Topic Forum

A linear algebra "puzzle" (not really, but ... ) - Gram Schmidt on space of 2x2 matrices

A linear algebra "puzzle" (not really, but ... ) - Gram Schmidt on space of 2x2 matrices

by Admin MathWit -
Number of replies: 0
Let \(\mathcal B=\left\{m_1=\left(\begin{array}{cc} 1 & 1 \\ 1 & 1\end{array}\right), m_2=\left(\begin{array}{cc} -1 & 1 \\ 1 & 1\end{array}\right), m_3=\left(\begin{array}{cc} 1 & 3 \\ 3 & 1\end{array}\right)\right\}\) be a basis for a subspace \(\mathbb S\) of \(M_{2\times 2}(\mathbb R)\). Use Gram-Schmidt procedure on \(m_1, m_2, m_3\) to find an orthonormal basis for \(\mathbb S\). 


Solution(wrong):

 Let \(n_1=m_1\). \(n_2=m_2-\frac{\langle m_2, m_1\rangle}{\langle m_1, m_1\rangle}m_1 = m_2-\frac{\text{Tr}(m^T_1m_2)}{\text{Tr}(m^T_1 m_1)}m_1= \frac{1}{2}\left(\begin{array}{rr}-3 & 1 \\ 1 & 1\end{array}\right).\) \(n_3=m_3-\frac{\langle m_3, m_1\rangle}{\langle m_1, m_1\rangle}m_1 - \frac{\langle m_3, m_2\rangle}{\langle m_2, m_2\rangle}m_2= m_3-\frac{\text{Tr}(m^T_1m_3)}{\text{Tr}(m^T_1 m_1)}m_1-\frac{\text{Tr}(m^T_2m_3)}{\text{Tr}(m^T_2 m_2)}m_2=\frac{1}{2}\left(\begin{array}{rr} 1 & -1 \\ -1 & -5 \end{array}\right).\) But as you can check that \(\langle n_1, n_3\rangle = -3\), not 0, as it supposed to be. What's wrong? Can you provide a correction solution?