Diagonalization 2: Explaining Theorem 2.5

In summary, Theorem 2.5 in Diagonalization 2 is a mathematical theorem that states that any square matrix with distinct eigenvalues can be diagonalized by finding its eigenvectors and using them to construct a diagonal matrix. It is a fundamental theorem in diagonalization and linear algebra, as it allows for the simplification of complex matrices and easier calculations. However, it can only be applied to square matrices with distinct eigenvalues and is used in real-world applications such as physics, engineering, and economics.
  • #1
jeff1evesque
312
0
I am reading yet another theorem and was wondering If I could get more clarification on it.

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
 
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  • #2
If it were invertible/nonsingular, then the only solution would be the zero vector. But by definition v is a nonzero vector.
 
  • #3
jeff1evesque said:
I am reading yet another theorem and was wondering If I could get more clarification on it.

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
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 [itex]Av= \lambda v[/itex] is equivalent to [itex]Av- \lambda v= 0[/itex] which is equivalent to [itex](A- \lambda I_n)v= 0[/itex]. If [itex]A- \lambda I_n[/itex] were invertible, we could solve the the equation by multiplying both sides by that inverse: [itex]v= (A- \lambda I_n)^{-1})0[/itex] 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 [itex]Av= \lambda v[/itex].
 

1. What is Theorem 2.5 in Diagonalization 2?

Theorem 2.5 in Diagonalization 2 is a mathematical theorem that states that any square matrix with distinct eigenvalues can be diagonalized by finding its eigenvectors and using them to construct a diagonal matrix.

2. How does Theorem 2.5 relate to diagonalization?

Theorem 2.5 is a fundamental theorem in diagonalization because it provides a method for diagonalizing a square matrix with distinct eigenvalues. This allows us to simplify complex matrices and make calculations easier.

3. What is the importance of Theorem 2.5 in linear algebra?

Theorem 2.5 is important in linear algebra because it allows us to transform a difficult matrix into a simpler diagonal matrix, making it easier to perform calculations and understand the properties of the original matrix.

4. Can Theorem 2.5 be applied to any type of matrix?

Theorem 2.5 can only be applied to square matrices with distinct eigenvalues. It cannot be applied to matrices with repeated eigenvalues or non-square matrices.

5. How is Theorem 2.5 used in real-world applications?

Theorem 2.5 is used in various fields such as physics, engineering, and economics to solve problems involving matrices. For example, it can be used to find the principal components of a data set or to analyze the stability of a system in control theory.

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