There is a $2\times2$ matrix $A$ composed of reals $a$ and $b$ such that $\det(A) = a^2 + b^2$, namely

$$A=\begin{bmatrix} a & -b \\ b & a \\ \end{bmatrix}.$$

It would seem there is not an analogy for $3\times3$ matrices. That is, I don't think there is a matrix $A$ composed of three reals $a$, $b$, and $c$ such that $\det(A) = a^3 + b^3 + c^3$.

However, matrices with even dimensions give an opportunity for cross-terms to cancel. Is there then, for example, a $4\times4$ matrix $A$ composed of reals $a$, $b$, $c$ and $d$, such that $$\det(A) = a^4+b^4+c^4+d^4$$

Formalized statement: For all reals $a$, $b$, $c$ and $d$, I am looking for a $4\times4$ matrix $A$ such that $A_{ij} = \{\pm a,\pm b,\pm c,\pm d\}$ and $\det(A) = a^4+b^4+c^4+d^4$. Is there such a matrix?

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    $\begingroup$ $\begin{bmatrix} a^4+b^4+c^4+d^4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$ probably doesn't work for you, but... $\endgroup$ – Watson Feb 3 '18 at 16:28
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    $\begingroup$ Or $\begin{bmatrix} a^4 & -1 & -1 & -1 \\ b^4 & 1 & 0 & 0 \\ c^4 & 0 & 1 & 0 \\ d^4 & 0 & 0 & 1 \end{bmatrix}$. $\endgroup$ – alex.jordan Feb 3 '18 at 16:32
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    $\begingroup$ You should maybe ask about some $A \in M_4(\Bbb R[a,b,c,d])$ such that for any coefficient $A_{ij}$, at most one of the variables $a,b,c$ or $d$ appears in $A_{ij}$ with non-zero coefficients (or even that $A_{ij} \in \{\pm a, \pm b,\pm c,\pm d\}$ for any $i,j$). $\endgroup$ – Watson Feb 3 '18 at 16:35
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    $\begingroup$ The quaternion $a^2 + b^2i + c^2j + d^2k$ has norm $a^4 + b^4 + c^4 + d^4$ and the norm can be found as the determinant of the matrix obtained by mutiplying this quaternion by each element of the basis $\{1, i, j, k\}$. $\endgroup$ – Tob Ernack Feb 3 '18 at 16:41
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    $\begingroup$ @Doubt The most close to a demanded matrix which generates highly symmetrical form of determinant it seems to be a quaternion matrix en.wikipedia.org/wiki/Quaternion#Matrix_representations ... but its determinant is, as they write in Wikipedia, the forth power of quaternion norm i.e. $(a^2+b^2+c^2+d^2)^2$ $\endgroup$ – Widawensen Feb 4 '18 at 14:47

To start with the answer notice that expression $a^4+ b^4+ c^4+d^4$ requires $a,b,c,d$ to appear in the matrix exactly four times in such a way that they not repeating themselves in every row and every column (this follows from the construction of determinant where entries from different rows and columns are taken for factors in products)

Let be presented such example of matrix with the preservation of entries indexes (what is important for the following reasoning) and neglecting of signs.

$ F= \begin{bmatrix}a_{11} & b_{12} & c_{13} & d_{14}\\d_{21} & a_{22} & b_{23} & c_{24}\\c_{31} & d_{32} & a_{33} & b_{34}\\b_{41} & c_{42} & d_{43} & a_{44}\end{bmatrix}$ so $x_{ij}$ are equal to themselves up to the sign..

The matrix $F$ was constructed in such a way that with increased number of row the entries are shifted horizontally by shift $+1$.

Calculating determinant we have

$\det(F)= \\ \begin{equation*}a_{11} a_{22} a_{33} a_{44} - a_{11} a_{22} b_{34} d_{43} - a_{11} a_{33} c_{24} c_{42} - a_{11} a_{44} b_{23} d_{32} + a_{11} b_{23} b_{34} c_{42} + a_{11} c_{24} d_{32} d_{43} - a_{22} a_{33} b_{41} d_{14} - a_{22} a_{44} c_{13} c_{31} + a_{22} b_{34} b_{41} c_{13} + a_{22} c_{31} d_{14} d_{43} - a_{33} a_{44} b_{12} d_{21} + a_{33} b_{12} b_{41} c_{24} + a_{33} c_{42} d_{14} d_{21} + a_{44} b_{12} b_{23} c_{31} + a_{44} c_{13} d_{21} d_{32} - b_{12} b_{23} b_{34} b_{41} + b_{12} b_{34} d_{21} d_{43} - b_{12} c_{24} c_{31} d_{43} + b_{23} b_{41} d_{14} d_{32} - b_{23} c_{31} c_{42} d_{14} - b_{34} c_{13} c_{42} d_{21} - b_{41} c_{13} c_{24} d_{32} + c_{13} c_{24} c_{31} c_{42} - d_{14} d_{21} d_{32} d_{43}\end{equation*}$

Some things are visible in this form:
we have here sum of $24$ products and if we neglect indexes we see what kinds of products they are:

  • four products type of $a^4$ (namely $a^4, b^4, c^4,d^4$)
  • four products type of $a^2c^2$ (namely two $a^2c^2$ and two $b^2d^2$)
  • 16 products type of $a^2bd$ (namely four $ a^2bd$, four $ b^2ac$, four $ c^2bd$, four $ d^2ac$)

I retain the names of these types for the rest of the answer.

Now look at the sign of products type $a^4$ - in order to have four $+$ signs for fourth powers ( I mean $+a^4, +b^4, +c^4,+d^4$) we need two numbers from the set $(a_{11},b_{12},c_{13},d_{14})$ to have minus signs - let it be $b_{12}$ and $d_{14}$.

Indeed the matrix $G=\begin{bmatrix}a & -b & c & -d\\d & a & b & c\\c & d & a & b\\b & c & d & a\end{bmatrix}$

has determinant $\det(G)= \ a^{4} - 2 a^{2} c^{2} + b^{4} - 2 b^{2} d^{2} + c^{4} + d^{4} $.

It has almost desired form of determinant ( happily all products of type $a^2bd$ have annihilated each other), but still products of type $a^2c^2$ are present.

Notice now that from the general form of the determinant it is visible that any manipulation of signs $a_{ij},b_{ij},c_{ij},d_{ij}$ will not annihilate them as in order to preserve $a^4+ b^4+ c^4+d^4$ only even number of changing the signs is allowable, but for $- a_{11} a_{33} c_{24} c_{42} - a_{22} a_{44} c_{13} c_{31}$ such even changing can't make the signs of listed products to have opposite signs.

So we will not get rid of expressions type $a^2c^2$ from the final form of determinant.

Of course starting from the matrix presented above the rows or columns can be permuted without substantial change of the form of determinant for new matrix what gives quite large number of possible matrices of that particular family which can be named "shift $+1$ in rows".

However presented type of matrix is not the only one (the whole list of such matrix types I added below as 'additional remarks')

The other type would be with the shift in following rows equal $+2$ (with necessary additional appropriate exchange of entries inside blocks $2 \times 2$ of a matrix $ 4 \times 4$ according to the rule "one entry of type in a single row and column").

It leads to the form known from quaternions (see link in wikipedia) where determinant has form $(a^2+b^2+c^2+d^2)^2$, here also besides desired $a^4+ b^4+c^4+d^4$ we have products of the type $a^2c^2$ which are irreducible by manipulations with signs of entries.

Hence the above reasoning leads to the conclusion that there is no matrix as stated in requirements of the question in at least these two structures.

Additional remarks about types of matrices with expression $a^4+ b^4+c^4+d^4$ in the determinant.

It seems that there are four types of such matrices up to the row/column permutation and signs of entries. Let '$a$' be a value which we call "leading" in the set of four allowable values (the signs we neglect). We can always permute columns in the matrix that values '$a$' will be located on the main diagonal.
Other entries in the first row denote as $b,c,d$.
So the starting form of a matrix looks as

$ \begin{bmatrix}a & b & c & d\\*& a & * & *\\* & * & a & *\\* & * & * & a\end{bmatrix}$

It can be shown that position 'b' in the second row determines the structure of the whole matrix in two cases and in the third generates two additional types. Therefore we can denote these types as $(b_{23}), (b_{24}), (b_{21},c_{31}), (b_{21}, c_{41})$. You can check that it is really the case: starting from the initial conditions and preserving the rule "in one column/row only a single type of value can appear" (it is some kind of independent exercise to put missing values into the structure starting from the initial conditions)

Look closer at these types:

  • type $(b_{23})$ which has initial form $ \begin{bmatrix}a & b & c & d\\*& a & b & *\\* & * & a & *\\* & * & * & a\end{bmatrix}$ generates unambigiously matrix $\begin{bmatrix}a & b & c & d\\d & a & b & c\\c & d & a & b\\b & c & d & a\end{bmatrix}$ which was considered above.

    A matrix with its structure is alike circulant matrix.

  • type $(b_{21},c_{31})$ generates $\begin{bmatrix}a & b & c & d\\b & a & d & c\\c & d & a & b\\d & c & b & a\end{bmatrix}$ which resembles quaternion matrix, blocks $ 2 \times 2$ are visible as substructures in this matrix.
  • type $(b_{21},c_{41})$ generates $\begin{bmatrix}a & b & c & d\\b & a & d & c\\d & c & a & b\\c & d & b & a\end{bmatrix}$ which is a little deviated from quaternion matrix, call it quasi-quaternion matrix..
  • finally type $(b_{24})$ generates $\begin{bmatrix}a & b & c & d\\c & a & d & b\\b & d & a & c\\d & c & b & a\end{bmatrix}$ which has the structure of centrosymmetric matrix.

From these $4$ types with permutations and changing the signs all possible matrices can be generated.
Now it is sufficient to check that expressions of type $a^2c^2$ are present in the determinant form, what leads to the conclusion similar to that for the circulant type considered above.

  • $\begingroup$ @Doubt I've extended a little my answer, It becomes a small research paper, I hope you will find in it some useful clues..anyway try to calculate determinants for specific 4 types of matrices and investigate them... $\endgroup$ – Widawensen Feb 9 '18 at 11:47
  • $\begingroup$ Your effort is well worth an upvote. $\endgroup$ – Piquito Feb 9 '18 at 11:51
  • $\begingroup$ @Piquito Thank you, at last I know that someone has read my answer :) $\endgroup$ – Widawensen Feb 9 '18 at 11:59
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    $\begingroup$ Finally I've checked (having used SymPy) all types of matrices for determinant. As we can generate always determinant with the expression $a^4+b^4+c^4+d^4$ at the same time in all determinants also products of type $a^2c^2$ are present. $\endgroup$ – Widawensen Feb 9 '18 at 12:15
  • $\begingroup$ @Widawensen Now two persons have read your answer. +1, thank you for contributing so much effort $\endgroup$ – Ewan Delanoy Feb 9 '18 at 14:39

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