Eigenvectors and Eigenvalues
We have seen a few notions we can attach to a linear tranformation built from the kernel and range. Here we introduce a different but very important one.
Definition. Let $T: V \to V$ be a linear transformation. A vector $v \in V$ is called an eigenvector for $T$ if $v \neq 0$ and \(T(v) = \lambda v\) for some scalar $\lambda \in k$. The scalar $\lambda$ is called the eigenvalue of the eigenvector $v$.
Given a scalar $\lambda \in k$, the $\lambda$-eigenspace is \(E_{\lambda}(T) := \lbrace v \in V \mid T(v) = \lambda v \rbrace\)
Lemma. Eigenspaces are subspaces.
Proof. (Expand to view)
We check $E_\lambda$ is closed under linear combinations of two vectors. If $T(v_1) = \lambda v_1$ and $T(v_2) = \lambda v_2$, then \(T(c_1v_1+c_2v_2) = c_1T(v_1) + c_2T(v_2) c_1 \lambda v_1 + c_2 \lambda v_2 = \lambda(c_1v_1 + c_2v_2)\) So $c_1v_1 + c_2v_2 \in E_\lambda$. ■
Eigenvectors are important because they allow use to understand
the structure of the linear transformation.
Example. Let \(D = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}\) be a diagonal matrix. Then, each standard basis vector $\mathbf{e}_i$ is eigenvector for $D$ with eigenvalue $\lambda_i$.
Proposition. Suppose that $T : V \to V$ is a linear transformation with $\dim V < \infty$. If $V$ has a basis of eigenvectors of $T$, $T$ has a matrix representation which is a diagonal matrix.
Proof. (Expand to view)
Let $v_1,\ldots,v_d$ be a basis of $V$ with \(T(v_i) = \lambda_i v_i\) for each $i$. Then the matrix representation of $T$ using the basis $v_1,\ldots,v_d$ in both the domain and codomain is \(D = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & \cdots & 0 & \lambda_n \end{pmatrix}\) ■
If we are allowed to choose two different bases for $V$, one to use in the domain and one for the codomain, then we can find diagonal matrix representation for any $T : V \to V$. However, this will not guarantee that we have found any eigenvectors. We need the matrix representation of $T$ to be diagonal for the same basis on each side.
After choosing a basis, we get a matrix representation $A$ for $T$. Changing to a different basis, the same in the source and target, replaces $A$ with \(S^{-1}AS\) where $S$ is the change of basis matrix.
Definition. A $n\times n$ matrix $A$ is diagonalizable if there exists an invertible $n \times n$ matrix $S$ with $S^{-1} A S$ diagonal.
Proposition. Let $T: V \to V$ be a linear transformation with $\dim V < \infty$. Then $V$ has a basis of eigenvectors if and only a(ny) matrix representation in the same basis for both domain and codomain for $T$ is diagonalizable.
Proof. (Expand to view)
As before, we can take the matrix representation of $T$ using our basis of eigenvectors in $V$ for both the domain and codomain to get a diagonal matrix representation.
Assume $A$ is some matrix representation of $T$ with the basis $v_1,\ldots,v_d$ used for the domain and codomain and we know that $S^{-1}AS$ is diagonal. Then \(w_i := \sum_{j=1}^d S_{ij}v_j\) is a new basis for $V$. The change of basis matrix from $v_1, \ldots, v_d$ to $w_1,\ldots,w_d$ is the matrix $S$ and $T$ in the new basis is $S^{-1}AS$ which is diagonal. Thus $w_1,\ldots,w_d$ is a basis for eigenvectors for $V$. ■
If we can find eigenvectors with different eigenvalues, we get linear independence automatically.
Lemma. Let $v_1,\ldots,v_s$ be eigenvectors for $T: V \to V$ with $\lambda_i \neq \lambda_j$ if $i \neq j$. Then, $v_1,\ldots,v_s$ is linearly independent.
Proof. (Expand to view)
If $v_1,\ldots,v_s$ are linearly dependent, we can write at least one as a linear combination of the other vectors. Up to relabeling we have, \(v_{t+1} = \sum_{j=1}^t c_i v_i\) with $v_1,\ldots,v_t$ linearly independent. Now applying $T$ we get \(\lambda_{t+1} v_{t+1} = \sum_{j=1}^t \lambda_i c_i v_i\)
If $\lambda_{t+1} = 0$, then we have relation so at least one $\lambda_i = 0$ since $v_1,\ldots,v_t$ is linearly independent and not all $c_i = 0$. This is a contradiction.
If $\lambda_{t+1} \neq 0$, then we can divide by $\lambda_{t+1}$ to get \(v_{t+1} = \sum_{j=1}^t \frac{\lambda_i}{\lambda_{t+1}} c_i v_i\) which gives another way to write $v_{t+1}$ as linear combination of $v_1,\ldots,v_t$. So \(c_i = \frac{\lambda_i}{\lambda_{t+1}} c_i\) for all $i$. If $\lambda_i \neq \lambda_{t+1}$, then $c_i = 0$ which contradicts $v_{t+1} \neq 0$. ■
Our next goal will be to understand the eigenspaces and the set of eigenvalues of a linear transformation. We will need a helpful tool: determinants.