Eigenvalues/Eigenvectors

May 2010
5
0
If say, a matrix A is s.t. A(0,3,5) = (0,6,10), then obviously, the eigenvalue is 2.
But how do you explain why (5,3,0) is also an eigenvector with eigenvalue of 2?
 
Oct 2009
4,261
1,836
If say, a matrix A is s.t. A(0,3,5) = (0,6,10), then obviously, the eigenvalue is 2.
But how do you explain why (5,3,0) is also an eigenvector with eigenvalue of 2?

What's there to explain? (5,3,0) is an eigenvector of A coorresponding to the e.v. 2 if A(5,3,0)=2(5,3,0)=(10,6,0)...one has to check whether this is true.

Tonio
 
May 2010
5
0
What's there to explain? (5,3,0) is an eigenvector of A coorresponding to the e.v. 2 if A(5,3,0)=2(5,3,0)=(10,6,0)...one has to check whether this is true.

Tonio
But how does (0,3,5) having an e.v. = 2 show that (5,3,0) also has e.v. = 2?

Is it due to some sort of row/column swap property? or maybe it has something to do with the magnitude of the vectors?
 
Oct 2009
4,261
1,836
But how does (0,3,5) having an e.v. = 2 show that (5,3,0) also has e.v. = 2?


It doesn't...who told you it does??

Tonio

Is it due to some sort of row/column swap property? or maybe it has something to do with the magnitude of the vectors?
.
 
May 2010
5
0
uggh, sorry, I changed the question slightly so I could try work out the question myself for a different set of vectors. The original question was:

Suppose that for some 3x3 matrix A, we have A(1,0,-1) = (2,0,-2), and A(1,2,0) = (2,4,0).

a) Find one eigenvalue of A.
b) Explain why (0,2,1) is an eigenvector for this eigenvalue and hence find A(0,2,1).
 
Oct 2009
4,261
1,836
uggh, sorry, I changed the question slightly so I could try work out the question myself for a different set of vectors. The original question was:

Suppose that for some 3x3 matrix A, we have A(1,0,-1) = (2,0,-2), and A(1,2,0) = (2,4,0).

a) Find one eigenvalue of A.
b) Explain why (0,2,1) is an eigenvector for this eigenvalue and hence find A(0,2,1).

You changed the question...""slightly""??! I suppose "slightly" must have another meaning in the southern hemisphere...(Giggle)

Anyway, since \(\displaystyle (0,2,1)=(1,2,0)-(1,0,-1)\) and \(\displaystyle A\) is linear then a linear combination of eigenvectors corresponding to one single eigenvalue is

again an eigenvector corresponding to the same eigenvalue, and we check:

\(\displaystyle A(0,2,1)=A(1,2,0)-A(1,0,-1)=(2,4,0)-(2,0,-2)=(0,4,2)=2\cdot (0,2,1)\)

Tonio
 
May 2010
5
0
Haha ^^. I didn't know that writing part of the question altered the entire meaning of the question.

I completely understood that. Thank you so much!
 
May 2009
1,176
412
No - he's right. It is a change of basis (a simple column swap).

If \(\displaystyle A =\left( \begin{array}{ccc}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \end{array} \right)\) then define \(\displaystyle A^{\prime} = \left( \begin{array}{ccc}

c_1 & b_1 & a_1 \\
c_2 & b_2 & a_2 \\
c_3 & b_3 & a_3 \end{array} \right)
\).

Then \(\displaystyle A^{\prime}\left( \begin{array}{c}5\\ 3\\ 0\end{array}\right) = \left(\begin{array}{c}10\\ 6\\ 0\end{array}\right)\) as required...

Unless I am missing something?
 
Oct 2009
4,261
1,836
No - he's right. It is a change of basis (a simple column swap).

If \(\displaystyle A =\left( \begin{array}{ccc}
a_1 & b_1 & c_1 \\
a_2 & b_2 & c_2 \\
a_3 & b_3 & c_3 \end{array} \right)\) then define \(\displaystyle A^{\prime} = \left( \begin{array}{ccc}

c_1 & b_1 & a_1 \\
c_2 & b_2 & a_2 \\
c_3 & b_3 & a_3 \end{array} \right)
\).

Then \(\displaystyle A^{\prime}\left( \begin{array}{c}5\\ 3\\ 0\end{array}\right) = \left(\begin{array}{c}10\\ 6\\ 0\end{array}\right)\) as required...

Unless I am missing something?

Yes, that mixing the matrix'x columns is mixing the basis of the vector space which was chosen to represent the operator, and then you also must change the representation of each vector according to the new basis...or else you must check whether any given equality is true wrt the same basis.

For example, \(\displaystyle \begin{pmatrix}1&\!\!\!-5&3\\1&7&\!\!\!-3\\3&0&2\end{pmatrix}\begin{pmatrix}0\\3\\5\end{pmatrix}=\begin{pmatrix}0\\6\\10\end{pmatrix}=2\cdot \begin{pmatrix}0\\3\\5\end{pmatrix}\) , but \(\displaystyle \begin{pmatrix}1&\!\!\!-5&3\\1&7&\!\!\!-3\\5&3&0\end{pmatrix}\begin{pmatrix}5\\3\\0\end{pmatrix}\neq 2\cdot \begin{pmatrix}5\\3\\0\end{pmatrix}\) .

Tonio