Trying to check a property of trace in matrices. Check please

Question

So here many websites says $$\operatorname{trace}(AB) =\operatorname{trace}(A)\cdot\operatorname{trace}(B).$$ But when I took two examples randomly and check it doesn't give equal answer. So it should be equal or not equal all the time? I am adding a picture to make it clear

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"So many websites". Examples? The equality is almost always false. — Martin Argerami, Jun 07 '23 at 02:00
The formula is not true in general. On the right side, only diagonal elements of $A$ and $B$ play a role while on the left side also non-diagonal elements of $A$ and $B$ can appear. Could you give an example, where you have read, that this formula is true? — Samuel Adrian Antz, Jun 07 '23 at 02:00
Your supposed identity is false. Similar statements that are true are $|AB|=|A| |B|$ and $\mathbb{trace}(AB)=\mathbb{trace}(BA)$. — Eric, Jun 07 '23 at 02:07
You may be mixed up with the determinant. It's true that det(AB)=det(A)det(B). The det is the product of the eigenvalues, whereas the trace is the sum. — rikhavshah, Jun 07 '23 at 02:07

Juliamisto · Accepted Answer · 2023-06-07T02:57:01.433

2

The trace of $AB$, in general, is not equal to the product of the trace of $A$ and the trace of $B$, if you mean $a_{1,1} + a_{2,2} + \dots + a_{n,n}$ by "the trace of $A = [a_{i,j}]_{n \times n}$". However, it is true that the trace of $AB$ equals that of $BA$.

I would like to list some facts for your reference:

The determinant of $AB$ equals the product of the determinant of $A$ and the determinant of $B$ (in which $A$ and $B$ are $n \times n$ matrices). Hence the determinant of $AB$ equals that of $BA$, since the multiplication of numbers is indeed commutative.
The trace of $A + B$ equals the sum of the trace of $A$ and the trace of $B$. The trace of $A + B$ equals that of $B + A$, by the way.
The determinant of $A + B$, in general, is not equal to the sum of the determinant of $A$ and the determinant of $B$. The determinant of $A + B$, however, equals that of $B + A$.

edited Jun 07 '23 at 02:57

answered Jun 07 '23 at 02:16

Juliamisto

1,470

Multiplication of numbers is commutative? – Nawal Allu Arjun Jun 07 '23 at 02:30
Ohh got it. Yes numbers are multiplication commutative because in determinant it is just a number hut matrices are not multiplicative commutative – Nawal Allu Arjun Jun 07 '23 at 02:40
@NawalAlluArjun $\det {(AB)} = \det {(A)} \det {(B)} = \det {(B)} \det {(A)} = \det {(BA)}$. Yes, that is what I wanted to say. – Juliamisto Jun 07 '23 at 02:58
Thank you so much sir for this easiest explanation . I am learning all the things on my own with books after graduation – Nawal Allu Arjun Jun 07 '23 at 04:08
Can we connect this property with any equality? I couldn't recall but I guess Schwartz or minkowiski? – Nawal Allu Arjun Jun 07 '23 at 04:10
As I can't ask more questions right now, so i am posting another question here There are two matrices A and B and both of them aren't null Matrix so AB = O then determinant of A and B will be 0 (definitely) – Nawal Allu Arjun Jun 07 '23 at 04:23
@NawalAlluArjun You are right. If $A$, $B$ are two non-zero matrices such that $AB$ is zero, then $\det {(A)} = \det {(B)} = 0$. If, for example, $\det {(A)}$ were not $0$, then $A^{-1}$ would exist. Hence $B = (A^{-1} A) B = A^{-1} (AB) = 0$, a contradiction! – Juliamisto Jun 07 '23 at 04:38
@NawalAlluArjun Sorry, what property do you wish to connect with some equality? – Juliamisto Jun 07 '23 at 04:39
Minkowiski or Schwartz inequality? – Nawal Allu Arjun Jun 07 '23 at 05:27
@NawalAlluArjun Sorry, I do not know such kinds of connections. – Juliamisto Jun 07 '23 at 05:29
Is there any chat form where people can discuss more related to topics? Thanks – Nawal Allu Arjun Jun 07 '23 at 05:31
@NawalAlluArjun Do you think that the chat room fits your need? – Juliamisto Jun 07 '23 at 05:35
I don't know it fits or not, but I wanted to talk, involve in discussion. Because most of people on this site are very helpful. So inverse of A and B will not exist. – Nawal Allu Arjun Jun 07 '23 at 05:37
@NawalAlluArjun Yes, neither $A^{-1}$ nor $B^{-1}$ exists, since they both have been assumed to be nonzero matrices. "Just ask; do not ask to ask." People here are helpful, since this is a place in which thinking is highly encouraged. – Juliamisto Jun 07 '23 at 05:38

Trying to check a property of trace in matrices. Check please

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