I have been given a $3×3$ symmetric matrix and I am required to write this in the form $BDB'$ where $D$ is a diagonal matrix and $B'$ is transpose of $B$.
From my knowledge of eigen vectors and diagonalization of matrices, I know that a diagonalizable matrix $A$ can be expressed as $A= PDP^-1$ where $P$ is the invertible matrix formed by arranging the eigen vectors of $A$ as column vectors and $D$ is the diagonal matrix formed by taking the eigen values of $A$ as diagonal elements.
But the above question asks me to have transpose matrix in place of inverse matrix. How can I get that? Please suggest.
(Real) symmetric matrices are automatically diagonalisable with real eigenvalues. This means it should be possible to find a basis of eigenvectors, which we do by smooshing together bases for the eigenspaces (e.g. if you have a two-dimensional eigenspace, then you simply pick two linearly independent eigenvectors from that eigenspace when forming your basis of eigenvectors).
Moreover, symmetric matrices have orthogonal eigenspaces, meaning that eigenvectors for distinct eigenvectors will automatically be orthogonal. If we take orthonormal bases for each eigenspace (if the eigenspace is $1$-dimensional, then just take a unit-length eigenvector), then the basis we get from smooshing these vectors together is an orthonormal basis for $\Bbb{R}^3$.
Now, if you diagonalise with respect to an orthonormal basis, the change-of-basis matrices will be orthogonal, meaning that the inverse of the change-of-basis matrix will be the same thing as the transpose of the change-of-basis matrix.
So, basically, you just need to perform the usual process of diagonalisation, just ensuring that your basis of eigenvectors is orthonormal.