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## Main Question or Discussion Point

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

(a.) T is one-to-one.

(b.) T is onto.

(c.) rank(T) = dim(V).

Thanks a lot,

JL

**Theorem 5.2:**Let A be in [tex]M_n_x_n(F)[/tex]. Then a scalar [tex]\lambda[/tex] is an eigenvalue of A if and only if det(A - [tex]\lambda[/tex][tex]I_n[/tex]) = 0.**Proof:**A scalar [tex]\lambda[/tex] is an eigenvalue of A if and only if there exists a nonzero vector v in [tex]F^n[/tex] such that Av= [tex]\lambda[/tex]v, that is, (A - [tex]\lambda[/tex][tex]I_n[/tex])(v) = 0. By theorem 2.5, this is true if and only if A - [tex]\lambda[/tex][tex]I_n[/tex] is not invertible. However, this result is equivalent to the statement that det(A - [tex]\lambda[/tex][tex]I_n[/tex]) = 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 - [tex]\lambda[/tex][tex]I_n[/tex] is not invertible."Thanks a lot,

JL