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12.23 Diagonalization

Invariant subspaces

Example: \([T]_{\mathcal{B}}=\begin{pmatrix}1 & 2 & & \\ 2 & -3 & & \\ & & 0 & 1\\ & & 1 & -1\end{pmatrix}\) where \(\mathcal{B}:v_1,v_2,v_3,v_4\)

\(w_1=\langle v_1,v_2\rangle,w_2=\langle v_3,v_4\rangle\)

\(T_1=T|_{w_1}:w_1\rightarrow w_1\), \(T_2=T|_{w_2}:w_2\rightarrow w_2\)​

\(V=W_1\oplus W_2\), \(T(w_1+w_2)=Tw_1+Tw_2=T_1w_1+T_2w_2\) and moreover

\(\text{Null}T=\text{Null}T_1\oplus\text{Null}T_2\), \(\text{Range}T=\text{Range}T_1\oplus\text{Range}T_2\),

Definition

Given \(T\in \text{End}(V)\) a subspace \(W\hookrightarrow V\) is said to be invariant under \(T\) or \(T\)- invariant if \(Tw\in W\), \(\forall w\in W\)

Notation

\(T(w)=\{Tw:w\in W\}\)

Example

\(\{0\}\) and \(V\) are \(T\)- invariant, \(\forall T\). Also \(\text{Null}T\) and \(\text{Range}T\) are \(T\)- invariant.

Proof

Take \(v\in \text{Null}T\), then \(Tv=0\in \text{Null}T\). Take \(w\in \text{Range}T\), then \(Tw\in \text{Range}T\)

Examples for Particular \(T\)

  1. \(T=R_{90\degree}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\) (rotation on \(\R^2\) by \(90\degree\) with center \(0\)) has no non-trivial invariant subspaces.

    Since \(0\subsetneq W\subsetneq\mathbb{R}^2\), then \(\dim W=1\), then \(W=\lang w\rang\), \(w\neq 0\). But \(Tw\neq \lambda w\)​

  2. \(T=\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\) image​

  3. \(T=\begin{pmatrix}1 & 1\\ 0 & 1\end{pmatrix}\), \(V=\lang v_1,v_2\rang\)​

    \(Tv_1=v_1\) first column, so \(W_1=\langle v_1\rangle\) is \(T\)- invariant

    Is there any other \(W_2:\dim W_2=1\) is \(T\)- invariant?

    Let's \(W_2=\lang av_1+bv_2\rang\), if \(W_2\) is \(T\)- invariant, then \(T(av_1+bv_2)=\lambda(av_1+bv_2)\)​

    Since \(\begin{cases}T(av_1+bv_2)=aTv_1+bTv_2=av_1+bv_1+bv_2=\left(a+b\right)v_1+bv_2\\ \lambda\left(av_1+bv_2\right)=\lambda av_1+\lambda bv_2\end{cases}\)​

    Then \(\begin{cases}\lambda a=a+b\\ \lambda b=b\end{cases}\Rightarrow\lambda=1,b=0\). Hence \(W_2=W_1\)

  4. \(T=\begin{pmatrix}1 & 1\\ 0 & 2\end{pmatrix}\), \(W=\langle v_1\rangle\) is \(T\)- invariant

    Is there any others? Let's \(W_2=\langle av_1+bv_2\rangle\), then \(T(av_1+bv_2)=\lambda(av_1+bv_2)\)

    Since \(\begin{cases}T(av_1+bv_2)=aTv_1+bTv_2=av_1+bv_1+2bv_2=\left(a+b\right)v_1+2bv_2\\ \lambda\left(av_1+bv_2\right)=\lambda av_1+\lambda bv_2\end{cases}\)

    Then \(\begin{cases}\lambda a=a+b\\ \lambda b=2b\end{cases}\Rightarrow\lambda=2,b=a\text{ or }b=0\). Thus \(W_2=\langle v_1+v_2\rangle\text{ or }W_1\)

    All the things are if and only if.

    Let's check \(T(v_1+v_2)=Tv_1+Tv_2=v_1+v_1+2v_2=2v_1+2v_2=2(v_1+v_2)\)​

Remark

\([T]_\mathcal{B}\) is block diagonal iff \(V=W_1\oplus W_2\oplus\ldots\oplus W_{k}\) with \(W_i\) that is \(T\)- invariant

Eigenvalues and Eigenvectors

Definition

Let \(T\in \text{End}(V)\). A \(\lambda\in \mathbb{F}\) is an eigenvalue of \(T\) if \(\exists 0\neq v\in V\) such that \(Tv=\lambda v\)

Remark: We ask for \(v\neq 0\) because \(T\cdot0=0=\lambda\cdot 0,\forall \lambda\)

Example

\(T=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) we already know that if \(\mathbb{F}=\R\), there are no eigenvalues. However if \(\mathbb{F}=\mathbb{C}\), then

Let \(V=\lang e_1,e_2 \rang\), then \(v=ae_1+be_2\).

Then \(\begin{cases}Tv=T(ae_1+be_2)=ae_2-be_1\\ \lambda v=\lambda ae_1+\lambda be_2\end{cases}\Rightarrow\begin{cases}\lambda a=-b\\\lambda b=a\end{cases}\Rightarrow\lambda^2=-1\Rightarrow\lambda=\pm i\)​

In fact for \(\lambda_1=i,v_1=e_1-ie_2\neq0\) and \(Tv_1=\lambda_1v_1\), for \(\lambda_2=-i,v_2=e_1+ie_2\neq 0\) and \(Tv_2=\lambda_2v_2\)​

Conclusion: \(\lambda_1=i\) and \(\lambda_2=-i\) are two different eigenvalues for \(T=\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) over \(\mathbb{C}\)​

Proposition

Let \(T\in \text{End}(V)\) and \(\lambda \in \mathbb{F}\). Then \(\lambda\) is an eigenvalue of \(T\) iff \(T-\lambda I\) is not invertible

Proof

\(\exists v\neq 0\) such that \(Tv=\lambda v\) iff \(Tv-\lambda v=0\quad(v\neq0)\) iff \((T-\lambda I)(v)=0\quad(v\neq0)\) iff \(\text{Null}(T-\lambda I)\neq 0\) iff \(T-\lambda I\) is not injective iff \(T-\lambda I\) is not invertible

Definition

Let \(T\in \text{End}(V)\) and \(\lambda\) an eigenvalue of \(T\). Any \(0\neq v\in V\) is an eigenvector of \(T\) with eigenvalue \(\lambda\) if \(Tv=\lambda v\)

Remark

All eigenvectors associated to the eigenvalue \(\lambda\) are the all \(v\neq 0\) such that \(v\in \text{Null}(T-\lambda I)\)

Notation

The eigenspace corresponding or associated to the eigenvalues \(\lambda\) is \(V_\lambda=\text{Null}(T-\lambda I)\)

Remark

\(V_0=\text{Null}T\), every \(v\neq 0\) such that \(Tv=0\) is a eigenvector of eigenvalue \(0\)

Proposition

Let \(T\in \text{End}(V)\) and \(\lambda_1,...,\lambda_k\) be different eigenvalues. If \(v_1,...,v_k\) are eigenvectors associated corresponding to \(\lambda_1,...,\lambda_k\) then \(v_1,...,v_k\) is linear independent

Proof

Suppose \(v_1,...,v_k\) is linear dependent. Let \(j\) be the largest index such that \(v_1,...,v_{j-1}\) is linear independent.

Hence \(v_j\in \lang v_1,...,v_{j-1}\rang\), that is \(v_j=a_1v_1+...+a_{j-1}v_{j-1}\).

On one hand, \(\lambda_{j}v_{j}=Tv_{j}=a_1Tv_1+\cdots+a_{j-1}Tv_{j-1}=a_1\lambda_1v_1+\cdots+a_{j-1}\lambda_{j-1}v_{j-1}\)

On the other hand, \(\lambda_{j}v_{j}=a_1\lambda_{j}v_1+...+a_{j-1}\lambda_{j}v_{j-1}\)

Then we subtract these two: \(0=a_1\left(\lambda_{j}-\lambda_1\right)v_1+a_2\left(\lambda_{j}-\lambda_2\right)v_2+\ldots+a_{j-1}\left(\lambda_{j}-\lambda_{j-1}\right)v_{j-1}\)

Since \(v_1,...,v_{j-1}\) is linearly independent and \(\lambda_{j},\lambda_1,\lambda_2,\ldots,\lambda_{j-1}\) are different, then \(a_1=...=a_{j-1}=0\)

Then \(v_j=0\) which is a contradiction to \(v_j\neq 0\)

Corollary

Let \(T\in \text{End}(V)\), then \(T\) has at most \(n=\dim V\) different eigenvalues.

Proof

Suppose we have at least \(n+1\) different eigenvalues, then we have \(n+1\) linearly independent list, which is impossible in \(V\) since \(\dim V=n\)

Theorem

Let \(T\in \text{End}(V)\) where \(V\) is a complex vector space. Then \(T\) has an eigenvalue always

Proof

Define \(n=\dim V\), let \(v\neq 0\). Consider \(v,Tv,T^2v,\ldots,T^{n}v\), since \(\dim V=n\), then this list is linearly dependent.

Thus \(0=a_0v+a_1Tv+...+a_nT^nv\) with not all zero.

Let \(P(x)=a_0+a_1x+...+a_nx^n\), now on the one hand \(p(T)(v)=0\)

On the other hand, Fundamental theorem of algebra, \(p(x)=a_{n}(x-\lambda_1)\cdot...\cdot(x-\lambda_{n})\) for some \(\lambda_1,...,\lambda_n\in \mathbb{C}\) (may be same)

Then \(p(T)=a_{n}(T-\lambda_1)\cdot...\cdot(T-\lambda_{n})\) and \(0=p(T)(v)=a_{n}\left(T-\lambda_1I\right)\cdot\ldots\cdot\left(T-\lambda_{n}I\right)\left(v\right)\)​

Since \(v\neq 0,v\in \text{Null}p(T)\). Thus \(p(T)\) is not injective also not invertible. In particular \(\exists i\) such that \((T-\lambda_{i}I)\) is not invertible.

Hence \(\lambda_i\) is an eigenvalue of \(T\)

Example

\(T=\begin{pmatrix}1&1\\0&1\end{pmatrix},\mathbb{F}=\mathbb{C}\)

\(T\) has only one eigenvalue \(\lambda=1\). There is only one up to multiple eigenvector associated to it

Another way to conclude that there are no \(2\) independent eigenvector associated to \(\lambda=1\)

So \(V=\lang v_1,v_2\rang\) where \(\mathcal{B}:v_1,v_2\). Then \([T]_{\mathcal{B}}=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\) since this is identity and similar only to itself.

True or False

If \(T\in \text{End}(V)\) has eigenvalues \(1,2,...,n\) and \(\dim V=n\), then \(\exists \mathcal{B}\) such that \([T]_{\mathcal{B}}=\begin{pmatrix}1 & & \\ & \ddots & \\ & & n\end{pmatrix}\)

True

\(1\rightarrow \exists v_1:Tv_1=v_1\) \(2\rightarrow \exists v_2:Tv_2=2v_2\), by proposition \(v_1,...,v_n\) is linearly independent.

Hence it is a basis.