Let $$ A = \begin{pmatrix} 5 & 7 \\ -2 & -4 \end{pmatrix}. $$ We will find an orthogonal matrix $P$ and an upper triangular matrix $T$ such that $P^tAP=T$ applying the proof of Schur's decomposition. We 're following the steps below

• We find the eigenvalues of $A$
  The eigenvalues of a matrix are precisely the solutions to the equation $$ \det(\lambda I - A) = 0 \leftrightarrow \lambda^2-\lambda -6=0 $$
  Hence the roots of the quadratic equation are the eigenvalues $\lambda_1=-2,\lambda_2=3$
• We find the eigenvectors
  For each eigenvalue $\lambda_i$ , solving the system $$ (A-\lambda_i I)X_i=0$$ So we have that for $\lambda_1=-2$ $$(A+2I)=0 \leftrightarrow \begin{pmatrix} 7 & 7 \\ -2 & -2 \end{pmatrix}\begin{pmatrix} x_1 \\x_2\end{pmatrix}=\begin{pmatrix} 0\\ 0\end{pmatrix}\rightarrow X_1=(1,-1)$$
  Analogously for $\lambda_2=3$ the eigenvector $X_2=(7,-2)$
• We get an orthonormal set of eigenvectors using Gram-Schmidt orthogonalization
  Consider the above two eigenvectors which are linearly independent but are not orthogonal
  
  First we take $w_1=X_1=(1,-1)$ . Therefore $$ w_2= X_2 - \frac{w_1\cdot X_2}{\Vert w_1\Vert^2}w_ $$ that is, $$ w_2=(\frac{5}{2},\frac{5}{2}) $$ and finally the orthonormal set is $\{w_1/\Vert w_1\Vert,w_2/\Vert w_2\Vert\}=\{(\frac{1}{\sqrt{2}},\frac{-1}{\sqrt{2}}),(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\}$
  So $$P =\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}.$$ Then $$T=P^tAP=\begin{pmatrix} -2 & 9 \\ 0 & 3 \end{pmatrix}.$$