THE LINEAR IMAGE of a SPHERE IS an ELLIPSOID in This Note, We

THE LINEAR IMAGE OF A SPHERE IS AN ELLIPSOID

R. CLARK ROBINSON

In this note, we explain why the image of a sphere by a linear map is an ellipsoid in dimensions three and above; in dimension two, the image of a circle by a linear map is an ellipse. Let A be a real n × n real matrix. The unit sphere in n-dimensions is the set of vectors of length one, n U = {x ∈ R : |x| = 1}. In dimension two, U is the unit circle. We show that the image A U is an ellipsoid. We want to explain why this is true and determine the axes of the ellipsoid and their lengths in terms of eigenvalues and eigenvectors. These quantities are used to define Lyapunov exponents in higher dimensions. We summarize the result at the end of the section with the precise statement of the theorem. The quantity |Av|/|v| is the amount the vector v is stretched when acted on by A. The square of this length, |Av|2 = (Av)T Av = vT AT Av, is related to the matrix AT A. This matrix is symmetric so has real eigen- T T T T T T values λj:(A A) = A (A ) = A A. The corresponding eigenvectors vj can be taken as an orthonormal basis, i.e., length one and mutually perpendicular: vk · vj = 0 for k 6= j and vj · vj = |vj|2 = 1. Also, 0 ≤ |Avj|2 = (Avj)T (Avj) j T T j j T j = (v ) A Av = (v ) λjv j 2 = λj|v | = λj. T Therefore, the eigenvalues λj of A A are all nonnegative. (Such a symmetric matrix AT A is called positive semi-definite.) This first step has T shown that the eigenvalues λj of A A are real and nonnegative. Their j p corresponding eigenvectors {v } form an orthonormal basis. Let sj = λj. j j 2 j If we define P to be the matrix such that Pv = sjv , then P v = s2vj = AT Avj and P is a positive semi-definite symmetric matrix. Since j √ P2 = AT A, P can be interpreted as the square root of AT A, P = AT A. The eigenvectors vj turn out to be the directions that determine the j j various amounts of stretching of vectors by A; their images Av = sju give the axes of the ellipsoid.

Date: May 8, 2002. 1 2 R. CLARK ROBINSON

The second step is to show that the scaled images of the vectors vj by j j T A, u = (1/sj)Av , are the eigenvectors of the matrix AA with the same eigenvalues λj. For simplicity, we have assumed that all the λj > 0; if sj = 0 for 1 ≤ j ≤ k, then we pick these uj as unit vectors which are perpendicular to ui for k < i ≤ n. We omit the details below for this possibility. The proof of the second step is a straight forward calculation which follows from the deﬁnitions regrouping the matrix product: 1 (AAT )uj = (AAT ) Avj sj 1 = A(AT A)vj sj 1 j = Aλjv sj j = λju .

The third step is to show that the vectors {uj} form an orthonormal basis, i.e., uk · uj = 0 for k 6= j and uj · uj = |uj|2 = 1. Again this is a straight forward calculation: 1 1 uk · uj = (uk)T uj = ( Avk)T ( Avj) sk sj 1 k T T j 1 k T j = (v ) (A A)v = (v ) λjv sksj sksj ( λ 0 for k 6= j = j vk · vj = sksj 1 for k = j.

The fourth step is to show that the image A U is an ellipsoid. To determine the image we express any vector x ∈ U in terms of the orthonormal j 1 n basis {v } rather than the standard basis: x = x1v + ··· + xnv ∈ U and 2 2 2 |x| = x1 + ··· xn = 1. The image point is 1 n y = Ax = x1Av + ··· + xnAv 1 n = x1s1u + · + xnsnu .

Thus, the component of the image point y = Ax has components yj = xjsj in terms of the basis {uj},

2 2 y1 yn 2 2 2 + ··· + 2 = x1 + ··· + xn = 1, s1 sn and y = Ax is on the ellipsoid

2 2 1 n y1 yn {y1u + · + ynu : 2 + ··· + 2 = 1} s1 sn ELLIPSOID 3

j p with axes along the vectors u , and with length of the axes sj = λj, i.e., 2 2 1 n y1 yn A U = {y1u + ··· + ynu : 2 + ··· + 2 = 1}. s1 sn The fifth and final step is to write the matrix A as the product of an orthogonal matrix and a positive semi-definite symmetric matrix. As we said at the end of step one, there is a positive semi-definite symmetric matrix P such that P2 = AT A. If we let Q = AP−1, then QT Q = (P−1)T AT AP−1 = P−1P2P−1 = I, so Q is an orthogonal matrix. (In terms of the basis given above, Qvj = uj.) Thus any square matrix A can be written as A = QP where Q is orthogonal and P is positive semi-definite symmetric. By a similar argument, it can also be written as A = P0Q0 where P0 is positive semi-definite symmetric and Q0 is orthogonal. We can further decompose the matrix A. Write A = QP as above. Then the positive semi-definite symmetric matrix P can be conjugated to a non-negative diagonal matrix Λ by an orthogonal matrix Q1. Then A = −1 QQ1 ΛQ1 = Q2ΛQ1 where Q1 and Q2 are orthogonal, and Λ is a nonnegative diagonal matrix. We summarize the results given above in the following theorem. Theorem. Let A be a real n×n matrix. Then, (a) the eigenvalues of AT A T and AA are the same, {λ1, . . . , λn}. Moreover, all the λj are real and p 1 n λj ≥ 0. Let sj = λj. Let {v ,..., v } be the corresponding eigenvectors T j j 1 n T of A A and sju = Av . Then {u ,..., u } are the eigenvectors for AA , and these two sets of eigenvectors can be chosen to be orthonormal, vk ·vj = 0 = uk · uj for k 6= j and |vj| = 1 = |uj|. Finally, the image of the n-sphere j j by A, A U, is an ellipsoid with semi-axes given by the vectors sju = Av . j j (b) Also, if we define the matrix P by P v = sjv , then P is a positive semi-definite symmetric matrix. If we define the matrix P by If we let Q = AP−1, then Q is an orthogonal matrix. Finally, A = QP. Thus, any matrix A can be written as the product of an orthogonal matrix and a positive semi-definite symmetric matrix.

Department of Mathematics, Northwestern University, Evanston IL 60208 E-mail address: [email protected]