Schur's Decomposition

Theorem

• Let $V$ be a finite-dimensional Complex Inner Product Space.
• Let $T$ be a Linear Operator on $V$ with matrix $A$
• Then, there is a Orthonormal Basis such that there is a Upper Triangular matrix $M$
• Then, there is a Unitary basis $U$ such that: $A=UM{U}^{\ast }$

Intuition

• If i show a matrix is normal, then I know it is Diagonalizable, and I know the eigenvectors are perpendicular to eachother This process can be used to find Similar Matrixes so that further computations are easier

Proof

We show this with PMI.

Base Case

When $dim(V)=1,[T{]}_{\alpha }=[{\alpha }_u]$ which is Upper Triangular.

N+1 Case

• With $dim(V)=n+1$
• Assume that if $W$ is a complex inner product space of dimension $n$, then there exists a Orthonormal Basis such that the matrix of $T$ is Upper Triangular
• Recall that since $V$ is a fin-dim inner product space, there exists an eigenvalue, eigenvector pair such that $T\alpha =c\alpha$
• Let $W=span\{\frac{\alpha }{||\alpha ||}\}$. Let $W^{\perp }=span\{v_1,\dots ,v_n\}$
• Let $\beta =\{\frac{\alpha }{||\alpha ||},v_1,\dots ,v_n\}$
• Then, $[T{]}_{\beta }=\left[\begin{array}{cc}1& V\\ 0& W\end{array}\right]$
• Then, let $S_W:W^{\perp }\to W^{\perp }$ be defined using the matrix $W$
• Then, by our inductive hypothesis, this is a Linear Map from a $n$ dimension vector space to a $n$ dimensional vector space
• Therefore, there exists a basis ${\beta }^{\prime }=\{{\alpha }_1,\dots ,{\alpha }_n\}$ such that $[S_W{]}_{\beta }^{{\beta }^{\prime }}$ is Upper Triangular
• Note then that $\{\frac{\alpha }{||\alpha ||},{\beta }^{\prime }\}=\{\frac{\alpha }{||\alpha ||},{\alpha }_1,\dots ,{\alpha }_n\}$ is an orthonormal basis for $V$
• Let $P$ be the change of basis matrix at $[S_W{]}_{{\beta }^{\prime }}=PW{P}^{-1}$
• Then, $\left[\begin{array}{cc}1& 0\\ 0& P\end{array}\right]$ for $\beta$ to $P^{\prime }$
• We want to show that we can get to this Upper Triangular matrix. $\left[\begin{array}{cc}x& VP\\ 0& [S_w{]}_{{\beta }^{\prime }}\end{array}\right]$
• We know that by Change of Basis, $[T{]}_{\beta }={\hat{P}}^{-1}[T{]}_{\beta }\hat{P}$
• $⟹[T{]}_{{\beta }^{\prime }}=\hat{P}[T{]}_{\beta }{\hat{P}}^{-1}$

=\left[\begin{array}{cc} 1 & 0\ 0 & P\ \end{array}\right] \left[\begin{array}{cc} C & V\ 0 & W\ \end{array}\right] \left[\begin{array}{cc} 1 & 0\ 0 & P^{-1}\ \end{array}\right]

- $$ =\left[\begin{array}{cc} 1 & 0\\ 0 & P\\ \end{array}\right] \left[\begin{array}{cc} C \cdot 1 + V \cdot 0 & C \cdot 0 + VP^{-1} \\ 0 \cdot 1 + W \cdot 0 & 0 \cdot 0 + WP^{-1} \end{array}\right]

• Remember that $V$ is a vector, $P$ is a matrix and $W$ is a matrix

=\left[\begin{array}{cc} 1 & 0\ 0 & P\ \end{array}\right] \left[\begin{array}{cc} C \cdot 1 + V \cdot 0 & C \cdot 0 + VP^{-1} \ 0 \cdot 1 + W \cdot 0 & 0 \cdot 0 + WP^{-1} \end{array}\right]

- $$ =\left[\begin{array}{cc} c & VP^{-1}\\ 0 & [S_{w}]_{\beta'}\\ \end{array}\right]

• Note that this is Upper Triangular