12.30

Diagonalizable operators

Definition

Let \(T\in\text{End}(V),\lambda\in\mathbb{F}\). The eigenspace of \(T\) corresponding to \(\lambda\) is \(V_{\lambda}=\text{Null}(T-\lambda I)\)

Remark

• \(\lambda\) is an eigenvalue of \(T\) iff \(V_\lambda\neq 0\)
• \(v\in V_\lambda\) is an eigenvector of \(T\) iff \(v\neq 0\)​
• \(T|_{V_{\lambda}}:V_{\lambda}\to V_{\lambda},T|_{V_{\lambda}}=\lambda I\)​

Proposition

If \(\lambda_1,...,\lambda_k\) are distinct eigenvalues of \(T\), then \(V_{\lambda_1}+...+V_{\lambda_k}\) is a direct sum

Proof

Directly from this

Given \(w=w_1+...+w_k\) with \(w_i\in V_{\lambda_i}\), since \(w_1,...,w_k\) is linearly independent, then they are unique

Definition

An operator \(T\in \text{End}(V)\) is diagonalizable if there exists a basis \(\mathcal{B}\) of \(V\) such that \([T]_{\mathcal{B}}\) is diagonal.

That is if \(\lambda _1,...,\lambda _k\) are all distinct eigenvalues of \(T\), then \(T\) is diagonalizable iff \(V_{\lambda_1}+\cdots+V_{\lambda_{k}}=V\) (\(=V_{\lambda_1}\oplus\cdots\oplus V_{\lambda_{k}}\)​)

Examples

1. Assume \(e_1,e_2\) is a basis of \(V\) and let \(T:e_1\mapsto e_2\), \(e_2\mapsto e_1\), \([T]=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\)

   \(T\) is diagonalizable for \(\mathbb{F}\) any field

   In fact if \(\mathcal{B}:e_1+e_2,e_1-e_2\), then \([T]_{\mathcal{B}}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}\) two eigenvalues

2. Assume \(e_1,e_2\) is a basis of \(V\) and let \(T:e_1\mapsto e_2\), \(e_2\mapsto-e_1\), \([T]=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\)

   If \(\mathbb{F}=\R\), then \(T\) is not diagonalizable (since no eigenvalue)

   If \(\mathbb{F}=\mathbb{C}\), then \(T\) is diagonalizable.

   In fact if \(\mathcal{B}:ie_1+e_2,-ie_1+e_2\), then \([T]_{\mathcal{B}}=\begin{pmatrix}i & 0\\ 0 & -i\end{pmatrix}\) two eigenvalues

3. Assume \(e_1,e_2\) is a basis of \(V\) and let \(T:e_1\mapsto e_2\), \(e_2\mapsto e_1+e_2\), \([T]=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\)

   \(T\) is not diagonalizable for \(\mathbb{F}\) any field since it has only one eigenvalue, then corresponding eigenvector has one dimension, it can not be the basis.

   By definition of diagonalizable, it is not diagonalizable.

Remark

\(T\) is diagonalizable iff \([T]_\mathcal{B}\), for a given basis \(\mathcal{B}\), is similar to a diagonal matrix.

\(\Rightarrow)\exists\mathcal{B}^{\prime}\text{ such that }[T]_{\mathcal{B}^{\prime}}=D\). Hence the matrix \(P\) of change of basis satisfies:

\(D = [T]_{\mathcal{B}'} = P^{-1}[T]_\mathcal{B}P\)

\(\Leftarrow)\)If \(P^{-1}[T]_\mathcal{B}P = D\), let \(\mathcal{B}' = P(\mathcal{B})\). Now \([T]_{\mathcal{B}'} = D\).

Notation

A matrix \(A\in M_{n\times n}\left(\mathbb{F}\right)\) is diagonalizable if it is similar to a diagonal matrix.

Theorem

Let \(T \in \text{End}(V)\) and let \(\lambda_1, \dots, \lambda_k\) be all the distinct eigenvalues of \(T\). The following are equivalent:

1. \(T\) is diagonalizable;

2. \(V\) has a basis of eigenvectors of \(T\);

3. There exist \(1\)-dimensional \(T\)- invariant subspaces \(U_1, \dots, U_n\) such that \(V = U_1 \oplus \cdots \oplus U_n\);

4. \(V = V_{\lambda_1} \oplus \cdots \oplus V_{\lambda_k}\);

5. \(\dim V = \dim V_{\lambda_1} + \cdots + \dim V_{\lambda_k}\).

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Proof

(a) \(\implies\) (b): If a basis \(v_1, \dots, v_n\) exists such that \([T] = \begin{pmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{pmatrix}\).

Then: \(T(v_1)=\lambda_1v_1+0v_2+\cdots+0v_{n}\) \(T(v_2)=0v_1+\lambda_2v_2+\cdots+0v_{n}\) \(\vdots\) \(T(v_{n})=0v_1+\cdots+0v_{n-1}+\lambda_{n}v_{n}\)

Thus, \(v_1, \dots, v_n\) are eigenvectors.

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(b) \(\implies\) (a) If \(v_1, \dots, v_n\) is a basis of eigenvectors of corresponding eigenvalues \(\mu_1, \dots, \mu_n\), then:

\(T(v_1) = \mu_1 v_1, \dots, T(v_n) = \mu_n v_n\). Hence, \([T] = \begin{pmatrix} \mu_1 & 0 & \cdots & 0 \\ 0 & \mu_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mu_n \end{pmatrix}\)

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(b) \(\implies\) (c)

Given a basis of \(V\), \(v_1, \dots, v_n\), made of eigenvectors of \(T\), let \(U_i = \langle v_i \rangle\). Now: \(V = U_1 + \cdots + U_n\).

Given \(v \in V\), since \(v_1, \dots, v_n\) is a basis of \(V\), we can write \(v=a_1v_1+\cdots+a_{n}v_{n}\), with \(a_{i}v_{i}\in U_{i}\). Thus, \(v \in U_1 + \cdots + U_n\). So, \(V = U_1 + \cdots + U_n\).

Moreover, the sum is direct. In fact, if \(0=u_1+\cdots+u_{n}\), with \(u_{i}\in U_{i}\),

Since \(u_{i}=\mu_{i}v_{i}\), then \(0=\mu_1v_1+\cdots+\mu_{n}v_{n}\Rightarrow\mu_1=\cdots=\mu_{n}=0\Rightarrow u_1=\ldots=u_{n}=0\)

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(c) \(\implies\) (b)

Given \(U_1, \dots, U_n\) such that \(V = U_1 \oplus \cdots \oplus U_n\) (with \(U_i\) being 1-dimensional).

Let \(0 \neq v_i \in U_i\). Since \(U_i\) is \(T\)-invariant, then \(T(v_i) \in U_i\) (which is 1-dimensional). So, \(T(v_i) = \lambda_i v_i\).

Hence, all \(v_1, \dots, v_n\) are eigenvectors of \(T\).

Since \(V = U_1 \oplus \cdots \oplus U_n\), \(v_1, \dots, v_n\) generate \(V\). Moreover, since the sum is direct, the set \(v_1, \dots, v_n\) is linearly independent.

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(b) \(\implies\) (d)

Let \(v_1, \dots, v_n\) be a basis of eigenvectors of \(T\). Assume that \(v_1, \dots, v_{i_1} \in V_{\lambda_1}\), \(v_{i_1+1}, \dots, v_{i_2} \in V_{\lambda_2}\), ..., \(v_{i_{k-1}+1},\dots,v_{i_{k}=n}\in V_{\lambda_{k}}\).

Since \(v_1, \dots, v_n\) is a basis, any \(v \in V\) can be written as \(v=a_1v_1+\cdots+a_{n}v_{n}=w_1+\cdots+w_{k}\), where \(w_i \in V_{\lambda_i}\). That is, \(V = V_{\lambda_1} + \cdots + V_{\lambda_k}\).

Now assume \(0=w_1+\cdots+w_{k}=a_1v_1+\cdots+a_{n}v_{n}\).

Since \(v_1,...,v_n\) is a basis, then \(a_1=\cdots=a_{n}=0\implies w_1=\cdots=w_{k}=0\)

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(d) \(\implies\) (e)

\(\dim V = \dim \left(V_{\lambda_1} \oplus \cdots \oplus V_{\lambda_k}\right) = \dim V_{\lambda_1} + \cdots + \dim V_{\lambda_k}\).

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(e) \(\implies\) (b)

Let \(v_1^1, \dots, v_{i_1}^1\) be a basis of \(V_{\lambda_1}\), \(v_1^2, \dots, v_{i_2}^2\) be a basis of \(V_{\lambda_2}\), ... let \(v_1^k, \dots, v_{i_k}^k\) be a basis of \(V_{\lambda_k}\).

Let \(\mathcal{B} = \{v_1^1, \dots, v_{i_1}^1, v_1^2, \dots, v_{i_2}^2, \dots, v_1^k, \dots, v_{i_k}^k\}\) (all of them).

Claim: \(\mathcal{B}\) is a basis of \(V\). In fact, they are linearly independent. If the length list is less than \(\dim V\), it cannot diagonalize

If \(0 = \left(a_1^1 v_1^1 + \cdots + a_{i_1}^1 v_{i_1}^1\right) + \cdots + \left(a_1^k v_1^k + \cdots + a_{i_k}^k v_{i_k}^k\right)\), then: \(0 = a_1^1 v_1^1 + \cdots + a_{i_1}^1 v_{i_1}^1 = \cdots = a_1^k v_1^k + \cdots + a_{i_k}^k v_{i_k}^k = 0\).

This implies: \(a_1^1 = \cdots = a_{i_1}^1 = 0\), \(a_1^2 = \cdots = a_{i_2}^2 = 0\), ... \(a_1^k = \cdots = a_{i_k}^k = 0\).

This suffices

Corollary

Let \(T\in \text{End}(V)\) with \(\dim V=n\). If \(T\) has \(n\) distinct eigenvalues, then \(T\) is diagonalizable

Proof

\(n\geq\dim(V_{\lambda_1}\oplus...\oplus V_{\lambda_{n}})\geq n\Rightarrow\dim(V_{\lambda_1}\oplus\ldots\oplus V_{\lambda_{n}})=n,V=V_{\lambda_1}\oplus\ldots\oplus V_{\lambda_{n}}\)​

Example

Let \(A=\begin{pmatrix}1 & & & \star\\ & 2 & & \\ & & \ddots & \\ & & & n\end{pmatrix}\).

The eigenvalues : \(\det(A-\lambda I)=0\).

\(A-\lambda I=\begin{pmatrix}1-x & & & \\ & 2-x & & \star\\ & & \ddots & \\ & & & n-x\end{pmatrix}\).

\(\det(A-\lambda I)=(1-x)(2-x)\cdots(n-x)\).

Thus, the eigenvalues are \(1, 2, \dots, n\).

Hence exist \(P^{-1}AP=\begin{pmatrix}1 & & & \\ & 2 & & \\ & & \ddots & \\ & & & n\end{pmatrix}\) such that diagonal

Diagonalizing

Let \(\mathcal{B} = \{e_1, e_2, e_3\}\) be a basis of \(V\) and let \(T\) such that \([T]_\mathcal{B} = A\), where \(A=\begin{pmatrix}1 & 1 & 1\\ & 2 & 3\\ & & 3\end{pmatrix}\).

We know it is diagonalizable.

1. Distinct eigenvalues: \(1, 2, 3\)

2. Determine \(V_{\lambda_1}, \dots, V_{\lambda_n}\).

1. For \(\lambda_1 = 1\): \(V_{\lambda_1} = \text{Null}(T - I)\): \(A-I=\begin{pmatrix}0 & 1 & 1\\ & 1 & 2\\ & & 2\end{pmatrix}\sim[T-I]_{\mathcal{B}}\implies\text{Null}(T-I)=\langle(1,0,0)\rangle\).

2. For \(\lambda_2 = 2\): \(V_{\lambda_2} = \text{Null}(T - 2I)\): \(A-2I=\begin{pmatrix}-1 & 1 & 1\\ & 0 & 1\\ & & 1\end{pmatrix}\sim[T-2I]_{\mathcal{B}}\implies\text{Null}(T-2I)=\langle(1,1,0)\rangle\).

3. For \(\lambda_3 = 3\): \(V_{\lambda_3} = \text{Null}(T - 3I)\): \(A-3I=\begin{pmatrix}-2 & 1 & 1\\ & -1 & 1\\ & & 0\end{pmatrix}\sim[T-3I]_{\mathcal{B}}\implies\text{Null}(T-3I)=\langle(1,1,1)\rangle\).

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Conclusion

\(\{(1, 0, 0), (0, 1, 0), (1, 1, 1)\}\) is a basis of \(V\) consisting of eigenvectors of \(T\).

Or: \(\dim V_{\lambda_1} + \dim V_{\lambda_2} + \dim V_{\lambda_3} = 3 = \dim V\).

Let \(\mathcal{B}' = \{(1, 0, 0), (1, 1, 0), (1, 1, 1)\}\). The change of basis matrix \(P\) is \(P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}\).

The inverse of \(P\) is \(P^{-1} = \begin{pmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{pmatrix}\).

\(P^{-1}AP=\begin{pmatrix}1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 3\end{pmatrix}\).

Thus, \(P^{-1} A P\) is diagonal.