$$A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1\\ 1 & 1 & -1 \end{bmatrix}$$

This matrix can be transformed into a superior trangiular matrix through left multiplication by a lower triangular matrix $L$ or by an orthogonal matrix $Q$. Find the matrix $L$ and the matrix $Q$. Solve $Ax= b$ with $b=\begin{bmatrix} 1 \\ -1 \\ 2 \end{bmatrix}$

What I know how to do is to make $A = LU$ and $A=QR$ which are the known LU and QR decompositions. However, this exercise asks me to left multiply by $L$ and left multiply by $Q$ to obtain a superior triangular matrix. What am I missing?

  • $\begingroup$ Note that if $A = QR$ with $Q$ an orthogonal matrix, then $R = Q^T A$. Similarly the inverse of an invertible lower triangular matrix $L$ is again lower triangular. $\endgroup$ – hardmath Apr 14 at 20:33

Note that the inverse of an upper(lower) triangular matrix - if it exists - is again an upper(lower) triangular matrix. So if $A = LU$ we have $L^{-1}A = U$ and similarly for $A = QR$ we have $Q^{-1}A = Q^tA = R$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.