Homework Help: Quick eigenvector help

1. May 14, 2008

rock.freak667

1. The problem statement, all variables and given/known data

The matrix,A,given by
$$A = \left( \begin{array}{ccc} 7 & -4 & 6\\ 2 & 2 & 2 \\ -3 & 4 & -2 \ \end{array} \right)$$

has eigenvalues 1,2,4 . Find a set of corresponding eigenvectors.

Hence find the eigenvalues of B, where

$$B = \left( \begin{array}{ccc} 10 & -4 & 6\\ 2 & 5 & 2 \\ -3 & 4 & 1 \ \end{array} \right)$$

and state a corresponding set of eigenvectors.
2. Relevant equations

3. The attempt at a solution

Well I easily found the eigenvectors
$\lambda=1$ corresponds to
$$\left( \begin{array}{c} -1\\ 0 \\ 1\ \end{array} \right)$$

$\lambda=2$ corresponds to
$$\left( \begin{array}{c} -4\\ 1 \\ 4\ \end{array} \right)$$

$\lambda=4$ corresponds to
$$\left( \begin{array}{c} 2\\ 3 \\ 1\ \end{array} \right)$$

Well for the one with B, just solve det(b-$\lambda$I)=0 to get the e.values... but it says to state a set of e.vectors meaning that I am not supposed to work them out.
The only thing I can really say about A and B is that in B all the elements in the main diagonal are the elements in the main diagonal of A with 3 added to them

2. May 14, 2008

slider142

In other words, B = A + 3I. Can you find a simple way to get B's eigenvalues and eigenvectors without explicitly working through B? Use the definitions of eigenvectors and eigenvalues.

3. May 14, 2008

rock.freak667

Well I could say that

$Det(B-\lamda I)=0$ and put B=A+3I, then say that I can reduce A to a triangular matrix such that the diagonal elements(Eigenvalues of A)+(3-$\lambda$)=0 and solve from there. And get the e.values to be 7,4,5. But how would the e.vectors be altered?

Last edited: May 14, 2008
4. May 14, 2008

slider142

The determinant is just a way to extract the eigenvalues mechanically. The original definition should state that L is an eigenvalue of A iff Av = Lv for some vector v (an eigenvector of A), or equivalently, A - LI = 0. Thus, to find eigenvalues L, we are looking for solutions of the equation B - LI = 0, which is equivalent to A + (3 - L)I = 0. We already know the solutions of this equation, specifically we know that L - 3 = L' where L' is an eigenvalue of A. No messing around with matrix forms or determinants necessary. Similarly, you can use the original equation to see how the eigenvectors of A compare to the eigenvectors of B. Can you see the way from here? :)

Last edited: May 14, 2008
5. May 14, 2008

rock.freak667

I got the same answer but it all amounts to the same thing I guess.thanks

But then the e.vectors would be unchanged?

6. May 14, 2008

slider142

Indeed, you get the equation Av = (L - 3)v, where L is an eigenvalue of B and v is an eigenvector of B, implying that whenever v is an eigenvector for A, it is also an eigenvector for B, albeit with a different eigenvalue.

Last edited: May 14, 2008