Eigenspace

Topics: Eigenvalues and Eigenvectors - Algebra

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(definition)

Let $T:V\to V$ and let $\lambda$ be an eigenvalue of $T$. We define:

$$E_{\lambda }=\{x\in V\mid T(x)=\lambda x\}$$

$$E_{\lambda }=\ker (T-\lambda I_V)$$

…and call this set the eigenspace that corresponds to the $\lambda$ eigenvalue.

(note)

$E_{\lambda }$ is a vector subspace that contains $0$ and the eigenvectors of $T$ that correspond to $\lambda$.

The number of linearly independent eigenvectors of $T$ is the dimension of $E_{\lambda }$.

(theorem)

Let $T$ be a linear operator on $V$ a vector space with finite dimension.

If $\lambda$ is an eigenvalue of $T$ with $k$ multiplicity, then:

$$1\le \dim (E_{\lambda })\le k$$

Example

Let $T:P_2(\mathbb{R})\to P_2(\mathbb{R})$ be defined $T(f)=f^{\prime }$.

What’s $\dim (E_{\lambda })$? To find out, let’s follow the next steps:

First step §

We build the associated matrix of $T$ under the $\beta$, the canonical basis:

$$[T{]}_{\beta }=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 2\\ 0& 0& 0\end{array}\right)$$

Second step §

We find the eigenvalues of this transformation:

$$\det ([T{]}_{\beta }-\lambda I)=\det \left(\begin{array}{ccc}-\lambda & 1& 0\\ 0& -\lambda & 2\\ 0& 0& -\lambda \end{array}\right)=-{\lambda }^3$$

$$⟹-{\lambda }^3=0$$

With that, we get that the eigenvalue is $\lambda =0$ with a multiplicity of 3.

Third step §

We’ll find the eigenvectors that correspond to $\lambda$:

$$E_{\lambda }=\ker ([T{]}_{\beta }-\lambda I)=\ker (T)$$

$$E_{\lambda }=B=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 2\\ 0& 0& 0\end{array}\right)$$

We equal $Bx=0$ for some $x\in {\mathbb{R}}^3$ (a coordinate vector of some $v\in V$):

$$Bx=\left(\begin{array}{ccc}0& 1& 0\\ 0& 0& 2\\ 0& 0& 0\end{array}\right)\left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$$

We solve and we get that (solve it)

Fourth step §

The eigenspace is the space that only contains the eigenvectors that correspond to $\lambda$: Find a basis for the eigenspace, then it’s easy to find the dimension