While sitting in a determinant class, our Professor, while describing determinants, coined an observation which made me wonder. He stated that the determinant of a square matrix can be represented as a function in the following way -Let M be the set of all square matrices, then a function, say X, is defined from M → R which returns the determinant of any square matrix C ∈ M i.e. X : M → R which returns |M|As soon as he said this, I began to wonder about the injectivity, surjectivity and bijectivity of this function.I was told that for a given C ∈ M, there exists only one unique determinant | M | ∈ R . But the converse is not true according to me. This can be proved by the following argument -Let's say that | x 5 x x | = − 6On solving for x, we obtain two values i.e 2 and 3, therefore the matrix is not unique for the given determinant i.e. for one image (−6), there exist multiple preimages. Hence, according to me, the function so defined is not injective, and consequently not bijective as well.Now checking for surjectivity, we'll have to look for the equality of the codomain and the range of the function. In my opinion, for infinite square matrices, there will be infinite unique determinants belonging to R . Hence the function should be surjective, but only for real entries into the matrix.For complex entries, I don't think the function remains surjective, as the range will not coincide with the codomain (i.e. R ). However, in both cases, I feel that the function is neither injective not bijective.

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While sitting in a determinant class, our Professor, while describing determinants, coined an observation which made me wonder. He stated that the determinant of a square matrix can be represented as a function in the following way -Let M be the set of all square matrices, then a function, say X, is defined from   M  →      R   which returns the determinant of any square matrix   C  ∈  M i.e.  X  :  M  →      R  which returns |M|As soon as he said this, I began to wonder about the injectivity, surjectivity and bijectivity of this function.I was told that for a given   C  ∈  M, there exists only one unique determinant       |    M      |    ∈      R  . But the converse is not true according to me. This can be proved by the following argument -Let&#039;s say that      |                            x                          5                                      x                          x                      |    =  −  6On solving for x, we obtain two values i.e 2 and 3, therefore the matrix is not unique for the given determinant i.e. for one image (−6), there exist multiple preimages. Hence, according to me, the function so defined is not injective, and consequently not bijective as well.Now checking for surjectivity, we&#039;ll have to look for the equality of the codomain and the range of the function. In my opinion, for infinite square matrices, there will be infinite unique determinants belonging to       R  . Hence the function should be surjective, but only for real entries into the matrix.For complex entries, I don&#039;t think the function remains surjective, as the range will not coincide with the codomain (i.e.       R  ). However, in both cases, I feel that the function is neither injective not bijective.

Raul Garrett · Accepted Answer

You are right in saying that the function is not injective. To show surjectivity, note that for any real       k                  1        n              I has determinant k (Where I is the   n  ×  n identity matrix)

Rishi Hale · Accepted Answer

Clearly there are infite matrices with   det  (  A  )  =  0. (Just scale any singular matrix to get an infinite number). Therefore you are right about injectivity and bijectivity. About surjectivity, you must have a matrix A for every   t  ∈      R  , that has exactly that value das determinant.You can just use a   1  ×  1 matrix, to set that determinant.This holds also true for any complex values

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