- #1
the_amateur
- 13
- 0
Given a square matrix A, the condition that characterizes an eigenvalue, λ, is the existence of a nonzero vector x such that A x = λ x; this equation can be rewritten as follows:
[itex]Ax=\lambda x[/itex]
[itex](A-\lambda )x[/itex] = 0
[itex](A-\lambda I )x[/itex] = 0 -------------------- 1
After the above process we find the determinant of [itex]A-\lambda I [/itex] and then equate it to 0.
det([itex]A-\lambda I [/itex]) = 0 -------------------- 2
Then from the above characteristic equation we find the Eigen values.
My question is how does the equation 1 imply the above condition 2
[itex]Ax=\lambda x[/itex]
[itex](A-\lambda )x[/itex] = 0
[itex](A-\lambda I )x[/itex] = 0 -------------------- 1
After the above process we find the determinant of [itex]A-\lambda I [/itex] and then equate it to 0.
det([itex]A-\lambda I [/itex]) = 0 -------------------- 2
Then from the above characteristic equation we find the Eigen values.
My question is how does the equation 1 imply the above condition 2