Diagonalization 2: Explaining Theorem 2.5

[jeff1evesque]

I am reading yet another theorem and was wondering If I could get more clarification on it.

Theorem 5.2: Let A be in $$M_n_x_n(F)$$. Then a scalar $$\lambda$$ is an eigenvalue of A if and only if det(A - $$\lambda$$$$I_n$$) = 0.

Proof: A scalar $$\lambda$$ is an eigenvalue of A if and only if there exists a nonzero vector v in $$F^n$$ such that Av= $$\lambda$$v, that is, (A - $$\lambda$$$$I_n$$)(v) = 0. By theorem 2.5, this is true if and only if A - $$\lambda$$$$I_n$$ is not invertible. However, this result is equivalent to the statement that det(A - $$\lambda$$$$I_n$$) = 0.

Theorem 2.5: Let V and W be vector spaces of equal (finite) dimension, and let T: V --> W be linear. Then the following are equivalent:
(a.) T is one-to-one.
(b.) T is onto.
(c.) rank(T) = dim(V).

Question: Could someone explain how the following sentence is true: "By theorem 2.5, this is true if and only if A - $$\lambda$$$$I_n$$ is not invertible."

Thanks a lot,


JL

 

[Random Variable]

If it were invertible/nonsingular, then the only solution would be the zero vector. But by definition v is a nonzero vector.

 

[HallsofIvy]

I am reading yet another theorem and was wondering If I could get more clarification on it.

Theorem 5.2: Let A be in $$M_n_x_n(F)$$. Then a scalar $$\lambda$$ is an eigenvalue of A if and only if det(A - $$\lambda$$$$I_n$$) = 0.

Proof: A scalar $$\lambda$$ is an eigenvalue of A if and only if there exists a nonzero vector v in $$F^n$$ such that Av= $$\lambda$$v, that is, (A - $$\lambda$$$$I_n$$)(v) = 0. By theorem 2.5, this is true if and only if A - $$\lambda$$$$I_n$$ is not invertible. However, this result is equivalent to the statement that det(A - $$\lambda$$$$I_n$$) = 0.

Theorem 2.5: Let V and W be vector spaces of equal (finite) dimension, and let T: V --> W be linear. Then the following are equivalent:
(a.) T is one-to-one.
(b.) T is onto.
(c.) rank(T) = dim(V).

Question: Could someone explain how the following sentence is true: "By theorem 2.5, this is true if and only if A - $$\lambda$$$$I_n$$ is not invertible."

Thanks a lot,


JL

I'm not sure why "theorem 2.5" is appended, this is a question about the sentence in theorem 5.2 only isn't it?

In any case, the equation $Av= \lambda v$ is equivalent to $Av- \lambda v= 0$ which is equivalent to $(A- \lambda I_n)v= 0$. If $A- \lambda I_n$ were invertible, we could solve the the equation by multiplying both sides by that inverse: $v= (A- \lambda I_n)^{-1})0$ and, since any linear transformation of the 0 vector is the 0 vector, we must have v= 0, contradicting the fact that there was a non-zero vector satisfying $Av= \lambda v$.