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Prove that a positive definite matrix has positive determinant and positive trace.

In order to be a positive determinant the matrix must be regular and have pivots that are positive which is the definition. Its obvious that the determinant must be positive since that is what a positive definite is, so how can I prove that?

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  • $\begingroup$ What do you mean by regular? Are you talking about Sylvester's criterion? $\endgroup$
    – EuYu
    Oct 8, 2012 at 5:36
  • $\begingroup$ Nope, sorry for the confusion. By regular I mean they reduce to upper triangular form without any row interchange. $\endgroup$
    – diimension
    Oct 8, 2012 at 5:54
  • $\begingroup$ You should probably specify which of the equivalent definition of positive definite matrix you refer to $\endgroup$
    – enzotib
    Oct 8, 2012 at 5:59

4 Answers 4

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All eigenvalues of a positive definite matrix are real and positive.

The determinant is the product of the eigenvalues, hence real and positive.

The trace is the sum of the eigenvalues, hence real and positive.

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  • $\begingroup$ We haven't learned eigenvalues yet though . $\endgroup$
    – diimension
    Oct 8, 2012 at 5:53
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It's rather awkward to define positive-definiteness using "regularity" and pivots. In particular, I'm not even sure all positive-definite matrices (as normally defined) fit under this category. But for this particular proof, you can form an $LU$ decomposition in which $L$ is a lower triangular matrix composed with only row addition matrices. Then $\det(A) = \det(U)$. Since $U$ is upper triangular with the pivot entries of $A$ as it's diagonal entries, it follows that $\det(U)$ is a product of $A$'s pivots. But all of $A$'s pivots are positive, so it follows that $\det(U) = \det(A)$ is also positive.

The trace part is not even true. So the matrices as you've defined them cannot actually be positive-definite. The below matrix is "regular" and has positive pivots $$\begin{pmatrix}1 & -2 & 2 \\ 2 & -1 & 1 \\2 & 2 & -1\end{pmatrix} \rightarrow \begin{pmatrix}1 & -2 & 2 \\ 0 & 3 & -3 \\0 & 6 & -5\end{pmatrix}\rightarrow \begin{pmatrix}1 & -2 & 2 \\ 0 & 3 & -3 \\0 & 0 & 1\end{pmatrix}$$ Incidentally the trace is negative.

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    $\begingroup$ Thank you a lot !! One last question, how were you able to come up with that proof? Is it from experience, you know the math so well, or you thought it up yourself because this is my first semester doing proofs (i never learned it before) and I am struggling at it. Any pointers? $\endgroup$
    – diimension
    Oct 8, 2012 at 6:16
  • $\begingroup$ It's familiarity I guess. The definition of "regular" you used is typically a criterion to determine if a matrix has an $LU$ decomposition without pivoting. So that seemed like the natural direction to proceed. $\endgroup$
    – EuYu
    Oct 8, 2012 at 6:22
  • $\begingroup$ Thanks a lot ! But for the trace that you just edited is positive though since the diagonals add up to be positive. $\endgroup$
    – diimension
    Oct 8, 2012 at 6:26
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    $\begingroup$ I'm not sure what you mean. The trace of the matrix is $1 - 1 - 1 = -1$. $\endgroup$
    – EuYu
    Oct 8, 2012 at 6:27
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    $\begingroup$ Right. That becomes quite a different problem. The definition you gave is the characterization of positive-definite matrices using the cholesky decomposition. The determinant should just follow from the fact that $D$ has positive diagonal entries. I don't have an immediate proof for the trace right now. $\endgroup$
    – EuYu
    Oct 8, 2012 at 6:58
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One more proof to show that the determinant of a +ve definite matrix is +ve:

let A be a +ve definite matrix. Now we know that A can be written as :

A = $(V)D(V^T)$ ,which is the single value decomposition form of the +ve definite matrix A.Here $V^T$ and V are the orthogonal vectors such that $ (V^T)V $ = 1 .Thus we can also write :

$|V^T|$ |V| = 1 ---(1).

Also D is the diagonal matrix with eigen values of A as its diagonal elements.

Now, A = $(V)D(V^T)$ .

or $|A| = |V| |D| |V^T| $ .

or |A| = |D| (from (1) ) ---(2) .

|D| = product of eigen values ----(3).

Thus if we prove that all the eigen values of a +ve definite matrix are +ve then we are done. $(x^T)A(x)$ > 0 .

Ax=bx ,where b is the eigen value.

thus $(x^T)bx > 0 $.

$b(x^T)x > 0 $.

$(x^T)x > 0 $ (always) .

thus b>0 .

thus from (2) and (3) :

|A| = |D| > 0

or |A| > 0 .

hence proved.

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  • $\begingroup$ besides the trace is the sum of eigen values and we proved that the eigen values of a +ve definite matrix are +ve..so,the trace is also +ve.... $\endgroup$
    – under root
    May 10, 2013 at 7:26
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We will solve it by assuming a function which is +ve definite and then using continuity definition of ϵ-δ.

So,if you know ϵ-δ continuity definition of a function,then only consider this solution,otherwise skip it.

A is +ve definite.lets define a function :

f(x)= |xA + (1-x)I| for 0<=x<=1 -----(1), here I is the identity matrix.

clearly,f(x) is never 0.

Now we can easily see [xA+(1-x)I] > 0 for 0<=x<=1.

Also,f is continuous on our assumed interval of x.

let us choose a point b in [0,1].

then |x-b| < δ (where δ>0) (assume) -----(2).

now, |f(x)-f(b)| = |xA+(1-x)I - bA-(1-b)I| = |(x-b)A+(b-x)I|.

|f(x)-f(b)| = |(x-b)A+(b-x)I| <= |(x-b)A| +|(b-x)I| (triangle inequality).

|f(x)-f(b)| <= |(x-b)A| +|(b-x)I| < |x-b| |A| |b-x| |I|.

|f(x)-f(b)| <= |x-b| |A| |b-x| |I| < $δ^2 |A|$ (using (2) ) [note: |I| = 1].

|f(x)-f(b)| <$ δ^2 |A|$ ----(3).

Now,since f is continuous ,thus |A| > 0 is a must as then only the definition of continuity holds.

thus we get |A| > 0 . and then we can choose δ = $[ϵ/|A|]^{1/2}$ and get :

|f(x)-f(b)| < ϵ .

Hence proved |A| > 0.

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