Show the Linear Map is Diagonalizable

In summary: This means that $v$ is an eigenvector for both $\lambda_1$ and $\lambda_2$, which is a contradiction since we assumed they were distinct eigenvalues. Therefore, the intersection of the null spaces must be trivial.Finally, using the fact that the null spaces are invariant and the intersection is trivial, we can conclude that $V$ can be decomposed into a direct sum of the null spaces of $T-\lambda I$ for each distinct eigenvalue. In other words, $V = ker(T-\lambda_1 I) \bigoplus ker(T-\lambda_2 I) \bigoplus ... \bigoplus ker(T-\lambda_n I)$, which is the definition
  • #1
joypav
151
0
Problem:
Suppose $V$ is a complex vector space of dimension $n$, where $n > 0$, and suppose that $T$ is a linear map from $V$ to $V$. Suppose that if $\lambda$ is any eigenvalue of $T$, then $ker(T−\lambda I)^2 = ker(T−\lambda I)$. Prove that $T$ is diagonalizable.

Here's what I think I need to use:
if $ker(T−\lambda I)^2 = ker(T−\lambda I)$ then $ker(T−\lambda I) \cap Range(T-\lambda I) = \left\{0\right\}$

and

$T$ is diagonalizable if $V = ker(T-\lambda_1 I) \bigoplus ker(T-\lambda_2 I) \bigoplus ... \bigoplus ker(T-\lambda_n I)$
$\lambda_1, \lambda_2, ... ,\lambda_n$ distinct eigenvalues of $T$.If someone could help me out I would appreciate it!
 
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  • #2


Hello there,

You are correct in identifying the key concepts that need to be used to prove that $T$ is diagonalizable. I will walk you through the proof step by step.

First, we know that if $\lambda$ is an eigenvalue of $T$, then $ker(T-\lambda I)^2 = ker(T-\lambda I)$. This means that the null space of $T-\lambda I$ does not change when we apply the map $T-\lambda I$ a second time. In other words, the null space of $T-\lambda I$ is invariant under $T-\lambda I$.

Now, let's consider the Jordan canonical form of $T$. This form tells us that $T$ can be written as a block diagonal matrix with Jordan blocks corresponding to each distinct eigenvalue of $T$. Each Jordan block has the form:
$\begin{pmatrix}
\lambda & 1 & 0 & \cdots & 0\\
0 & \lambda & 1 & \cdots & 0\\
\vdots & \vdots & \ddots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda & 1\\
0 & 0 & \cdots & 0 & \lambda
\end{pmatrix}$

Notice that the null space of $T-\lambda I$ is the same as the null space of the Jordan block, which is just the span of the last column. This means that the null space of $T-\lambda I$ is invariant under $T-\lambda I$, just like we saw earlier. This is true for each distinct eigenvalue, so we can conclude that $ker(T-\lambda I)$ is invariant under $T$ for each eigenvalue $\lambda$.

Next, we need to show that the intersection of the null spaces of $T-\lambda I$ is trivial. This is where we use the fact that $ker(T-\lambda I)^2 = ker(T-\lambda I)$. Suppose there are two distinct eigenvalues $\lambda_1$ and $\lambda_2$ such that $ker(T-\lambda_1 I) \cap ker(T-\lambda_2 I) \neq \{0\}$. Then, there exists a non-zero vector $v$ such that $(T-\lambda_1 I)v = 0$ and $(T-\lambda_2 I)v
 

FAQ: Show the Linear Map is Diagonalizable

1. What does it mean for a linear map to be diagonalizable?

A linear map is considered diagonalizable if it can be represented by a diagonal matrix. This means that the map can be simplified and its operations can be easily performed by just multiplying the input vector by a diagonal matrix.

2. How can I determine if a linear map is diagonalizable?

To determine if a linear map is diagonalizable, you can use the diagonalization theorem. This theorem states that a linear map is diagonalizable if and only if it has n distinct eigenvectors, where n is the dimension of the vector space.

3. What are the advantages of having a diagonalizable linear map?

A diagonalizable linear map has several advantages. It simplifies the operations of the map, making it easier to understand and analyze. It also allows for more efficient computations, as the diagonal matrix can be easily multiplied with input vectors. Additionally, diagonalizable linear maps have a clearer geometric interpretation, making it easier to visualize their effects on vector spaces.

4. Can any linear map be diagonalizable?

No, not all linear maps are diagonalizable. For a linear map to be diagonalizable, it must have n distinct eigenvectors, where n is the dimension of the vector space. If a map does not have enough distinct eigenvectors, it cannot be represented by a diagonal matrix and therefore cannot be diagonalizable.

5. How can I show that a linear map is diagonalizable?

To show that a linear map is diagonalizable, you can use the diagonalization theorem and find n distinct eigenvectors, where n is the dimension of the vector space. You can also show that the map's matrix representation is diagonal by finding a diagonal matrix that is similar to the original matrix representation. Additionally, you can use the characteristic polynomial to find the eigenvalues of the map, and then use those eigenvalues to find the corresponding eigenvectors.

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