ds. This pap er fo cuses on showing that determinants should b e banished from much of the theoretical part of linear algebra. Determinants are also useless in the com- putational part of linear algebra. For example, Cramer's rule for solving systems of linear equations is already worthless for 10 10 systems, not to mention the much larger systems often encountered in the real world. Many computer programs eciently calculate eigenvalues numerically|none of them uses determinants. To emphasize the p oint, let me quote a numerical analyst. Henry Thacher, in a review (SIAM News , Septemb er 1988) of the Turb o Pascal Numerical Metho ds Toolbox , writes, I nd it hard to conceive of a situation in whichthenumerical value of a determinant is needed: Cramer's rule, b ecause of its ineciency,iscom- pletely impractical, while the magnitude of the determinant is an indication of neither the condition of the matrix nor the accuracy of the solution. 2. Eigenvalues and Eigenvectors The basic ob jects of study in linear algebra can b e thought of as either linear transformations or matrices. Because a basis-free approach seems more natural, this pap er will mostly use the language of linear transformations; readers who prefer the language of matrices should have no trouble making the appropriate translation. The term linear op erator will mean a linear transformation from a vector space to itself; thus a linear op erator corresp onds to a square matrix (assuming some choice of basis). Notation used throughout the pap er: n denotes a p ositiveinteger, V denotes an n-dimensional complex vector space, T denotes a linear op erator on V ,andI denotes the identity op erator. A complex number is called an eigenvalue of T if T I is not injective. Here is the central result ab out eigenvalues, with a simple pro of that avoids determinants. @ det @ 3 Theorem 2.1 Every linear op erator on a nite-dimensional complex vector space has an eigenvalue. Pro of. Toshow that T (our linear op erator on V ) has an eigenvalue, x any non- 2 n zero vector v 2 V . The vectors v; T v; T v; : : : ; T v cannot b e linearly indep endent, b ecause V has dimension n and wehave n +1 vectors. Thus there exist complex numbers a ;:::;a , not all 0, such that 0 n n a v + a Tv + + a T v =0: 0 1 n Make the a's the co ecients of a p olynomial, which can b e written in factored form as n a + a z + + a z = c(z r ) :::(z r ); 0 1 n 1 m where c is a non-zero complex numb er, each r is complex, and the equation holds j for all complex z .We then have n 0= (a I + a T + + a T )v 0 1 n = c(T r I ) :::(T r I )v; 1 m which means that T r I is not injective for at least one j .Inotherwords, T has j an eigenvalue. Recall that a vector v 2 V is called an eigenvector of T if Tv = v for some eigenvalue . The next prop osition|which has a simple, determinant-free pro of| obviously implies that the numb er of distinct eigenvalues of T cannot exceed the dimension of V . Prop osition 2.2 Non-zero eigenvectors corresp onding to distinct eigenvalues of T are linearly indep endent. Pro of. Supp ose that v ;:::;v are non-zero eigenvectors of T corresp onding to 1 m distinct eigenvalues ;:::; .We need to prove that v ;:::;v are linearly inde- 1 m 1 m p endent. To do this, supp ose a ;:::;a are complex numb ers such that 1 m a v + + a v =0: 1 1 m m Apply the linear op erator (T I )(T I ) :::(T I ) to b oth sides of the 2 3 m equation ab ove, getting a ( )( ) :::( )v =0: 1 1 2 1 3 1 m 1 Thus a = 0. In a similar fashion, a = 0 for each j , as desired. 1 j @ det @ 4 3. Generalized eigenvectors Unfortunately, the eigenvectors of T need not span V .For example, the linear 2 op erator on C whose matrix is 0 1 0 0 has only one eigenvalue, namely 0, and its eigenvectors form a one-dimensional 2 subspace of C .We will see, however, that the generalized eigenvectors (de ned b elow) of T always span V . Avector v 2 V is called a generalized eigenvector of T if k (T I ) v =0 for some eigenvalue of T and some p ositiveinteger k . Obviously, the set of generalized eigenvectors of T corresp onding to an eigenvalue is a subspace of V . The following lemma shows that in the de nition of generalized eigenvector, instead of allowing an arbitrary p ower of T I to annihilate v ,we could have restricted th attention to the n power, where n equals the dimension of V . As usual, ker is an abbreviation for kernel (the set of vectors that get mapp ed 0). Lemma 3.1 The set of generalized eigenvectors of T corresp onding to an eigen- n value equals ker(T I ) . n Pro of. Obviously,every elementofker(T I ) is a generalized eigenvector of T corresp onding to . To prove the inclusion in the other direction, let v be a n generalized eigenvector of T corresp onding to .Weneedtoprove that (T I ) v = 0. Clearly,we can assume that v 6= 0, so there is a smallest non-negativeinteger k k such that (T I ) v =0.Wewillbedoneifwe show that k n. This will b e proved byshowing that 2 k 1 v; (T I )v; (T I ) v; : : : ; (T I ) v (3.2) are linearly indep endentvectors; we will then have k linearly indep endent elements in an n-dimensional space, which implies that k n. Toprove the vectors in (3.2) are linearly indep endent, supp ose a ;:::;a are 0 k 1 complex numbers such that k 1 a v + a (T I )v + + a (T I ) v =0: (3.3) 0 1 k 1 k 1 k 1 Apply (T I ) to b oth sides of the equation ab ove, getting a (T I ) v =0, 0 k 2 which implies that a =0.Now apply (T I ) to b oth sides of (3.3), getting 0 k 1 a (T I ) v = 0, which implies that a =0: Continuing in this fashion, wesee 1 1 that a = 0 for each j , as desired.