Finding Eigenvectors of a 3x3 matrix

in this video we're going to learn how to find the eigenvectors of a 3x3 matrix in a previous video we've already learned how to find the eigenvalues so we have to just given here but we're going to recall that the eigenvalues are given by a times some vector x equals lambda times the same vector X where X is nonzero and here lambda is the eigen value and this X is the eigenvector of the matrix a now we're going to rearrange this equation to have a minus lambda I times X equals 0 and we know that we can solve those kinds of equations using the Gaussian elimination hmm now we have this eigenvalue lambda equals minus 3 here so our a minus lambda I times the vector X is in this case minus 2 minus minus 3 is 1 then we just rewrite 2 minus 3 2 then a 1 minus minus 3 is 4 and we have minus 6 minus 1 minus 2 and 0 minus minus 3 is just 3 times X which is X 1 X 2 X 3 and that is equal to 0 the 0 vector now we can use Gaussian elimination to find our vector X 2 minus 6 0 and minus 1 minus 2 3 0 so we're going to reduce that by adding the third row to the first one and adding twice the third row to the second one and that leaves us with just zeros and both the first and the second row and then the third one stays the same from that we can read them - 1 X 1 minus 2 X 2 Plus 3 X 3 is equal to 0 now suppose that X 1 is equal to some number 8 and X 2 is equal to some B then if we rearrange this equation we're going to have X 3 equals 1/3 X 1 plus 2/3 X 2 and therefore our eigenspace can be written as for the lambda equals minus 3 can be written as a times some vector plus b times some another vector now X 1 is equal to a therefore we're going to have 1 here as 1 times a is just 8 then X 2 has no ay in it so we're going to put 0 here and X 3 has a third a and then for B X 1 and so now B x2 is equal to B and then X 3 is as 2/3 therefore our eigenvectors are given by x equals one zero on the third and then x equals to 0 1 and 2/3 let's for lambda equals minus 3 now we're going to move on to lambda equals 5 and again our a minus lambda I times the vector x equals -2 minus 5 is minus 7 then we rewrite 2 minus 3 & 2 1 minus 5 is minus 4 minus 6 minus 1 minus 2 and minus 5 and we multiply that by our vector X to obtain the zero vector now we're going to use the Gaussian elimination again so we have minus 7 2 minus 3 0 2 minus 4 - 6 0 minus 1 minus 2 minus 5 and 0 until it is that we're going to subtract 7 times the second row from the first row and then for the second row we're going to add the third row twice and that will leave us with 0 16 32 0 + 0 - 8 - 16 0 and the last row stays the same - 1 minus 2 minus 5 and 0 for the first row if we add to it twice the second row that will leave us with straight zeros and the first row second one stays the same 9:16 zero and the third one stays the same again now we're left with those two equations - 8 X 2 minus 16 X 3 equals 0 and minus 1 X 1 minus 2 X 2 minus 5 X 3 equals 0 as well we can reduce down to get x 2 equals -2 X 3 and then if we just multiply that by minus 1 that will give us X 1 plus 2 X 2 plus 5 X 3 equals 0 now suppose again that X 2 this time is equal to some a and from that we have that X 3 is equal to minus 1 over half 8 and then from the other equation we have X 1 if we rearrange it for us we have X 1 equals -2 X 2 minus 5 X 3 and that will give us minus 2 a plus 5 over 2 a and that is just 1/2 a therefore our eigenspace peace is given by a 4 lambda equals 5 a time's the vector we have x1 is a half a so we have a half here x2 as a so we just have 1 here and X 3 is minus 1/2 so that is here therefore our eigen vector s here we're going to use red for that x equals 1/2 1 or minus 1/2 or it's going to look nicer X is just 1 2 and minus 1 both of them satisfy the equation ax equals lambda X thank you very much

Channel: Mathematics Aberystwyth

Views: 43,208

Rating: 4.2095809 out of 5

Keywords: Eigenvector

Id: 05hNXHv6idw

Length: 10min 59sec (659 seconds)

Published: Thu Feb 12 2015