Why is this always the case A=[itex]\begin{array}{cc}1 & 6

  • #1
Why is this always the case...

A=[itex]\begin{array}{cc}
1 & 6 \\
4 & 3 \\
\end{array}[/itex]

[itex]λ_{1}[/itex]=7

[itex]λ_{2}[/itex]=-3

A=[itex]\begin{array}{cc}
1-7 & 6 \\
4 & 3-7 \\
\end{array}[/itex]

A=[itex]\begin{array}{cc}
-6 & 6 \\
4 & -4 \\
\end{array}[/itex]

Why is [itex]A_{11}[/itex] and [itex]A_{12}[/itex]
always a multiple of
[itex]A_{21}[/itex] and [itex]A_{22}[/itex]?

Is this a feature of Eigenvalues or is this done on purpose to make solving the eigenvectors easier?
 

Answers and Replies

  • #2
Fredrik
Staff Emeritus
Science Advisor
Gold Member
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Why is [itex]A_{11}[/itex] and [itex]A_{12}[/itex]
always a multiple of
[itex]A_{21}[/itex] and [itex]A_{22}[/itex]?
You seem to mean [itex](A-\lambda I)_{11}[/itex], [itex](A-\lambda I)_{12}[/itex] and so on.

[itex]Ax=\lambda x[/itex] implies [itex](A-\lambda I)x=0[/itex]. So if [itex]A-\lambda I[/itex] is invertible, x=0. An eigenvector of A with eigenvalue [itex]\lambda[/itex] is by definition a non-zero x that satisfies [itex]Ax=\lambda x[/itex]. So A has an eigenvector with eigenvalue [itex]\lambda[/itex] if and only if [itex]A-\lambda I[/itex] is not invertible. For a matrix to be not invertible, its rows must be linearly dependent.
 
  • #3


Clear as day now, thank you
 

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