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There is vast area of physics where we have to use matrices.It is not only to do the mathematical problems in physics but also to produce a physical realization of an operation. I think matrices carry a huge amount of physics in symmetry operations. Again a matrix can be described by two numbers one is determinant and another one is trace. My question is what are the physical significances of DETERMINANT as well as TRACE?

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closed as too broad by Emilio Pisanty, stafusa, ACuriousMind Oct 23 '17 at 23:42

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Well, this is different on a case by case scenario. Matrices are indeed representations of operators (that are widely used in, say, Quantum Mechanics) but still this will be quite a narrow description of the why. $\endgroup$ – gented Oct 23 '17 at 15:23
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    $\begingroup$ A matrix certainly cannot be described only by its determinant and its trace. For instance $$ P_{12}=\left(\begin{array}{ccc} 0&1&0\\ 1&0&0\\ 0&0&1\end{array}\right)\, \qquad P_{13}= \left(\begin{array}{ccc} 0&0&1\\ 0&1&0\\ 1&0&0\end{array}\right) $$ have the same determinant and trace and are obviously distinct. $\endgroup$ – ZeroTheHero Oct 23 '17 at 15:24
  • $\begingroup$ My question is not about the use of matrices in Quantum Mechanics or somewhere. $\endgroup$ – DIPANJAN HAZRA Oct 23 '17 at 15:33
  • $\begingroup$ What is your question then? Your title is " Why the use of matrices is so important in physics". Moreover, you claim "Again a matrix can be described by two numbers one is determinant and another one is trace" which is incorrect. $\endgroup$ – ZeroTheHero Oct 23 '17 at 15:36
  • $\begingroup$ The trace and the determinant are but two of a number of quantities invariant under conjugation of a matrix by a unitary transformation, i.e. under a change of basis. This is hardly enough to completely specify a matrix. For instance in $3\times 3$ there is another invariant (see math.stackexchange.com/a/807183/160660) . $\endgroup$ – ZeroTheHero Oct 23 '17 at 19:28
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Well, there is not much to tell. The more physical you can get with determinant is the following: the determinant represents the "volume distorsion", which means it only tells you by how much your linear transformation will change the volume of a parallelogram. For instance, the matrix $2\mathbb{1}_{2\times2}$ tells you that the square of area 1 will be stretched to a square of area 4 (because $\mathrm{det}(2\mathbb{1}_{2\times2})=4$). More generally, this is true for any dimension, and also for every kind of linear transformation. Think of the jacobian when performing a coordinate transformation in an integral. The measure has to change since the coordinate transformation may change the volume of an infinitesimal n-dimensional parallelogram. Also, that is why we use in quantum mechanics SO(n) or SU(n) as symmetry groups; their determinant is 1 so that the volume is "conserved" when we rotate things.

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I can think about $2\times 2$ matrices to describe a set of two linear equations. These systems of two linear equations are extremely common in physics.

The trace and the determinant are strongly related to th eigenvalues of this matrix, and their values determine wether the solution is stable, inestable, oscillating, and so on.

In particular, any 1D system of a second order differential equations (and those are equations of motion) can be seen as two first order linear differential equations together. In this case, knowing the trace and determinant allows you to check wheter a solution is an attractor, repeller, spiral... etc.

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Here is a useful relation between determinant and trace. The linear Lie Group transformations done in physics are of the form $M=e^\Theta$ where both $M$ and $\Theta$ are matrices. Then it is true that: $$det(M)=1 \quad if \ and \ only \ if \quad Trace(\Theta)=0$$ This is how the det=1 condition, signified by the "S" in SO(n), SU(n), and SL(n), turns into a constraint on the coordinates and generators of the group.

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