positive definite matrix

positive definite matrix

A positive definite matrix is a symmetric matrix with all positive eigenvalues. Note that as it’s a symmetric matrix all the eigenvalues are real, so it makes sense to talk about them being positive or negative.

Positive Definite Matrices - What Are They, and What Do They Want?

A matrix is positive definite if it’s symmetric and all its eigenvalues are positive.

The thing is, there are a lot of other equivalent ways to define a positive definite matrix. One equivalent definition can be derived using the fact that for a symmetric matrix the signs of the pivots are the signs of the eigenvalues. So, for example, if a 4 × 4 matrix has three positive pivots and one negative pivot, it will have three positive eigenvalues and one negative eigenvalue. We can apply this fact to positive definite matrices to derive the next equivalent definition.

A matrix is positive definite if it’s symmetric and all its pivots are positive.

Pivots are, in general, way easier to calculate than eigenvalues. Just perform elimination and examine the diagonal terms. No problem. In practice this is usually the way you’d like to do it. For example, in that matrix from the introduction:

$$\left[ \begin{matrix} 1 & 2 \\ 2 & 1\\ \end{matrix} \right]$$

If we perform elimination (subtract 2× row 1 from row 2) we get

$$\left[ \begin{matrix} 1&2\\ 0&-3 \end{matrix} \right]$$

The pivots are 1 and −3. In particular, one of the pivots is −3, and so the matrix is not positive definite. Were we to calculate the eigenvalues we’d see they are 3 and −1.

Another way we can test for if a matrix is positive definite is we can look at its n upper left determinants.

$$A_k = \dfrac{det(A_k)}{det(A_{k-1})}$$

where A_k is the upper left k x k submatrix. All the pivots will be positive if and only if det(A_k) > 0 for all 1<=k<= n. So, if all upper left k x k determinants of a symmetric matrix are positive, the matrix is positive definite.

Example - Is the following matrix positive definite?

$$\left[ \begin{matrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{matrix}\right]$$

$$\left| \begin{matrix} 2 \end{matrix} \right| =2$$

$$\left|\begin{matrix} 2&-1\\ -1&2 \end{matrix}\right| =3$$

$$\left| \begin{matrix} 2&-1&0\\ -1&2&-1\\ 0&-1&2 \end{matrix}\right| =8-2-2=4$$
2,3,4>0 => 正定

If x is an eigenvector of A then x!=0 and Ax =$\lambda$ x. In this case x^TAx =$\lambda$x^Tx. If $\lambda$ > 0, then as x^Tx> 0 we must have x^TAx> 0.

The thing about positive definite matrices is x^TAx is always positive, for any non-zero vector x, not just for an eigenvector. In fact, this is an equivalent definition of a matrix being positive definite.

A matrix is positive definite if x^TAx > 0,for all vectors x != 0.

if A and B are positive definite then so is A + B.

Our final definition of positive definite is that a matrix A is positive definite if and only if it can be written as A = R^TR, where R is a matrix, possibly rectangular, with independent columns. Note that, using our energy-based definition, it’s easy to prove that if A = R^TR then A is positive definite.

$$x^TAx=x^TR^TRx=(Rx)^T(Rx)=||Rx||^2$$
If the columns of R are linearly independent then Rx!=0 if x!=0,and so $x^TAx>0$.

A matrix A is positive definite if and only if it can be written as $A=R^TR$ for some possibly rectangular matrix R with independent columns.

Note that we say a matrix is positive semidefinite if all of its eigenvalues are non-negative.

参考文献

1. https://www.math.utah.edu/~zwick/Classes/Fall2012_2270/Lectures/Lecture33_with_Examples.pdf