DIGONLIZTION Definition: n n x n matrix, "", is said to be diagonalizable if there exists a nonsingular matrix "X" and a diagonal matrix "D" such that X X D. Theorem: n n x n matrix, "", is diagonalizable only if "" has "n" linearly independent eigenvectors. Let "" have "n" linearly independent eigenvectors "x ", with corresponding eigenvalues," λ ",,,.., n. Construct the matrix "X" by using the eigenvectors as its columns. X x x... x n x λ x X λ x λ x... λ n x n XD λ x λ x... λ n x n Here "D" is a diagonal nxn matrix with eigenvalue " λ i " as the element "d ii " of "D". X XD Page of 9
Since the columns of "X" are linearly independent, "X" is nonsingular and is thus invertable. X X D Example # : Factor the matrix into a product X D X, if possible. λ I λ λ λ λ λ λ x x x X D X Page of 9
The eigenvectors are linearly independent. Therefore, X ( x x x ) is nonsingular and thus "" is diagonalizable with diagonal form λ D λ λ. Example # : Compute: using the factorization : XDX. ( XD X ) XD X XD X X DX XD IDX XD X Continue the process to obtain XD X. Page of 9
Example # : Factor the matrix 4 into a product XD X, if possible. "" is triangular so its main diagonal entries are its eigenvalues. λ 4, multiplicity of "". λ 5, distinct. If "" is diagonalizable, then the eigenspace associated with the repeated eigenvalue of mutiplicity "" must be -dimensional. Otherwise, the matrix "" is said to be defective and not diagonalizable because the eigenvectors do not span R. 4 5 4I x is the only free variable. So we get only "" eigenvector not "" as required to span the subspace of dimension equal to the multiplicity of the eigenvalue. The matrix "" is thus defective and therefore not diagonalizable. Page 4 of 9
Theorem: Let the n x n matrix, "", have "p" distinct eigenvalues : ( λ, λ,..., λ p ). (a) For ( p), the dimension of the eigenspace for λ is less than or equal to the multiplicity of the eigenvalue, λ. (b) The matrix, "", is diagonalizable if and only if the sum of the distinct eigenspaces equals "n", and this occurs if and only if the dimension of the eigenspace for each λ equals the multiplicity of λ. (c) If "" is diagonalizable and β is a basis for the eigenspace corresponding to λ for each "", then the total collection of vectors in the sets: β, β,..., β p forms an eigenvector basis for R n. Example # 4: Show that the matrix 5 which has λ λ and λ, where "λ " is an eigenvalue of multiplicity "" is diagonalizable. 7 4 5 6 8 I 4 4 6 8 8 4 Page 5 of 9
In this example, we have "" free variables associated with the eigenvalue of multiplicity "". Thus its eigenspace is -dimensional and thus the matrix is not defective. Therefore, we can diagonalize "". Example # 5: Factor, where product. XDX into a XDX ( XD X ) ( X ) D X XD X If "D" has diagonal elements "d ii corresponding diagonal elements " λ i ". λ i ", then "D " has From Example #, we found that "" could be diagonalized. XDX Page 6 of 9
X X X XD Definition: The matrix exponential, e, of the matrix "" of dimension "nxn" is itself a matrix also of dimension "nxn" but is defined as an infinite series of nxn terms. e I I!...!... Theorem: If "" is diagonalizable, where then e Xe D X. XDX, e I XD X I XIX e XI X XD X Page 7 of 9
e XI X D e D I D e Xe D X Theorem: The i th diagonal element of e D is e λ i if e Xe D X. e D I D The i th diagonal element of D is ( λ i ). The i th diagonal element of D is λ i. Page 8 of 9
th The i diagonal element of D is λ i. th The i diagonal element of I D is λ i. But ( λ i ) e λ i. Therefore, the i th diagonal element of e D is e λ i. Example # 6: Compute e, where. λ λ x x Page 9 of 9
X D e D e Xe D X e e e ( e e ) ( e e ) ( e e ) ( e e ) Theorem: The solution to the Initial Value Problem (IVP) d dt y y where y y is y() t e t y. Let "x" be a nonzero vector that is independent of "t", but otherwise arbitrary. Page of 9
e t x I t t! t...! t... x d t et x d I t t! t..! t.. x d t et x d e t x Now let y e t x, then d dt y y. y e x Ix x y yt () e t y Page of 9
Example # 7: Solve the IVP where. d dt y y y 4, y yt () e t e t 4 λ λ x x X D e D e t e t Page of 9
yt () e t y Xe td X y yt () e t e t 4 yt () e t e t e t e t Example # 8: Solve the IVP. d dt y y y, where λ λ λ x Page of 9
The matrix "" is thus defective so we can not diagonalize "". But we can still find an explicit, closed-form solution to this IVP. Consider the following identity. t λt I tλi Exponentiate both sides. e t e λti tλ I e λti e tλ I e λt I I ( λt) I n e λt I e t I I ( λt) λt e λt I e λt Ie tλ I e λt e tλ I e t e λt e tλ I I t λ I e tλ I t λi...! Page 4 of 9
e t e λ t I t λ I t λi...! I e t( I) I t e t t..! Page 5 of 9
The null space of the matrix ( I) contains nonzero vectors, precisely the number of additional vectors that we need. Let x and x denote those vectors. Since ( I) x and ( I) x, then ( I) x and ( I) x for. x x However, x must also lie in Nul ( I) we can only choose one of the two. We pic x.. Thus, is all of R, so our third vector Note that Nul ( I) can be any vector that does not lie in the plane formed by x and x. Page 6 of 9
Choose x. x e t Ix t( I) x e t x e t Ix t( I) x e t x e t e t Ix t( I) x t ( I) x Ix t( I) x x Ix t( I) x t t t t Page 7 of 9
x t I x t I x t t x t I x t I x t t t t t t y c c c c c c c c c c 4 Page 8 of 9
yt () e t t 4 t 4t yt () e t 5t 4t t 4t t t t t t t t t Page 9 of 9