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The concept of diagonalization of matrices in linear algebra. It explains when a matrix is diagonal and provides an example of diagonalizing a matrix using eigenvectors and an invertible matrix. The document also introduces the diagonal matrix d and the transition matrix s, and shows how to find the matrix a10 and the exponential function ea. The document concludes with an application of diagonalization to systems of differential equations.
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Linear Algebra
Harold P. Boas
June 28, 2006
From last time: I (^) application of eigenvectors to systems of differential equations Today: I (^) diagonalization of matrices and applications
Suppose a linear operator L on R^3 is represented in a basis
[ u 1 , u 2 , u 3 ] by the matrix
This means that L u 1 = 2 u 1 and L u 2 = 3 u 2 and L u 3 = 5 u 3. In other words, the basis vectors are eigenvectors of the operator L. A square matrix A is diagonalizable if the linear transformation x 7 → A x can be represented in some basis by a diagonal matrix; in other words, if there is a basis consisting of eigenvectors of A; equivalently, if there is an invertible matrix S such that S−^1 AS is a diagonal matrix.
Diagonalize the matrix A =
. In other words, find an invertible matrix S and a diagonal matrix D such that S−^1 AS = D or A = SDS−^1. Solution. First find the eigenvalues and eigenvectors of A. The vector
is an eigenvector with eigenvalue 1, and the
vector
is an eigenvector with eigenvalue 2. The matrix
is the transition matrix from the eigenvector basis to the standard basis, and the matrix S−^1 AS is the diagonal matrix
If A =
, find A^10.
Solution. Since S−^1 AS = D =
, and D^10 =
we have A^10 = SD^10 S−^1 =
More. Since the exponential function is given by a power series (ex^ = 1 + x + (^21)! x^2 + (^31)! x^3 + · · · + (^) n^1! xn^ + · · ·), we also have e (A := I + A + (^21)! A^2 + · · · = SeD^ S−^1 = 3 2 − 2 − 1
e 0 0 e^2
− 3 e + 4 e^2 − 6 e + 6 e^2 2 e − 2 e^2 4 e − 3 e^2
We have two ways to solve the system of differential equations y ′^ =
y.
From yesterday, we can write the general solution as y (t) = c 1 et
Alternatively, we can write the general solution as y (t) = etA
c 1 c 2
= SetD^ S−^1
c 1 c 2
− 3 et^ + 4 e^2 t^ − 6 et^ + 6 e^2 t 2 et^ − 2 e^2 t^4 et^ − 3 e^2 t
c 1 c 2
In the second form,
c 1 c 2
y 1 ( 0 ) y 2 ( 0 )