3.3 Revision and Notation of Diagonalization

Let \(k\) field, \(V\) is \(k-\)vector space (k-vs). Suppose \(\dim _kV=m>0\)

Let \(f:V\to V\) \(k\)-linear. Given a basis \(\mathcal{B}\) of \(V\), we have \([f]_\mathcal{B}\in k^{m\times m}\) (\(M_m(k)\))

Diagnolization consists of finding a basis \(\tilde{\mathcal{B}}\) \(\exists\) a matrix of \(V\) such that \([f]_{\tilde{\mathcal{B}}}\) is diagonal

If \(\tilde{\mathcal{B}}\) exists, then \(\tilde{\mathcal{B}}\) is a basis of eigenvectors: \(\tilde{\mathcal{B}}=\{v_1,...,v_m\}\) where \(d_i\in k,\forall i,1\leq i\leq m\), then \(f(v_i)=d_iv_i\)

Remark: If \(v_i\) is an eigenvector with eigenvalue \(d_i\) and \(p(x)\in k[x]\), \(p(x)=\sum_{j=0}^{r}a_{j}x^{j}\) where \(a_j\in k\) and \(p(f):V\to V\)

Then \(p(f)(v_{i})=\sum_{j=0}^{r}a_{j}f^{j}\left(v_{i}\right)=\sum_{j=0}^{r}a_{j}d_{i}^{j} v_{i}=p\left(d_{i}\right)\cdot v_{i}\)

Let \(c\in k\), \(V_{c}=\left\lbrace v\in V:f\left(v\right)=cv\right\rbrace=\left\lbrace0_{v}\right \rbrace\cup\left\lbrace\underbrace{\text{eigenvectors of eigenvalues}}_{\text{empty if c is not an eigenvalue for f}}\right\rbrace\) when \(c\) is an eigenvalue for \(f\)

eigenvector \(=\) characteristic vector

eigenvalue \(=\) characteristic value

\(V_c\) is called the eigenspace corresponding to the eigenvalue \(c\)

Remark

\(\forall c\in k\), \(V_c\) is a subspace of \(V\)
\(c\) is an eigenvalue \(\iff\)\(V_{c}\supsetneq\left\lbrace0_{v}\right\rbrace\)\(\iff\)\(f-c\cdot Id_V:V\to V\) has non zero kernel(it is not a monomorphism (injective)) \(\;\iff \;\det\left\lbrack f-c\cdot Id_{V}\right\rbrack_{\mathcal{B}}=0\;\iff \;c\) is a root of the characteristic polynomial of \(f\) (characteristic polynomial)
\(V_{c}=\ker\left(f-c\cdot Id_{V}\right)\)
Recall that for any bases \(\mathcal{B},\mathcal{B'}\) of \(V\), \(\det\left\lbrack f-c\cdot Id_{V}\right\rbrack_{\mathcal{B}}=\det\left\lbrack f-c \cdot Id_{V}\right\rbrack_{\mathcal{B}^{\prime}}\;\)

A k-linear map \(h:V\to W\) (\(V,W\) are k-vectorspaces) is also called a morphism of k-vector

Morphism refers: reserving all the operations on vector space

If \(h\) is injective, then \(h\) is a monomorphism

If \(h\) is surjective, then \(h\) is a epimorphism

When \(W=V\) and suppose that \(\dim_kV<\infty\), then \(\dim_kV=\dim _k\ker h+\dim_k\text{Im}h\) (dimension theorem)

\(h:V\to V\), then \(h\) is monomorphism \(\iff\) \(\ker h=\{0_v\}\)\(\iff\)\(\dim_k\ker h=0\)\(\;\iff \;\dim_{k}V=\dim_{k}\text{Im}h\iff h\) is an epimorphism\(\iff\)\(h\) is an isomorphism

And \(h\) is an isomorphism \(\begin{cases} h\text{ epimorphism} \\h\text{ monomorphism} \end{cases}\)

We can quickly determine whether the kernel is zero or not by computing the determinant

The characteristic polynomial of \(f\) will be denoted \(\chi_{f}\left(x\right)=\det\left(\left\lbrack x\cdot Id-f\right\rbrack_{B}\right )\)

Example

\(k=\R\), \(V=\R^2\) let \(f:\R^2\to \R^2\) be the linear map defined by: \(E=\{e_1,e_2\}\) standard(canonical) basis

\(f(e_1)=-13e_1+30e_2\) and \(f(e_2)=-9e_1+20e_2\)

Answer

\([f]_{e}= \begin{pmatrix} -13 & -9 \\ 30 & 20 \end{pmatrix}\), then \(\chi_{f}(x)=\det \begin{pmatrix} x+13 & -9 \\ 30 & x-20 \end{pmatrix}=x^{2}-7x+10=\left(x-2\right)\left(x-5\right)\)

eignvalue: \(2,5\). Thus \(f\) is diagonalizable since \(V_{2}=\langle\left(3,-5\right)=v_{1}\rangle,V_{5}=\langle\left(1,-2\right)=v_{2} \rangle\) and \(\mathcal{B}=\{v_1,v_2\}\)

Thus \([f]_{e}= \begin{pmatrix} 2 & 0 \\ 0 & 5 \end{pmatrix}\)
\(k=\R\), \(V=\R^2\) let \(f:\R^2\to \R^2\) be the linear map defined by: \(E=\{e_1,e_2\}\) standard(canonical) basis

\(f(e_{1})=-e_{2}\) and \(f(e_{2})=2e_{1}\)

Answer

\([f]_{e}= \begin{pmatrix} 0 & 2 \\ -1 & 0 \end{pmatrix}\), then \(\chi_{f}(x)=\det \begin{pmatrix} x & -2 \\ 1 & x \end{pmatrix}=x^{2}+2\) has no root in \(\R\), so \(f\) is not diagonalizable

However, if \(k=\mathbb{C}\), then eigenvalue is \(\sqrt{2}i,-\sqrt{2}i\), \(f\) is diagonalizable

(no eigenvalue)
\(k=\mathbb{R}\), \(V=\R^2\) let \(f:\R^3\to \R^3\) be the linear map defined by: \(E=\{e_1,e_2,e_3\}\) standard(canonical) basis

\(f(e_{1})=3e_{1}+2e_{2}+2e_{3}\) and \(f(e_{2})=e_{1}+2e_{2}+2e_{3}\) and \(f(e_3)=-e_1-e_2\)

Then \([f]_{e}= \begin{pmatrix} 3 & 1 & -1 \\ 2 & 2 & -1 \\ 2 & 2 & 0 \end{pmatrix}\) and \(\chi_f(x)=(x-1)(x-2)^2\), thus \(f\) will be diagonalizable \(\iff\) \(\dim_kV_2=2\)

But \(V_{1}=\langle(1,0,2)\rangle,V_{2}=\langle(1,1,2)\rangle\). So \(f\) is not diagonalizable

(no enough eigenvectors)

So a linear map \(f:V\to V\) can be not diagonalizable because of different reasons:

There are not enough eigenvalue in \(k\) (Example 2)
The dimension of an eigenspace \(V_c\) is smaller than it should (Example 3)

Lemma

Let \(V\) be a k-vector space with \(\dim_kV=m\). \(f:V\to V\) be a k-linear map

Let \(c_1,...,c_r\) be the different eigenvalues of \(f\) in \(k\) and \(V_{c_i}\) be the space of eigenvectors of eigenvalues \(c_i\) for \(i=1,...,r\)

Then \(W=V_{c_1}+...+ V_{c_r}\) (this is a subspace of \(V\)) and \(\dim_{k}W=\sum_{i=1}^{r}\dim_{k}V_{c_i}\).

Morever, if \(\mathcal{B_i}\) is a basis of \(V_{c_i}\) for each \(i=1,..,r\), then \(\mathcal{B}=\mathcal{B_1}\cup\ldots\cup\mathcal{B_{r}}\) is a basis of \(W\)

Proof

Let \(B_{1}= \{ v_{11}, \dots, v_{1m_1}\}\) be a basis of \(V_{c_i}\),......, \(B_{r}= \{ v_{r1}, \dots, v_{rm_r}\}\) be basis of \(V_{c_r}\).

Assume \(W = V_{c_1}+ \dots + V_{c_r}\), then \(w \in W\): \(w = v_{c_1}+ \dots + v_{c_r}= a_{11}v_{i1}+ \dots + a_{rm_r}v_{rm_r}\).

Since \(V_{c_i}\cap V_{c_j}= \emptyset\) for \(i \neq j\) (eigenvalues are different), then \(B = \{ v_{11}, \dots, v_{1m_1}, \dots, v_{r1}, \dots, v_{rm_r}\}\) is linearly independent.

Then \(B = B_1 \cup \dots \cup B_r\) is linearly independent.

Thus it's a basis of \(W\). Then \(\dim W = \dim B = \sum_{i=1}^{n}\dim V_{c_i}\).

Theorem

Assume \(\dim_kV=m,f:V\to V\) k-linear. Let \(c_1,...,c_r\) be the different eigenvalues of \(f\) and \(V_{c_i}=\ker(f-c_iId_V)\). The following are equivalent:

\(f\) is diagonalizable
\(\chi_{f}(x)=\left(x-c_{1}\right)^{d_1}\ldots\left(x-c_{r}\right)^{d_{r}}=\prod_{i=1} ^{r}\left(x-c_{i}\right)^{d_{i}}\) and \(\dim_{k}V_{c_{i}}=d_{i},\forall i\)
\(\sum_{i=1}^{r}\dim_{k}V_{c_{i}}=\dim_{k}V\)

Proof

\(1\Rightarrow 2\)

Since \(f\) is diagonalizable, then \(\exists B\) such that \([f]_B\) is diagonal.

Then , so that \(\chi_{f}(x)=\prod_{i=1}^{r}\left(x-c_{i}\right)^{d_{i}}\) and \(d_i=\dim V_{c_i},\forall i,1\leq i\leq r\)

\(2\Rightarrow 3\)

Let \(W=V_{c_1}+\dots+V_{c_{r}}\). Then \(W\) is spanned by eigenvectors.

By lemma, \(\dim W = \sum_{i=1}^{r} \dim V_{c_i} = d_1 + \dots + d_r\)

We know \(\deg \chi_f (x) = \dim V = m\), then \(d_1 + \dots + d_r = \dim W = m\)
Thus \(\dim V = \sum_{i=1}^{r} \dim V_{c_i}\)

\(3\Rightarrow 1\)

Main idea: \(f\) is diagonalizable\(\iff\)\(V\) can be spanned by eigenvectors

Since \(\dim V = \sum_{i=1}^{r} \dim V_{c_i}\) and by lemma \(\dim W = \sum_{i=1}^{r} \dim V_{c_i}\)

Then \(\dim V = \dim W\). Since \(W \subseteq V\), then \(V\) is spanned by eigenvectors.

Thus, \(f\) is diagonalizable.

Example

\(k=\R\), \(V=\R^4\). \(f:\R^4\to \R^4\), \(\R\)-linear such that \([f]_{e}= \begin{pmatrix} 0 & 0 & 0 & 0 \\ a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \end{pmatrix}\)

Find all possible value of \(a,b,c\in \R\) such that \(f\) is diagonalizable?

\(f\) is diagonalizable\(\iff\)\(\dim_{\R}V_0=4\iff rank[f]_\mathcal{B}=0\iff a=b=c=0\)

\(\chi_f(x)=x^4\), thus eigenvalue is \(0\). But \(V_0=\ker f=\lang e_4\rang\)

‍