Alg1231W8T1 - Eigenvectors and Eigenvalues

Sunday, 9 September 2018

6:46 PM

 

 

Machine generated alternative text: The eigenvectors of a square matrix A are all for some scalar A, called the eigenvalue non-zero O vectors v such that For example; if A o -2 3 -5 -3 -3 5 2 then we can directly calculate that 2 1 1 0 -2 3 -5 -3 -3 5 2 2 5 1 1 So we have demonstrated that is an eigenvector with eigenvalue 1 Note: the Maple notation for the vector Similarly we can check that 2 3 1,2,3 > 1 1 1 1 is an eigenvector with eigenvalue is an eigenvector with eigenvalue

 

Machine generated alternative text: -0.87 -0.5 Consider the matrix This matrix has eigenvalues 2 and 3 with respective eigenvectors Note: 0.12 • the Maple notation for the vector is < 0.12, 0.34 > 0.34 • make sure VI corresponds to the eigenvalue 2 and v2 corresponds to the eigenvalue 3.

 

Machine generated alternative text: 0.97 0.26 Consider the matrix This matrix has eigenvalues —1 and 3 with respective eigenvectors 112

 

Machine generated alternative text: For a square matrix A, the statement is equivalent to the statement Say your friend claims that the matrix A equations 10 1 -1 2 11 1 -2 -1 9 "A has an eigenvalue " "(A — M)v = O has a non-zero solution v ' has eigenvalue = 9 and wants our help finding the eigenvector We proceed by attempting to solve the system of linear (A — 91)v = O. That is, we try to solve the system represented by the augmented matrix 1 1 —1 This matrix can be row-reduced to 1 o o Recalling our experience in Math1131 we can see this has a unique solution; namely v 1 Note: the Maple notation for the vector 2 1 , > 2 2 1 2 3 o —2 —2 1 o 3 Since an eigenvector MUST be non-zero our friend should double check their work Cl the eigenvector is O that 9 is not an eigenvalue of the matrix A. O , we conclude that [select all that apply]

 

Machine generated alternative text: Your friend from the last question claims that the matrix B 7 1 -1 2 8 1 -2 -1 has eigenvalue 6 5 (your friend swears they double-checked this time). We can check this claim by attempting to solve the system of linear equations represented by the augmented matrix This can be row-reduced to the matrix and so we can find a non-zero solution v = 1 Note: the Maple notation for the vector 2 is < 1 3 From this we can conclude that [select all that apply] 5 really is an eigenvalue of the matrix our friend might just pass this course 5 is not an eigenvalue of the matrik 2 1 —1 2 o o 2 3 1 2 2 o

 

Machine generated alternative text: For a square matrix A, the statement "(A — = O has a non-zero solution v ' is equivalent to the statement "det(A - M) = O The quantity det(A — M) is a polynomial in A, known as the characteristic polynomial of A, and its roots are the eigenvalues of A. This gives us a technique for finding the eigenvalues of t2 — Let's check our understanding 1 i) The characteristic polynomial of A 2 2 1 det(A - ti) 2 is the cubic 2 = det 2 VI ¯ and v 2 ¯ det(B - ti) - (I-t)A2 - 4 O . Ordered the same way, the associated eigenvectors are Ordered tl < t2, the roots ofthis roots are tl — 1 Note: the Maple notation for the vector is < ii) The characteristic polynomial of B 2 5 -1 o -1 5 0 1, 0 -3 Hint: try expanding the determinant along the last row or last column. Ordered tl < t2 < t3 this has roots (and hence B has eigenvalues) -3,

 

 

MATH1131 -
The determinant of a matrix

 

 

Created with Microsoft OneNote 2016.