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12.24 Triangularizable Operators

When does there \(\exists \mathcal{B}\) such that \([T]_\mathcal{B}\) is upper triangular?

Theorem

Let \(T\in \text{End}(V)\) and \(\mathcal{B}:v_1,...,v_n\) be a basis of \(V\). Then the following are equivalent statements.

  1. \([T]_\mathcal{B}\) is upper triangular

  2. \(Tv_j\in \lang v_1,...,v_j\rang\) for \(j=1,...,n\)​

  3. \(\lang v_1,...,v_j\rang\) is \(T\)- invariant for \(j=1,...,n\)​

Proof

\(1\Rightarrow 2\)

Since \([T]_{\mathcal{B}}=A\) is upper triangular, then \(A_{ij}=0\) if \(i>j\).

Then \(Tv_{j}=a_{1j}v_1+a_{2j}v_2+\cdots+a_{jj}v_{j}+a_{j+1}v_{j+1}+\cdots+a_{nj}v_{n}\)

Then \(Tv_{j}=a_{1j}v_1+a_{2j}v_2+\cdots+a_{jj}v_{j}\Rightarrow Tv_{j}\in\langle v_1,\ldots,v_{j}\rangle\) since the property of upper triangular

\(2\Rightarrow 3\)

Let \(w\in \lang v_1,...,v_j\rang\). So, \(w=b_1v_1+...+b_{j}v_{j}\). then \(Tw=b_1Tv_1+\cdots+b_{j}Tv_{j}\in \lang v_1,...,v_j\rang\)

Because for every \(i=1,...,j\) we have \(Tv_{i}\in\langle v_1,\ldots,v_{i}\rangle\) that \(Tv_i\) is a linear combinations of \(v_1,...,v_i\).

Then every \(Tv_i\) for \(i=1,...,j\) is a linear combinations of \(v_1,...,v_j\)

\(3\Rightarrow 1\)

Let \([T]_{\mathcal{B}}=A\), we have \(Tv_{j}=a_{1j}v_1+a_{2j}v_2+\cdots+a_{jj}v_{j}+a_{\left(j+1\right)j}v_{j+1}+\cdots+a_{nj}v_{n}\)

Since \(Tv_{j}\in\langle v_1,\ldots,v_{j}\rangle\), then \(a_{j+1}=\ldots=a_{nj}=0\). Hence \(A\) is upper triangular

Remark

Notice that if \([T]_\mathcal{B}\) is upper triangular, then in particular for \(j=1\) we have that \(Tv_1\in\langle v_1\rangle\Rightarrow Tv_1=a_{11}v_1\)

That is \(v_1\) is an eigenvector of \(T\) with eigenvalue \(a_{11}\)​

Example

  1. \(T=R_{90\degree}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\) (rotation on \(\R^2\) by \(90\degree\) with center \(0\))

    \(\mathbb{F}=\R\), Since it has no eigenvalues, then it is not triangularizable

    \(\mathbb{F}=\mathbb{C}\), there are two different eigenvalues: \(i,-i\). Two corresponding eigenvectors: \(v_1=ie_1+e_2,v_2=-ie_1+e_2\)

    \(\mathcal{B}:v_1,v_2\rightarrow[T]_{\mathcal{B}}=\begin{pmatrix}i&0\\0&-i\end{pmatrix}\) since \(Tv_1=iv_1,Tv_2=-iv_2\)

  2. \([T]_{\mathcal{B}}=\begin{pmatrix}0 & 4\\ -1 & 4\end{pmatrix},\mathcal{B}:e_1,e_2\). Does \(T\) have an eigenvalue/eigenvector?

    By definition we need \(\begin{pmatrix}0 & 4\\ -1 & 4\end{pmatrix}\begin{pmatrix}a\\ b\end{pmatrix}=\lambda\begin{pmatrix}a\\ b\end{pmatrix}\;\iff\;\begin{cases}4b=\lambda a\\-a+3b=\lambda b\end{cases}\)​

    Thus \(\lambda=2,a=2,b=1\)​

    Thus \(\lambda=2\) is an eigenvalue of \(T\) and \(2e_1+e_2=v_1\) is an associated eigenvector to \(\lambda=2\)​

    We can check \(Tv_1=2v_1\).

    Then we have one eigenvalue, but it is enough in dimension \(2\): \(\mathcal{B}':v_1,v_2\)​

    Then \([T]_{\mathcal{B}^{\prime}}=\begin{pmatrix}2&\star\\0&\star\end{pmatrix}\)​

Theorem

Let \(V\) be a finite-dimensional complex vector space. Then every \(T\in \text{End}(V)\) is triangularizable

Proof

Induction on \(\dim V\)

\(n=1\) is an upper triangular matrix then is triangularizable

Since this, then let \(\lambda\) be an eigenvalue of \(T\) , we know that \(T-\lambda I\) is not invertible, in particular it is not surjective

Hence \(U=\text{Range}(T-\lambda I)\) satisfies \(\dim U<\dim V\)

Claim: \(U\) is \(T\)- invariant

Proof1: \(w\in U,w=Tv-\lambda v\Rightarrow Tw=T\left(Tv\right)-\lambda\left(Tv\right)=\left(T-\lambda I\right)\left(Tv\right)\in U\)

Proof2: \(w\in U,Tw=Tw-\lambda w+\lambda w=\left(T-\lambda I\right)w+\lambda w\in U\)

Then by inductive hypothesis: \(T|_{U}:U\rightarrow U\) is triangularizable.

Let's \(\mathcal{B}^{\prime}:u_1,\ldots,u_{k}\) be a basis of \(U\) such that \([T|_{U}]_{\mathcal{B}}=\begin{pmatrix}\star & \cdots & \star\\ \vdots & \ddots & \vdots\\ 0 & \cdots & \star\end{pmatrix}\) where \(Tu_j=\lang u_1,...,u_j\rang,j=1...k\)

Finally, let \(\mathcal{B}:u_1,\ldots,u_{k},v_1,\ldots,v_{n-k}\) be a basis of \(V\).

Observe that \(Tv_i=Tv_i-\lambda_i+\lambda_i=(T-\lambda I)v_i+\lambda v_i=\alpha_1u_1+...+\alpha_ku_k+\lambda v_i\in \lang u_1,...,u_k,v_1,...,v_i\rang\)

Therefore \([T]_\mathcal{B}\) is upper triangular

Example

not invertible singular

\([T]_{\mathcal{B}}=\begin{pmatrix}3 & 1 & -1\\ 0 & 1 & 0\\ 1 & 1 & 1\end{pmatrix},\mathcal{B}:e_1,e_2,e_3\left(\mathbb{F}=\mathbb{C}\right)\)​

Step 1: Find the Eigenvalues of \(T\)

\(\det(T-\lambda I)=0\;\iff\;T-\lambda I\) is not invertible \(\iff \lambda\) is an eigenvalue of \(T\)

\(\det(T-\lambda I)=\begin{vmatrix}3-\lambda & 1 & -1\\ 0 & 1-\lambda & 0\\ 1 & 1 & 1-\lambda\end{vmatrix}=\left(1-\lambda\right)\left(\lambda-2\right)^2=0\;\iff\;\lambda=1,\lambda=2\)

  • Step 2: Choose an Eigenvalue and Construct \(U = \text{Range}(T - \lambda I)\)

\(\lambda=1\), since \(U=\text{Range}(T-I):[T-I]=\begin{pmatrix}2 & 1 & -1\\ 0 & 0 & 0\\ 1 & 1 & 0\end{pmatrix}=\begin{pmatrix}0 & 1 & 1\\ 0 & 0 & 0\\ 1 & 1 & 0\end{pmatrix},\text{Range}(T-I)=\langle2e_1+e_3,e_1+e_3\rangle=\langle e_1,e_3\rangle=\mathcal{B}_{u}\)​

\(T|_U:U\rightarrow U\), then \(Te_1=3e_1+e_3,Te_3=-e_1+e_3\), thus \([T|_{U}]_{\mathcal{B}_{u}}=\begin{pmatrix}3 & -1\\ 1 & 1\end{pmatrix}\)

\(\begin{vmatrix}3-u & -1\\ 1 & 1-u\end{vmatrix}=\left(3-u\right)\left(1-u\right)+1=4-4u+u^2=\left(u-2\right)^2\), then \(u=2\) is an eigenvalue

\(T|_{U}-2I=\begin{pmatrix}1 & -1\\ 1 & -1\end{pmatrix}\Rightarrow e_1+e_3\) is an eigenvector since \(\begin{pmatrix}1 & -1\\ 1 & -1\end{pmatrix}\begin{pmatrix}1\\ 1\end{pmatrix}=0\in\text{Null}(T|_{U}-2I)\)​

Now let \(u_1 = e_1 + e_3, \quad u_2 = e_3\). Then \(\mathcal{B}^{\prime}_{U}:\,u_1,u_2\quad\,\text{let's triangularizes}\,T|_{U}\)

Check:

\(T_U(u_1) = T_U(e_1 + e_3) = T(e_1) + T(e_3)\) \(= 3e_1 + e_3 - e_1 + e_3 = 2e_1 + 2e_3\) \(= 2(e_1 + e_3) = 2u_1\)

\(T_U(u_2) = T_U(e_3) = -e_1 + e_3\) \(= -(e_1 + e_3) + 2e_3 = -u_1 + 2u_2\)

\([T|_{U}]_{\mathcal{B}^{\prime}_{^{}U}}=\begin{pmatrix}2 & -1\\ 0 & 2\end{pmatrix}\)

Finally, let \(\mathcal{B}: u_1, u_2, e_2\quad [T]_{\mathcal{B}} \, \text{is upper triangular}\) why add e_2

Check: \(T(u_1) = T(e_1 + e_3) = 3e_1 + e_3 - e_1 + e_3 = 2e_1 + 2e_3 = 2u_1\)

\(T(u_2) = T(e_3) = -e_1 + e_3 = -(e_1 + e_3) + 2e_3 = -u_1 + 2u_2\)

\(T(e_2) = e_1 + e_2 + e_3 = u_1 + e_2\)

Thus \([T]_{\mathcal{B}} = \begin{pmatrix} 2 & -1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\) * ‍

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\([T]_{\mathcal{B}}=\begin{pmatrix}3 & 1 & -1\\ 0 & 1 & 0\\ 1 & 1 & 1\end{pmatrix},\mathcal{B}:e_1,e_2,e_3\left(\mathbb{F}=\mathbb{C}\right)\)

Step 1: Find the Eigenvalues of \(T\)

\(\det(T-\lambda I)=0\;\iff\;T-\lambda I\) is not invertible \(\iff \lambda\) is an eigenvalue of \(T\)

\(\det(T-\lambda I)=\begin{vmatrix}3-\lambda & 1 & -1\\ 0 & 1-\lambda & 0\\ 1 & 1 & 1-\lambda\end{vmatrix}=\left(1-\lambda\right)\left(\lambda-2\right)^2=0\;\iff\;\lambda=1,\lambda=2\)

If \(\lambda=2\), then \(T-2I=\begin{vmatrix}1 & 1 & -1\\ 0 & -1 & 0\\ 1 & 1 & -1\end{vmatrix}=\begin{vmatrix}0 & 0 & 0\\ 0 & 1 & 0\\ 1 & 0 & -1\end{vmatrix}\).

Since \(U=\text{Range}(T-2I)=\langle e_1+e_3,e_1-e_2+e_3\rangle=\langle e_1+e_3,e_2\rangle\)

And \(T\left(e_1+e_3\right)=2\left(e_1+e_3\right)\), \(Te_2=\left(e_1+e_3\right)+e_2\), then \(\left\lbrack T|_{U}\right\rbrack_{\mathcal{B}_{U}}=\begin{pmatrix}2 & 1\\ 0 & 1\end{pmatrix}\)

Then we can triangularize this matrix, we need to find eigenvalue, then eigenvector

Then \(\left\lbrack T|_{U}\right\rbrack_{\mathcal{B}_{U}}-uI=\begin{pmatrix}2-u & 1\\ 0 & 1-u\end{pmatrix}\), where \(\det(\left\lbrack T|_{U}\right\rbrack_{\mathcal{B}_{U}}-uI)=\left(2-u\right)\left(1-u\right)=0\), then \(u=1,u=2\)

  • \(u=1\)​

\(\left\lbrack T|_{U}\right\rbrack_{\mathcal{B}_{U}}-I=\begin{pmatrix}1 & 1\\ 0 & 0\end{pmatrix}\begin{pmatrix}a\\ b\end{pmatrix}=0\Rightarrow a+b=0\), then \(e_1+e_3-e_2\) is an eigenvector

Let \(u_1=e_1-e_2+e_3\) and \(u_2=e_2\) be the \(\mathcal{B}_{U'}\), then \(Tu_1=T\left(e_1+e_3\right)-Te_2=2\left(e_1+e_3\right)-\left(e_1+e_3\right)-e_2=e_1+e_3-e_2=u_1\)

\(Tu_2=Te_2=e_1+e_2+e_3=u_1+2u_2\), then \(\left\lbrack T|_{U}\right\rbrack_{\mathcal{B}_{U^{\prime}}}=\begin{pmatrix}1 & 1\\ 0 & 2\end{pmatrix}\)​

Then we choose \(u_1=e_1-e_2+e_3,u_2=e_2,e_3\), Since \(T(e_3)=-e_1+e_3=-u_1-u_2+2e_3\)​

Then we have \([T]_{\mathcal{B}_{U^{\prime}}}=\begin{pmatrix}1 & 1 & -1\\ 0 & 2 & -1\\ 0 & 0 & 2\end{pmatrix}\) * \(u=2\)​

\(\left\lbrack T|_{U}\right\rbrack_{\mathcal{B}_{U}}-2I=\begin{pmatrix}0 & 1\\ 0 & -1\end{pmatrix}\begin{pmatrix}a\\ b\end{pmatrix}=0\Rightarrow b=0\), then \(e_1+e_3\) is an eigenvector

Let \(u_1=e_1+e_3\) and \(u_2=e_2\) be the \(\mathcal{B}_{U'}\), then \(Tu_1=T\left(e_1+e_3\right)=2\left(e_1+e_3\right)=2u_1\)

\(Tu_2=Te_2=e_1+e_2+e_3=u_1+u_2\), then \(\left\lbrack T|_{U}\right\rbrack_{\mathcal{B}_{U^{\prime}}}=\begin{pmatrix}2 & 1\\ 0 & 1\end{pmatrix}\)

Then we choose \(u_1=e_1+e_3,u_2=e_2,e_3\), Since \(T(e_3)=-e_1+e_3=-u_1+2e_3\)

Then we have \([T]_{\mathcal{B}_{U^{\prime}}}=\begin{pmatrix}2 & 1 & -1\\ 0 & 1 & 0\\ 0 & 0 & 2\end{pmatrix}\)

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