# Linear Maps and Eigenvalues/Eigenvectors

• MHB
• joypav
In summary, the conversation discussed two problems from linear algebra, both involving the concept of eigenvalues. The proof for the first problem showed that if a linear transformation has a non-zero kernel at a certain power, then the corresponding eigenvalue is valid. The second problem can be solved using a similar approach.
joypav
A problem from matrix analysis. This proof seems to simple... is it correct?

Problem 1:
Suppose $T: V \rightarrow V$ is linear and $\mu \in F$. Show that if there exists a positive integer p such that $ker((T - \mu I)^p)$ $\ne \left\{ 0 \right\}$, then $\mu$ is an eigenvalue of T.

Proof:
$ker((T - \mu I)^p) \ne \left\{ 0 \right\} \implies \exists y \in ker((T - \mu I)^p), y \ne 0$
$\implies (T - \mu I)^p y = 0$

Consider,
$0 \ne y, (T - \mu I)y, (T - \mu I)^2 y , ... , (T - \mu I)^p y = 0$

Then, $\exists j, 0 < j \leq p$, the smallest positive integer so that
$(T - \mu I)^{j-1} y \ne 0$ and $(T - \mu I)^j y = 0$

Let $z=(T - \mu I)^{j-1} y$.
Then $z \ne 0$ and $(T - \mu I) z = (T - \mu I)^j y = 0$
$\implies (T - \mu I) z = 0$
$\implies \mu$ is an eigenvalue of V.Additionally,
Problem 2:
Suppose $T: V \rightarrow V$ is linear, $\mu \in F$, $x \in V, x \ne 0$, and $(T - \mu I)^p x = 0$ for some $p \in \Bbb{N}$. Show that one of the vectors
$x, (T - \mu I) x, (T - \mu I)^2 x, ... ,(T - \mu I)^{p-1} x$
is an eigenvector of T.

Can this problem be approached similarly?

Hi joypav,

Everything for Problem 1 looks good, nicely done. You're correct about Problem 2 as well; it can be done in a similar fashion to Problem 1.

## 1. What is a linear map?

A linear map is a mathematical function that maps between two vector spaces in a way that preserves the structure of both spaces. This means that the map must satisfy the properties of homogeneity and additivity.

## 2. What are eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are properties of a linear map that describe how the map transforms certain vectors. An eigenvector is a vector that, when multiplied by the linear map, is only scaled by a constant factor. This constant factor is known as the eigenvalue.

## 3. How are eigenvalues and eigenvectors calculated?

To calculate eigenvalues and eigenvectors, we first need to find the characteristic polynomial of the linear map. This polynomial is then solved to find the eigenvalues, which are the roots of the polynomial. The corresponding eigenvectors can then be found by solving the system of equations resulting from substituting each eigenvalue into the characteristic polynomial.

## 4. What is the significance of eigenvalues and eigenvectors?

Eigenvalues and eigenvectors are important because they provide a way to simplify the analysis of linear maps. They allow us to focus on the transformation of specific vectors rather than the entire vector space. They also have many applications in various fields including physics, engineering, and computer science.

## 5. Can a linear map have complex eigenvalues and eigenvectors?

Yes, a linear map can have complex eigenvalues and eigenvectors. This occurs when the vector space is over the field of complex numbers rather than real numbers. In this case, the eigenvalues and eigenvectors may have both a real and imaginary component.

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