# Quadratic forms, diagonal form, and whether an orthogonal transformation exists for a matrix,

a) Let

$$\begin{bmatrix} 3 & 2 & -2 \\ 2 & 3 & -2 \\ -2 & -2 & 5 \\ \end{bmatrix}$$

be a quadratic form. Write explicitly an orthogonal transformation, $O$, which transforms A into a diagonal form $a_1y_1^2 + a_2y_2^2 + a_3y_3^2$ with coefficients $a_i$ >0.

b) Whether such orthogonal transformation exists for $A_1$, $A_2$?

$$A_1= \begin{bmatrix} 3 & 4 & 3 \\ 4 & 3 & -3 \\ 3 & -3 & 5 \\ \end{bmatrix}$$ and $$A_2= \begin{bmatrix} 4 & 2 & -2 \\ 2 & 4 & -2 \\ -2 & -2 & 6 \\ \end{bmatrix}$$

If yes find such transform. If not provide the proof.

Edit: My main difficulty for this problem was relating matrices to their quadratic and diagonal forms, though most of this has already been addressed and cleared up in the comments and answers below. Thanks...

• What are your thoughts on the problem? Do you know how to find the eigenvalues of these matrices? Commented Dec 23, 2014 at 5:22
• Yes, @Omnomnomnom. I can compute the characteristic polynomial for each of these 3 matrices, which is given by p($\lambda$) = det[$A_i$ - $\lambda$$I]. I guess I'm more in need of help with what's being asked, i.e., what does it mean for a matrix to be a quadratic form, what does it mean to find an orthogonal transformation that transforms the matrix into a diagonal form with a_i>0. Is it very close to the problem of diagonalizing a matrix? If so, what is different about this problem than just diagonalizing the matrix? Thanks... Commented Dec 23, 2014 at 5:27 • (there's a little bit of reading material on Wiki and Wolfram but not much...) Commented Dec 23, 2014 at 5:28 • It would be better if you edited some of the information in your comment, Lebron, into the body of the question, so people could see straight off that what you don't know is how to relate a matrix to a quadratic form, etc. Commented Dec 23, 2014 at 6:22 • Ok - while do. Good point, @GerryMyerson :) Commented Dec 23, 2014 at 6:34 ## 1 Answer For a), you want to find the spectral-decomposition of that matrix. That is, find the eigen-decomposition, but select the eigenvectors to be orthonormal, so that the resulting matrix of eigenvectors gives you the orthogonal matrix O. For b), it suffices to check whether A_1 and A_2 are positive definite. The easiest approach to use here is Sylvester's criterion. Because the matrices are symmetric, we can put them into a diagonal form with an orthogonal transformation. However, the a_i (i.e. the eigenvalues) might not all be positive. Clarification of the question: When we think of A as a quadratic form, we consider the map x \mapsto x^TAx rather than the usual linear transformation x \mapsto Ax. An important distinction here is that when we transform A as a quadratic form, we get a new matrix S^TAS (for some invertible matrix S). When we transform A as a linear transformation, we get a new matrix S^{-1}AS (for some invertible matrix S). If we can choose S to be orthogonal, then S^T = S^{-1}, so that we can simultaneously transform it both as a linear map and as a quadratic form. In this case, we're looking for an orthogonal matrix O so that the matrix$$ OAO^T = OAO^{-1} $$is a diagonal matrix (consisting of the eigenvalues of A). • hmm...so, for your discussion on part a), it is really just the usual diagonalization problem, and I just have to find my orthonormal matrix that diagonalizes A? What about that diagonal form thing? It's just a linear a combination of terms...how would I check that my matrix indeed transforms A into that diagonal form? and regarding b) positive definite => positive eigenvalues and positive determinant. how would checking this information show whether the orthogonal transformation exists? Commented Dec 23, 2014 at 5:37 • I just answered your second question in an edit. A matrix is a diagonal form when it's diagonal. That is, we have$$ y^T \pmatrix{a_1\\&a_2\\&&a_3} y = a_1 y_1^2 + a_2 y_2^2 + a_3 y_3^2 $$and this function of the vector y can be represented by the diagonal matrix$$ \pmatrix{a_1\\&a_2\\&&a_3}$\$ Commented Dec 23, 2014 at 5:39
• Ok, got it. Thank you so much for demystifying this question for me, @Omnomnomnom. Have a great night :) Commented Dec 23, 2014 at 6:10
• Actually, @Omnomnomnom, I have one more quick question: I'm getting ready to write this up, and I was wondering...when I get the eigenvalues and eigenvectors for the first matrix, A, my set of eigenvectors is linearly independent. Now, should I used the Gram-Schmidt algorithm to turn this set into an orthogonal set, then scale the set of vectors to turn the set into an orthonormal set? Or is there a better way to find a orthonormal set of vectors? And, now that I'm writing this I'm realizing that - Gram Schmidt doesn't not preserve eigenvectors. The vectors that get created from it... Commented Dec 23, 2014 at 6:22
• Eigenvalues are numbers, not vectors, so it makes no sense, Lebron, to say eigenvalues are linearly independent. If a matrix is symmetric (as yours are) and its eigenvalues are distinct then its eigenvectors are guaranteed to be orthogonal, and no Gram-Schmidt is required. Of course we need an orthonormal set of eigenvectors; the hard case is the one in which the matrix has repeated eigenvalues. Do your matrices fall into this case? Commented Dec 23, 2014 at 6:26