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Quick eigenvector help

  • #1
rock.freak667
Homework Helper
6,230
31

Homework Statement



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

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

Hence find the eigenvalues of B, where

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

and state a corresponding set of eigenvectors.

Homework Equations





The Attempt at a Solution




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

[itex]
\lambda=2[/itex] corresponds to
[tex]
\left(
\begin{array}{c}
-4\\
1 \\
4\
\end{array}
\right)
[/tex]

[itex]
\lambda=4[/itex] corresponds to
[tex]
\left(
\begin{array}{c}
2\\
3 \\
1\
\end{array}
\right)
[/tex]


Well for the one with B, just solve det(b-[itex]\lambda[/itex]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
 

Answers and Replies

  • #2
1,013
65
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
rock.freak667
Homework Helper
6,230
31
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.
Well I could say that

[itex]Det(B-\lamda I)=0[/itex] 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-[itex]\lambda[/itex])=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:
  • #4
1,013
65
Well I could say that

[itex]Det(B-\lamda I)=0[/itex] 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+[itex]\lambda[/itex])=0 and solve from there. And get the e.values to be 7,4,5. But how would the e.vectors be altered?
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:
  • #5
rock.freak667
Homework Helper
6,230
31
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
1,013
65
I got the same answer but it all amounts to the same thing I guess.thanks

But then the e.vectors would be unchanged?
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:

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