Question

@MIT 18.06 isnt the null space of this 3x3 matrix [0,0,0;-2,4,0;4,8,-2] just zero???
@MIT 18.06 isnt the null space of this 3x3 matrix [0,0,0;-2,4,0;4,8,-2] just zero???
@MIT 18.06 Linear …

Answer 1

no, there are vectors other than the 0 vector that when you multiply the matrix and vector you get 0. For example, (1,0,0), (20,0,0) ,etc are all in the Null Space.

Its because the first row is all zeros.

Answer 2

ah im sry, those vectors arent in the null space. my bad >.<

Answer 3

so you are saying \[ \left[\begin{matrix}0 & 0 & 0\\ -2& 4& 0\\ 4 & 8 & -2\end{matrix}\right]\] times [1,0,0] ^T equals zero???

Answer 4

my bad my bad >.< [2,1,8]^T is in the null space. any multiple as well.

Answer 5

i tried to eyeball it and failed miserably >.<

Answer 6

In general, if you row reduce a matrix and end up with any all zero rows, the Null Space wont be trivial. There has to be a non zero vector in the null space.

Answer 7

I appreciate the effort. :) Anyways The original problem asked me to diagonalize a 3x3 matrix. I can find the eigenvalues just fine. When I remove an eigenvalue from the diagonal (make the matrix singular) I cant seem to find the corresponding eigenvector. I don't know why I keep getting it wrong

Answer 8

ah, so you are looking for a basis for the null space of:\[A-\lambda I\]
where lambda is the eigenvalue. if you dont mind, can you give me the original matrix and the eigenvalue you are working with? i'll write out a solutions and post it.

Answer 9

Ok this is the original matrix\[\left[\begin{matrix}3 & 0 & 0\\ -2 & 7 & 0\\ 4 & 8 & 1\end{matrix}\right]\]

Answer 10

oo, its diagonal, so you can already tell the eigenvalues are 3, 7, and 1.

Answer 11

How about this matrix??\[\left[\begin{matrix}1 & 0& 2 \\ 0 & 1& 0\\ 2& 0& 1\end{matrix}\right]\] I thought the eigenvalues were 3, 1 and 1 at first but the trace indicated that it should be 3,-1 and 1...i calculate the wrong eigenvectors for both matrices

Answer 12

The trace only tells you what the sum of the eigenvalues will be (as the determinant of the matrix tells you the product). Im scanning the solution to finding the first eigenvector corresponding to the eigenvalue of 3. one sec while i upload it here.

Answer 13

OK. Thank you. I was using an online matrix caluclator to check my work and it said that the eigen vector matrix was

0, 0, 0.2407717061715384;
0, 0.6000000000000001, 0.1203858530857692;
1, 0.8, 0.9630868246861536;

which threw me off. seeing your answer confirmed that I know what I am doing ...at least a little bit :)

Answer 14

cool :) the eigenvalues for the second matrix are 3, 1, and -1.

Answer 15

last question. Can you help me check the eigenvector for lambda = 1 ?

Answer 16

Ok well you were helpful. I'll call it a night.