Regarding Eigen values Of a Matrix

[the_amateur]

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:

$Ax=\lambda x$

$(A-\lambda )x$ = 0

$(A-\lambda I )x$ = 0 -------------------- 1




After the above process we find the determinant of $A-\lambda I$ and then equate it to 0.

det($A-\lambda I$) = 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

 

[lineintegral1]

Well, if the nullspace of the transformation,

$$A-\lambda I$$

is not the zero subspace, as you stated, then certainly the determinant of the transformation will be zero. Recall that this is a property of singular linear transformations. This is true because a singular transformation is non-invertible, which implies the determinant is zero.

 

[HallsofIvy]

If a matrix, A, is NOT invertible, then the equation Ax= b either has no solution or an infinite number of solutions (the set of solutions will be a subspace). Obviously, the equation Ax= 0 has the "trivial solution", x= 0, so if A is not invertible, the equation has an infinite number of solutions.

Conversely, if A IS invertible, then Ax= b has the unique solution, $x= A^{-1}b$. Obviously if A is invertible, then Ax= 0 has the unique solution x= 0.

So $(A- \lambda I)x= 0$ has only the trivial solution x= 0 if $A- \lambda I$ has non-zero determinant (and so has an inverse) and has an infinite number of soltions if $A- \lambda I$ has determinant 0 (and so is not invertible).

 

[epenguin]

If a matrix, A, is NOT invertible, then the equation Ax= b either has no solution or an infinite number of solutions (the set of solutions will be a subspace). Obviously, the equation Ax= 0 has the "trivial solution", x= 0, so if A is not invertible, the equation has an infinite number of solutions.

Conversely, if A IS invertible, then Ax= b has the unique solution, $x= A^{-1}b$. Obviously if A is invertible, then Ax= 0 has the unique solution x= 0.

So $(A- \lambda I)x= 0$ has only the trivial solution x= 0 if A has non-zero determinant (and so has an inverse) and has an infinite number of soltions if A has determinant 0 (and so is not invertible).

Shouldn't that last bit read
has only the trivial solution x= 0 if $(A- \lambda I)$ has non-zero determinant (and so has an inverse) and has an infinite number of soltions if $(A- \lambda I)x= 0$ has determinant 0 (and so is not invertible). ?

 

[HallsofIvy]

Yes, thanks. I confused myself by by using "A" in general and then using $A- \lambda I$ for the specific response about eigenvalues! I will edit my post.

 

[epenguin]

Yes, thanks. I confused myself by by using "A" in general and then using $A- \lambda I$ for the specific response about eigenvalues! I will edit my post.

If it had been anyone else it might not have mattered, but being you the student would go crazy for not seeing how to get your answer which he would not dream could be wrong.