Notes A2: Linear Algebra

Math 60 -- D. C. Smolarski, S.J.
Santa Clara University, Department of Mathematics and Computer Science

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Linear Algebra

Let A be a matrix, x a vector, and ka scalar constant. If it is true that
Ax = kx
then x is said to be an eigenvector of A and k the corresponding eigenvalue.

For example, 


          ( 1 0 ) (1)          (1)
	  ( 0 2 ) (0)    =  1  (0)
implies that 1 is an eigenvalue and (1 0)T the corresponding eigenvector of the matrix A, where 

A = (1 0)
    (0 2)
We also know that 2 is an eigenvalue and (0 1)T the corresponding eigenvector of the same matrix. (Both sides multiply to (0 2)T.) (The number of eigenvalues/vectors is the same as the dimension of the matrix.)

It is not always obvious what the correct eigenvalues and eigenvectors are for a matrix. Given 


       ( 1 2 1 )
A =    ( 2 3 4 )
       ( 3 4 5 )
an eigenvector is (.243, .591, .769)T, which corresponds to eigenvalue 9.025. (Both sides multiply to (2.195, 5.336, 6.939)T.)

Companion Matrix

Given a polynomial with leading coefficient 1, e.g., x4 + bx3 + cx2 + dx + e = 0, the companion matrix for this polynomial is the matrix with all zeroes except for (1) ones in the superdiagonal (diagonal above the major diagonal), and (2) the negatives of the coefficients in the last row (with the constant in the first column).

For example, with the polynomial given above, we have 


( 0  1  0  0 )
( 0  0  1  0 )
( 0  0  0  1 )
(-e -d -c -b )
as the companion matrix.

THEOREM: The roots of the polynomial are the same as the eigenvalues of the companion matrix.

This page is maintained by Dennis C. Smolarski, S.J. dsmolarski@math.scu.edu
© Copyright 2000 Dennis C. Smolarski, SJ, All rights reserved.
Last changed: 9 February 2000.