Jordan Conical Form

Defn

\(T:V\rightarrow V\) is diagonalizable if \(\exists\) a basis \(\beta\) of \(V\) st \([T]_\beta=D\) is a diagonal matrix

Theorem 5.1

\(T\) is diagonalizable iff \(\beta\) is a basis of \(V\) consisting of eigenvectors and \(\lambda_i=d_{i,i}\)

Defn

\(v\in V\) is an eigenvector of \(T\) if \(v\neq 0\) and \(\exists \lambda\in F\) st \(T(v)=\lambda v\)

Note. Eigenvalues do not depends on the choice of basis.

\([T]_\beta=[I]^\beta_\alpha[T]_\alpha[I]^\alpha_\beta\)

\(det(B-tI)=det(Q^{-1}AQ-tI)=det(Q^{-1}AQ-Q^{-1}(tI)Q)\) \(=det(Q^{-1}(A-tI)Q)=det(Q^{-1})det(Q)det(A-tI)=det(A-tI)\)

Thus, the characteristic polynomial are the same

Theorem 5.5

If eigenvalues are distinct, then the corresponding eigenvectors are linearly independent.

Corollary

If \(dim V=n\) and \(T:V\rightarrow V\) are distinct eigenvalues, then \(T\) is diagonalizable

Proof.

Proof by induction

if \(k=1\)

then \(\{v_1\}\) is linearly independent set as \(v_1\neq 0\)

Assume it is true for \(k=n\), prove it is true for \(k=n+1\)

Suppose \(\exists a_i\in F\) st \(\sum _{i=0}^{n+1}a_iv_i=0\)

\(c_1(\lambda_1-\lambda_{n+1})v_1+...+c_n(\lambda_n-\lambda_{n+1})v_n+0=0\)

Since \(v_1,...,v_n\) are linearly independent, thus \(c_i(\lambda_i-\lambda_{n+1})=0\)

Since, \(\lambda_i\neq\lambda_{n+1}\), thus, \(c_i\) is zero and thus \(c_{n+1}\) is zero

Thus, \(c_{n+1}k_{n+1}=0\)

Thus, \(\{v_1,...,v_k\}\) is linearly independent

The reverse is not true \(In\) for example

Defn

A polynomial \(p(t)\in P(n)\) splits over \(F\) if \(\exists c,b_1,...,b_n\in F\) st \(f(t)=c(t-b_1)(t-b_2)...(t-b_n)\)

Theorem 5.6

If \(T\) is diagonalizable, then its characteristic polynomial is split

Proof.

\(T\) diagonalizable \(\implies\exists\beta\) basis of \(v\) st \([T]_\beta=D\) so \(p(t)=det(D-tI)=...\) and thus \(p(t)\) is split.

Summary

\(\lambda_i\) are ll distinct \(\implies\) \(T\) is diagonalizable \(\implies\) \(p(t)\) is split

Theorem 5.7

Let \(\lambda\) be an eigenvalue of \(T\) having multiplicity \(m\) then \(1<dim(E_\lambda)\leq m\)

Theorem 5.9

Let \(\lambda_1,...,\lambda_k\) be distinct eigenvalues of \(T\)

\(T\) is diagonalizable \(\iff\) characteristic polynomial of \(T\) splits and multiplicity of \(\lambda_i\) equals the dimension of \(E_{\lambda_i}\)
If \(T\) is diagonalizable and \(\beta_i\) is a basis of \(E_{\lambda_i}\forall i=\{1,...,k\}\), then \(\beta=\beta1\cup...\cup\beta_k\) is a basis of \(V\) consisting of eigenvectors of \(T\), i.e., \([T]_\beta\) is a diagonal matrix.

Proof.

\(\implies\)

Let \(m_i\) be multiplicity of \(\lambda_i\), \(d_i=dim(E_{\lambda_i})\), \(n=dim V\)

Suppose \(T\) is diagonalizable. By Theorem \(5.6\) \(p(t)\) splits

Let \(\beta\) be a basis of \(V\) consisting of eigenvectors of \(T\)

Let \(\beta_i=\beta\cap E_{\lambda_i}\), \(n_i=|\beta_i|\)

Then \(n_i\leq d_i\) as \(\beta_i\subset E_{\lambda_i}\) and linearly independent

and \(d_i\leq m_i\) by theorem 5.7

Thus, \(\sum n_i\leq\sum d_i\leq \sum m_i=n\)

Thus, \(\sum m_i=\sum d_i,\sum(m_i-d_i)=0\)

Since, \(m_i\geq d_i\) (Theorem 5.7) thus \(m_i=d_i\)

\(\impliedby\)

Assume \(p(t)\) splits and \(m_i=d_i\)

Let \(\beta_i\) be a basis of \(E_{\lambda_i}\), \(\beta=\beta_1\cup...\cup\beta_k\)

Claim \(\beta\) linearly independent

Suppose \(\beta_i=\{V_{i,1},...,V_{i,n}\}\)

Thus, \(b=\sum_{i=1}^k\sum_{j=1}^n c_{ij}V_{ij}=\sum_{i=1}^kw_i=0\) where \(w_i\in E_{\lambda_i}\)

If any \(w_i\neq 0\) then this contradict with the fact that the eigenvectors of distinct eigenvalues are linearly independent.

Thus, \(w_i=0\)

Thus, \(\sum_{j=1}^n c_{ij}V_{ij}=0\)

Thus, \(c_{ij}=0\) as \(\beta_i\) is linearly independent

Thus, \(\beta\) is linearly independent

Claim \(|\beta|=n\)

\(|beta|=\sum|\beta_i|=n\)

So, \(\beta\) is a basis of \(V\) consisting of eigenvectors of \(T\)

\([T]_\beta\) is a diagonal matrix

Theorem 5.11

Let \(\lambda_1,...,\lambda_k\) are distinct eigenvalues of \(T\), Then \(V=E_{\lambda_1}\oplus...\oplus E_{\lambda_k}\iff T\) is diagonalizable

Theorem Cayley-Hamilton Theorem 5.23

If \(p(t)\) is the characteristic polynomial of \(T\) then \(p(T)=0\)

Theorem

\(T\) has a Jordan Conical Form if \(p(T)\) splits.

Theorem

\(T\) has a Rational Canonical Form