Nevermind, found the mistake, it's actually
[1 -1 |0]
[0 0 |0]
Hi, I'm new here, and couldn't find the 'Linear algebra' subforum, only noticed advanced algebra and this subforum. So I figured, since this is not a 2nd year subject I'd post here. My apologies if this post doesn't belong here.
In class I came to understand that the eigenvector is the vector v, in the equation (A-lamda*I)*v = 0, where A is an nxn matrix, lamda belong to R, is the eigenvalue of this matrix.
In an example we solved: for A =
[1 3]
[2 2]
where lamda = -1 and 4
For lamda = -1: its (A+I)*v=0 simplifies to:
[1 3/2 | 0 ]
[0 0 | 0 ] so converting this into vector form
[v1] v2*[-3/2]
[v2] = [ 1 ] So v2=t then the eigenvector is
[-3] * t for t belong to R
[2]
I understand for the above step it was taken from the linear simultaneous equation system:
1) v1+(3/2)v2=0
2) 0(v1) + 0(v2) = 0
Because of the 2nd row being all 0s, we have an infinite amount of solutions, which is why we had to add a "t belong to R" parameter to the eigenvector.
BUT for lamda = 4:
{ [1 3] - 4* [1 0] } * v = 0
{ [2 2] [0 1] }
it simplifies down to
[1 0 | 0]
[0 1 | 0]
If I was to follow the same method of linear equation system as above, how is something like this converted into the eigenvector
[1] * t for t belong to R?
[1]
since: v1+0(v2) = 0
0(v1)+v2 = 0 shouldn't v =
[0] instead of [1]
[0] instead of [1] ? And that it is a unique solution without even the need for parameter t?
The advanced algebra sub-forum includes Linear Algebra (it's in the description). This sub-forum is for pre-university algebra.
Solving for A - 4I = 0 leads to:
[-3 3][x]....[0]
[2 -2][y] = [0]
which leads to x = y.
So any non-zero scalar multiple of (1,1) is an eigenvector belonging to the eigenvalue 4.
[QUOTE=autumnwater;809869]Hi, I'm new here, and couldn't find the 'Linear algebra' subforum, only noticed advanced algebra and this subforum. So I figured, since this is not a 2nd year subject I'd post here. My apologies if this post doesn't belong here.
In class I came to understand that the eigenvector is the vector v, in the equation (A-lamda*I)*v = 0, where A is an nxn matrix, lamda belong to R, is the eigenvalue of this matrix.[quote]
Better to think "Av= lambda v". Of course that is the same as "(A- lambda I)v= 0" but is more fundamental.
An eigenvector for this matrix, with eigenvalue -1 must satisfyIn an example we solved: for A =
[1 3]
[2 2]
where lamda = -1 and 4
which is equivalent to the two equations x+ 3y= -x and 2x+2y= -y. Those both reduce to 3y= -2x, y= (-2/3)x so that a eigenvector must be of the form <x, (-2/3)t>= 3t<3, -2>. A vector is an eigenvector of A corresponding to eigenvalue -1 if and only if it is a multiple of <3, -2>.
For eigenvalue 4, the equivalent
which gives the two equations x+ 3y= 4x and 2x+ 2y= 4y. Those both reduce to y= x so that an eigenvector corresponding to eigenvalue 4 is <x, x>= x<1, 1>. A vector is an eigenvector of A corresponding to eigenvalue 4 if and only if it is a multiple of <1, 1>.
For lamda = -1: its (A+I)*v=0 simplifies to:
[1 3/2 | 0 ]
[0 0 | 0 ] so converting this into vector form
[v1] v2*[-3/2]
[v2] = [ 1 ] So v2=t then the eigenvector is
[-3] * t for t belong to R
[2]
I understand for the above step it was taken from the linear simultaneous equation system:
1) v1+(3/2)v2=0
2) 0(v1) + 0(v2) = 0
Because of the 2nd row being all 0s, we have an infinite amount of solutions, which is why we had to add a "t belong to R" parameter to the eigenvector.
BUT for lamda = 4:
{ [1 3] - 4* [1 0] } * v = 0
{ [2 2] [0 1] }
it simplifies down to
[1 0 | 0]
[0 1 | 0]
If I was to follow the same method of linear equation system as above, how is something like this converted into the eigenvector
[1] * t for t belong to R?
[1]
since: v1+0(v2) = 0
0(v1)+v2 = 0 shouldn't v =
[0] instead of [1]
[0] instead of [1] ? And that it is a unique solution without even the need for parameter t?