Regarding Eigen values Of a Matrix

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

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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
Homework Helper
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).

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epenguin
Homework Helper
Gold Member
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).
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.
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.