2
$\begingroup$

So we want to find the basis for the eigenspace of each eigenvalue λ for some matrix A .

Through making this question, I have noticed that the basis for the eigenspace of a certain eigenvalue has some sort of connection to the eigenvector of said eigenvalue. Now I'm not sure if they actually equal each other, because I have some trouble when it comes to eigenvalues with a geometric multiplicity of two or more.

Take the following example:

\begin{equation*} A = \begin{pmatrix} 0 & -1 & 0 \\ 4 & 4 & 0 \\ \ 2 & 1 & 2 \end{pmatrix} \end{equation*}

This matrix has a characteristic polynomial $−λ3+6λ2−12λ+8$. The root of this is $λ=2$, which has an algebraic multiplicity of 3. Then, I am calculating : $ E_2 : (A - 2I)(x_1,x_2,x_3)=(0, 0, 0) $ After solving it, I found 2 eigenvectors :

$ x_1 * (1, -2, 0) + x_3 * (0, 0, 1) $

I found this two vectors by using 1 pivot and 2 free variables.

The geometric multiplicity is equal to the number of free variables for this eigenvalue. Since the geometric multiplicity is different of the algebric multiplicity, this matrix is not diagonalizable.

First question : Is this correct ?

Second question : My teacher told me that a matrix is diagonalizable iff : $\sum_{λ}^{} dim E_λ(A) = n$ with n the dimension of a (n,n) square matrix. Is it the same rule as the geometric multiplicity ?

Third question : If a matrix has 2 eigenvalues, will it have 2 different eigenspaces ? I have trouble figuring out what is it ?

$\endgroup$
1
  • $\begingroup$ there is a "minimal polynomial" which is the polynomial $f(x)$ of lowest degree for which $f(A) = 0$ gives the zero matrix. For each eigenvalue, the degree of that term is the size of the largest Jordan block. So: is $A-2I = 0 \; ? \; $ IS $(A-2I)^2 = 0 \; ? \; $ $\endgroup$
    – Will Jagy
    Oct 24, 2020 at 18:40

1 Answer 1

1
$\begingroup$

There are several issues with your question.

Your first sentence mentions “the basis for the eigenspace”, but each eigenspace has infinitely many bases.

Then you talk about “the eigenvector of said eigenvalue”; again, every eigenvalue has infinitely many corresponding eigenvectors.

You say that the characteristic polynomial of $A$ is $-\lambda ^3+6 \lambda ^2-12 \lambda +8$ and that its only root is $2$; that is correct. And those $2$ eigenvectors are indeed eigenvectors. They are linearly independent and they form a basis of the eigenspace corresponding to the eigenvalue $2$.

Now, concerning your questions:

  1. Yes, that is correct.
  2. Yes, since the geometric multiplicity of $\lambda$ is $\dim E_\lambda$.
  3. If a matrix has $k$ eigenvalues, then it has $k$ distinct eigenspaces.
$\endgroup$
1
  • 1
    $\begingroup$ To be honest, I copied the example from a previous post of SE and did not read the first lines that were written by the other asker. I will edit them for clarity. Thanks a lot for your answer, you are amazing ! $\endgroup$ Oct 25, 2020 at 16:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .