Question

Show that the following matrix has only one distinct eigenvalue. What is it? (matrix in comments)

Answer 1

\[C=\left[\begin{matrix}-1 & 0&1 \\ -1 & 0&1\\-1 & 0&1\end{matrix}\right]\]

Answer 2

let me see if i we can work together as i've never done this yet i'm in linear algebra as of now

Answer 3

Great - always good to have others working on the same material!

Answer 4

\[\text{hint: the eigenvalues} \ \lambda \ \text{satisfy the equation:}\]

\[\det\left(A-\lambda I\right)=0\]

Answer 5

my books has the equatoion \[(\lambda I - A)x = 0\]

Answer 6

as i haven't done determinants yet and the last time i did them was 4 years ago i might just tick with this equation

Answer 7

hmm never seen it in that form

Answer 8

that might be an intermediate step

Answer 9

alright so i'll do it your way jim if i understand determinants right

Answer 10

gahd i hate the equation editor how do you make a 3x3

Answer 11

\[\det(A-\lambda I) = (-1-\lambda)(0-\lambda)(1-\lambda)+0+0-(1)(0-\lambda)(-1)-0-0=0\]

Answer 12

Fill in the blanks \[\left[\begin{matrix}? & ? & ?\\ ? & ? & ?\\? & ? & ?\end{matrix}\right]\]

Answer 13

Is that along the right lines? (sorry this is a bit confusing, helping both of us!)

Answer 14

hold on

Answer 15

looks right sorta

Answer 16

that looks right

Answer 17

if \[A=\left[\begin{matrix}a & b&c \\ d&e&f\\g&h&i\end{matrix}\right]\]

then det(A) = (aei + bfg + cdh - ceg - bdi - afh) ... according to my notes!

Answer 18

the characteristic polynomial is just \[-\lambda^3\]

thus the only eigenvalue is 0

Answer 19

When you take the determinant of \[\large A-\lambda I\], you should get \[\large -\lambda^3\]

Answer 20

my book shows expanding the matrix by the first two colums

Answer 21

which is what you have so if you correctly wrote it down, then yes that is correct

Answer 22

yes, if feldy expands and simplifies then it'll come to negative lambda cubed

Answer 23

Yep :) thanks!

Answer 24

... OK if the next part of the question is "Obtain a complete set of independent eigenvectors of C. Your answer should be a set of column vectors." ... How would one go about answering that?

Answer 25

Jim , got a calculator questiooon go on group chat

Answer 26

ok one sec

Answer 27

the eigenvectors that correspond to \[\large \lambda\] will be the vectors x that satisfy the equation

\[\large A\vec{x}=\lambda \vec{x}\]


But \[\large \lambda = 0\], so \[\large A\vec{x}=(0) \vec{x}\] which means that \[\large A\vec{x}=0\]
So solve this equation to find your eigenvector

Answer 28

Great, thanks so much!

Answer 29

Sorry - have been busy between when this was first posted, and now.... Just getting back to this question...
By using the Ax=0 equation, does that give you:\[\left[\begin{matrix}-1 & 0 &1\\-1 & 0 &1\\-1 & 0 &1 \end{matrix}\right] \left[\begin{matrix}x_1 \\ x_2 \\x_3\end{matrix}\right] = 0\]
\[ie, -x_1+x_3 = 0\]\[x_1 = x_3\]

.... How does this help in finding the eigenvectors??

Answer 30

Looks like you have two free variables

let \[x_3=t\text{ and }x_2=r\]

write out the solution in terms of \[t\text{ and }r\]

Answer 31

So my answer to this would be {[t,r,t]}??

Answer 32

no...have you ever found the null space of a matrix?

Answer 33

y...es? What do I do with it, once I found the null space? Thanks for the helpl (as always!)

Answer 34

this problem turned into finding the null space \[A\vec{x}=0\]