1
$\begingroup$

I'm working on a practice exam, and I am having a lot trouble finding the solution to this problem. The solution's are posted, however they seem to be completely computationally wrong. In the hours I have been working on this I have kind of lost all comprehension of the problem.

$$ A = \begin{pmatrix} 3&1 \\1&3 \\ \end{pmatrix} $$

Find S such that:

$$ S^T\cdot A\cdot S = \begin{pmatrix}1&0\\0&1\end{pmatrix} $$

I recognize that: $$ v^T\cdot A\cdot w $$

conforms to the axioms of an inner product and I am certain this comes into play, and I'm sure the Gram-Schmidt process needs to be applied somehow as well, but I can't fully grasp how to continue. Any help would be appreciated.

To be clear I see that we can diagonalize this matrix easily to: $$ \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{pmatrix}\cdot\begin{pmatrix} 3&1 \\1&3 \\ \end{pmatrix}\cdot \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{pmatrix} = \begin{pmatrix} 4&0\\0&2\end{pmatrix} $$

and I see: $$ K = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{pmatrix} = K^T = K^{-1} $$ and $$ K^2 = \begin{pmatrix} 1&0\\0&1\end{pmatrix} $$ But I'm not sure where to go from here.

$\endgroup$
2
  • $\begingroup$ Do you know about eigenvalues and eigenvectors and diagonalization and the spectral theorem? $\endgroup$ – mathreadler Apr 14 '16 at 10:28
  • $\begingroup$ This is the keyword in this kind of problems: en.wikipedia.org/wiki/Matrix_congruence You are trying to diagonalize $A$ with respect to congruence instead of matrix similitude. This is the same as to say that you are diagonalizing $A$ by seeing it as a bilinear form, instead of seeing it as a linear operator. There is a bit of theory about it, but in the end, all boils down to a standard diagonalization, carried over using orthogonal diagonalizing matrices $S$. $\endgroup$ – Giuseppe Negro Apr 14 '16 at 10:39
0
$\begingroup$

Some hints I hope help you on the right track.

  1. If you manage to find a diagonalization of a matrix A , that means that ${\bf TDT}^{-1} = {\bf A}$ for a diagonal matrix $\bf D$ and some matrix $\bf T$.
  2. If the matrix ${\bf A}$ is symmetric, one can show that we can find $\bf D$ and ${\bf T}$ so that ${\bf T}^{-1} = {\bf T}^T$ this is known as the spectral theorem.

Now you need to find out what to do to find a diagonalization. It has to do with finding eigenvalues : solution to $\det({\bf A}-\lambda {\bf I}) = 0$ and eigenvectors which are the vectors $\bf v$ satisfying the equations $({\bf A}-\lambda {\bf I}) {\bf v} = {\bf 0}$ for those $\lambda$. Once you have a diagonalization then you can figure out what to do to turn the diagonal matrix into an identity matrix.


EDIT

What is left after you found diagonalization is basically to find $P$ so that $$P\left(\begin{array}{cc}4&0\\0&2\end{array}\right)P^T = \left(\begin{array}{cc}1&0\\0&1\end{array}\right)$$

So just assume $P = \left(\begin{array}{cc}p_1&p_2\\p_3&p_4\end{array}\right)$ and see what equations you will be getting when you multiply stuff together and solve those. Then when you have found $P$ you can multiply both sides of the diagonalization equation with $P$ and $P^T$ respectively and see what the various products of $K$ and $P$ and their transposes become.

$\endgroup$
2
  • $\begingroup$ Diagonalization is very much the easy part. I'm also aware of the existence such a T but I'm not sure how to bridge the final gap and find it.I'm also confused as to how it will result in the identity matrix. $\endgroup$ – Probot Apr 14 '16 at 17:40
  • $\begingroup$ @Probot-can you think of a diagonal $P$ that will work? $\endgroup$ – David Wheeler Apr 15 '16 at 2:35

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.