0
$\begingroup$

There is a matrix $$A=\left[\begin{array}{cccc} 1& 1& 1& 0\\ 1&-1& 0& 1\\ 1& 1&-1& 0\\ 1&-1& 0&-1 \end{array}\right].$$ This matrix has column vectors $a_1$, $a_2$, $a_3$ and $a_4$ and $D$ (a $4\times4$-matrix) is a diagonal matrix with diagonal entries $d_1$, $d_2$, $d_3$ and $d_4$. The matrix $Q$ is a $4\times4$-matrix which is orthonormal.

The problem is actually to find $Q$ and $D$ from this information, where A = QD. What I have tried so far is to try out different types of matrix decomposition like LU(but this leads to forming upper triangular and lower triangular matrices which is not required and SVD). But I am not sure how to head on from this information, so if someone could point me in the right path.

$\endgroup$
7
  • $\begingroup$ Are $A$, $D$ and $Q$ such that $A=Q^TDQ$, by any chance? $\endgroup$ May 26, 2014 at 14:06
  • $\begingroup$ There is no real mention of that... It just says that A = QD... $\endgroup$ May 26, 2014 at 14:11
  • $\begingroup$ It would be convenient to put this in the question, because so far there are no restrictions on what kind of $D$ and $Q$ you should find. $\endgroup$ May 26, 2014 at 14:14
  • $\begingroup$ Just noticed that I missed it... Sorry... $\endgroup$ May 26, 2014 at 14:15
  • $\begingroup$ @aspiringProgrammer please see my edit :) $\endgroup$
    – rlartiga
    May 27, 2014 at 14:23

1 Answer 1

1
$\begingroup$

First notice that your matrix is invertible. Then if such decomposition exists: $$AD^{-1}=Q$$ So you are multiplying the columns of $A$ for scalar values so that don't make it orthogonal. The decomposition which is more near of what do you want is the QR Decomposition

That in general, but in your case A is actually orthogonal and the QR decomposition is: $$ Q = \left(\begin{array}{cccc} \frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}} & 0 \\ 0 & \frac{1}{\sqrt{2}} & 0 & -\frac{1}{\sqrt{2}}\end{array}\right) $$ $$ R = \left(\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0\\ 0 & 0 & \sqrt{2} & 0 \\ 0 & 0 & 0 & \sqrt{2} \end{array}\right) $$ Which coincides with what do you want, but only because as I said $A$ is orthogonal.

$\endgroup$

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.