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.

  • $\begingroup$ 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

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.


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.

  • $\begingroup$ 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

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