Expert: Paul Klarreich - 12/15/2011

Question
QUESTION: Hi, I'm a first year Science student from Ireland and I've been studying for my Algebra final. I've been practicing a problem shhet on eigenvalues/vectors of matrices and I was hoping you might be able to help. I have no problem finding the eigenvalues but when I try to find the corresponding eigenvectors, I end up with three vectors with parameters, for example (0, 2t-r, t). That should be vertical but I couldn't figure out how to do that.
The answers given for the problem have just numbers and no t's or r's. Although I noticed that if I let t be 1 and r -1 then it works. I hope I explained myself well enough. Any help would be greatly appreciated.

ANSWER: Questioner: Adam
Country: Wicklow, Ireland
Category: Advanced Math
Private: No
Subject: Eigenvectors
Question: Hi, I'm a first year Science student from Ireland and I've been studying for my Algebra final. I've been practicing a problem shhet on eigenvalues/vectors of matrices and I was hoping you might be able to help. I have no problem finding the eigenvalues but when I try to find the corresponding eigenvectors, I end up with three vectors with parameters, for example (0, 2t-r, t). That should be vertical but I couldn't figure out how to do that.
The answers given for the problem have just numbers and no t's or r's. Although I noticed that if I let t be 1 and r -1 then it works. I hope I explained myself well enough. Any help would be greatly appreciated.
...............................................
I don't see what's wrong -- If A is a matrix, and v1 is an eigenvector for it, it means  

A * v1 = k1 v1.  

Now k1 v1 is a vector in the same direction as v1.  

Then if v1 is an eigenvector, so is 2 v1, or 7 v1, or -4 v1, or r v1 for any r.

Why?  A * r v1 = r A * v1 = r k1 v1 = (rk1) v1.

Now if v2 is another, then:

A * (r v1 + s v2) = A * r v1 + A * s v2 =
r A * v1 +  s A * v2 =
r k1 v1 +  s k1 v2

Send along your work and we'll see what we can do.


---------- FOLLOW-UP ----------

QUESTION: Find the eigenvalues and corresponding eigenvectors for the matrix,
A= 3  1   1
  2  4   4
  -1 -1  -1

Hence find an invertible matrix B which diagonalizes A.

I found eigenvalues of 0,2 and 4. When I found the eigenvectors I got them in terms of parameters. For 0 I got (0, t, -t), for 2 I got (t, -t, 0) and for 4 I got (-2t, -3t, t). The answers given are the same but with 1 in place of all the t's. Can I just substitute in 1 at the end of the problem?

Answer
Questioner:    Adam
Private: Yes
Changed it to public     
Subject:  Eigenvectors.
Question: QUESTION: Hi


QUESTION: Find the eigenvalues and corresponding eigenvectors for the matrix,
A= 3  1   1
 2  4   4
 -1 -1  -1

Hence find an invertible matrix B which diagonalizes A.

I found eigenvalues of 0,2 and 4.

>>>>> Yes, I got them, too.


When I found the eigenvectors I got them in terms of parameters. For 0 I got (0, t, -t), for 2 I got (t, -t, 0) and for 4 I got (-2t, -3t, t). The answers given are the same but with 1 in place of all the t's. Can I just substitute in 1 at the end of the problem?

>>>>>>>>> See the final comment,

[ 3  1   1  ]
[ 2  4   4  ]
[ -1 -1  -1 ]

Det of:
[ 3-k   1    1  ]
[ 2    4-k   4  ]
[ -1   -1   -1-k ]

= (after much simplifying):

8k + k^3 - 6k^2

Set it = 0 and solve:

k(8k + k^2 - 6k) = 0

k(k - 2)(k - 4) = 0

k = 0, k = 2, k = 4

========================
With k = 0:

[ 3-0  1   1  ] x
[ 2  4-0   4  ] y
[ -1 -1  -1-0 ] z
=  <0,0,0>

[ 3  1   1  ] x
[ 2  4   4  ] y
[ -1 -1  -1 ] z
=  <0,0,0>

3x + y + z = 0    A
2x + 4y + 4z = 0  B
x + y + z = 0     C    <<<<< flipped all the signs.

A - C:  2x = 0;  x = 0

y + z = 0    A
y + z = 0    B    <<<< div out the 4
y + z = 0    C

y = -z, so we can take:

y = t, then  z = -t

EXACTLY AS YOU GOT.
.......................

With k = 2:

[ 3-2  1    1  ] x
[ 2   4-2   4  ] y
[ -1  -1   -1-2 ] z
=  <0,0,0>

[ 1  1    1  ] x
[ 2  2    4  ] y
[ -1  -1  -3 ] z
=  <0,0,0>

x + y + z = 0      A
2x + 2y + 4z = 0   B
x  + y  + 3z = 0   C   <<<<<< again.

C-A: 2z = 0  --->  z = 0

x + y = 0      A
x + y = 0      B  <<< div out the 2
x + y  = 0    C

Again, take  x = t,  then y = -t,

AS YOU DID.
.........................................

k = 4

[ 3-4  1    1   ] x
[ 2   4-4   4   ] y
[ -1  -1   -1-4 ] x
=  <0,0,0>

[ -1  1    1 ] x
[ 2   0   4  ] y
[ -1  -1  -5 ] x
=  <0,0,0>

-1x + y + z = 0  A
x      +2z = 0  B
x + y + 5z = 0  C

A+C : 2y + 6z = 0
      y + 3z = 0
A+B : y + 3z = 0

Take  z = t,  y = -3t.

B:  x + 2t = 0,  x = -2t

OKAY, AGAIN.
.....................................
Now, if you like, you can take t = 1, and you can take  t = 23, t = 419, or whatever.  "Whatever" means the value that gives you the answer in the back of the book.

Note: I don't usually work these out in detail.  I did this as a reminder of how this works (from 39 years ago).