# When is it true that $\lambda_{\min} \ge \det(A)$ if $A$ is a symmetric positive semi-definite matrix?

In one of my textbooks, there is this inequality for square matrices larger than $2 \times 2$: $$\tag{0} \lambda_{\min} \ge \det(A)$$

This inequality is written in the context of autocorrelation matrices, so I suppose that $A$ is symmetric and positive semi-definite. But even with these constraints, this inequality doesn't seem to be always true, as I can give a counterexample: $$\left[\matrix{2 && 0 && 0 \\ 0 && 2 && 0 \\ 0 && 0 && 2}\right]$$ This matrix has eigenvalues of 2, and determinant of 8, so the inequality is false.

Is this inequality well known for some kind of matrices? Or maybe it is true when all eigenvalues are less than one (I'm just guessing this...)?

$(0)$ is part of the proof for this inequality: $$\tag{1}\frac{\lambda_\max}{\lambda_\min} \le \frac{\mathrm{Tr}(A)}{\det(A)}.$$

And the proof is: $$\tag{2}\lambda_\max \le \mathrm{Tr}(A)$$ and the inequality in question: $$\tag{3} \lambda_{\min} \ge \det(A)$$

$(2)$ is true (I think), but $(3)$ is not always true, as I have given a counterexample. So maybe, $(1)$ isn't true either?

• Could you cite the place where you found this statement? What textbook? What page? What else do you know about the matrix? Jun 13 '18 at 20:53
• The determinant is the product of the eigenvalues, so if you know that all the eigenvalues are between $0$ and $1$, the statement is true. Jun 13 '18 at 20:56
• If you take $aI_n$ with $a \in (0, 1)$, then $\lambda_{min} = a > \det(A) = a^n.$ Jun 13 '18 at 20:56
• @BrianBorchers: unfortunately, it is in a non-english offline book. The inequality doesn't have any condition attached, it is just said "a matrix with N>2", this is true: <inequality here>. But I suppose it is symmetric positive semi definite, as the chapter is about using autocorrelation matrices.
– geza
Jun 13 '18 at 20:58
• @saulspatz: thanks, so my guess was right, thanks for proving it!
– geza
Jun 13 '18 at 21:00

A symmetric positive semi-definite real matrix is a correlation matrix if, and only if, all of its diagonal entries are equal to $1.$ Thus your proposed counterexample fails if it was assumed that it's a correlation matrix.

(I'm going to write a more leisurely answer and post it here tonight or tomorrow.)

• Hmm. Maybe I need to retract the accept, I was too fast. Your answer tells me, why my example is bad, but the answer doesn't say, what is the exact condition. Is there an easy proof that for autocorrelation matrices, the inequality always holds? Maybe I don't see something trivial. Of course, if all eigenvalues are less than or equal to 1, the inequality holds. But is this always true for autocorrelation matrices?
– geza
Jun 13 '18 at 21:35
• @geza : Can you give us a more complete statement of what was assumed? Jun 13 '18 at 22:16
• Unfortunately, absolutely nothing except N>2. Assumptions can be made because it is in a chapter which is around autocorrelation. This is not from a math book, but from a neural network book, so it is not that rigorous. Note, that I understand the possible reason of the inequality now (because of your help, thanks a bunch!), so my questions in the comment are just curiosity (but if we take it seriously, gimusi's answer is more correct, as it gives the exact condition, even if it is trivial for me now - but on the other hand, your answer was more helpful for me, that's why I accepted it)
– geza
Jun 13 '18 at 23:18
• It seems that the "autocorrelation" $E(xx^*)$ in signal processing looks more like a shifted version of covariance while the correlation in statistics is a scaled (standardized) covariance. The OP means the former.
– A.Γ.
Jun 14 '18 at 3:13
• I reopened the question, as Brian Borchers showed a matrix which has ones at the diagonal, and yet fails the inequality.
– geza
Jun 14 '18 at 21:33

Here's an example of an autocorrelation matrix that doesn't satisfy the proposed inequality (1).

$\frac{\lambda_{max}}{\lambda_{min}} \leq \frac{\mbox{tr}(A)}{\det(A)}$

>> C=[1 -0.70 0; -0.70 1 -0.70; 0 -0.70 1]
C =

1.00000  -0.70000   0.00000
-0.70000   1.00000  -0.70000
0.00000  -0.70000   1.00000

>> trace(C)/det(C)
ans =  150.00
>> lambda=eig(C)
lambda =

0.010051
1.000000
1.989949

>> lambda(3)/lambda(1)
ans =  197.99

• Thanks! I'll close this issue with this. Maybe, there is some other constrain (which is implied in the book, or I just don't see it) so that inequality is true, but the conclusion is that it is not true in general for autocorrelation matrices.
– geza
Jun 15 '18 at 3:03

Since $$\det(A)=\prod \lambda_i$$

we have that for $\lambda_{\min}=0$ the inequality is always true and for $\lambda_{\min}\neq 0$

$$\lambda_{\min} \ge \det(A)\iff \lambda_{\min}\ge\prod \lambda_i \iff 1\ge \frac{\prod \lambda_i}{\lambda_{\min}}$$

– geza
Jun 13 '18 at 21:21
• You are welcome! Bye
– user
Jun 13 '18 at 21:22
• How do you know that the product of the other eigenvalues is less than or equal to 1? Jun 13 '18 at 23:39
• @BrianBorchers it is the condition obtained and required to satisfy the inequality.
– user
Jun 14 '18 at 4:45

This is clearly not true for matrices that are symmetric and positive definite with diagonal elements that are one (the typical definition of an autocorrelation matrix.) Try for example:

>> A=[1 -0.5 0; -0.5 1 -0.4; 0 -0.4 1]
A =

1.00000  -0.50000   0.00000
-0.50000   1.00000  -0.40000
0.00000  -0.40000   1.00000

>> det(A)
ans =  0.59000
>> eig(A)
ans =

0.35969
1.00000
1.64031


Further restricting the matrix to be Toeplitz (as it would be if it were the autocorrelation matrix of a stationary stochastic process) doesn't help:

>> B=[1 -0.5 0; -0.5 1 -0.5; 0 -0.5 1]
B =

1.00000  -0.50000   0.00000
-0.50000   1.00000  -0.50000
0.00000  -0.50000   1.00000

>> eig(B)
ans =

0.29289
1.00000
1.70711

>> det(B)
ans =  0.50000
>>

• Thanks for the better counterexample! I think I'll reopen this question.
– geza
Jun 14 '18 at 21:31
• Either the author of the book is just wrong, or if you've misunderstood/misquoted what the author wrote, or there is some important context in the book that is missing. Until/unless you can provide more context, there's not going to be a positive answer to your question. Jun 14 '18 at 22:17