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Show that if $a,b,c,d \in \mathbb R$ and at least one is different from $0$, then

$$\begin{vmatrix} a & b & c & d\\ b & -a & d &-c \\ c & -d & -a &b \\ d & c & -b &-a \end{vmatrix} \neq 0$$

Ghost
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    What have you done so far? It seems straightforward... –  Nov 13 '16 at 20:30
  • I do not even know how to start XD. – Ghost Nov 13 '16 at 20:30
  • Do you know how to compute the determinant of a $4 \times 4$ matrix by expanding along a row or column? If so, just do it! If not... well, then I don't think you're really in a position to be solving the problem. – pjs36 Nov 13 '16 at 20:51
  • I know how to do it but is it not pointless? – Ghost Nov 13 '16 at 20:55
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    @OvyOvy A more clever solution is using quaternions, see the proof reference of my answer. But still just computing it is not pointless, I think. – Dietrich Burde Nov 13 '16 at 21:18

4 Answers4

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This is a well-known determinant arising from quaternions. It has the value $$ -(a^2+b^2+c^2+d^2)^2. $$ Over the real numbers the value can only be zero if all of $a,b,c,d$ are zero.

The quaternion algebra $\mathbb{H}$ can be represented as the set of the above $4\times 4$ matrices with the $4$ real parameters $a,b,c,d$.

The determinant has been already calculated by Dickson in A Matrix Defined by the Quaternion Group, page $245$, in the year $1902$ (three rows differ by a factor of $-1$, so the result there is $+(a^2+b^2+c^2+d^2)^2$).

Dietrich Burde
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There are several references you might find useful.

  • Determinant of block matrices. In particular, for $2\times 2$ real matrices $A,B,C,D$, the following formula holds WHEN $CD=DC$, $$ {\displaystyle \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(AD-BC).} $$ Here is a proof of this formula.

  • The matrix representation of quaternions.

[Added] To elaborate Artem's answer, here is another way to do it.

Denote the matrix in your question as $$ A=\left(\begin{matrix} a & b & c & d\\ b & -a & d &-c \\ c & -d & -a &b \\ d & c & -b &-a \end{matrix}\right). $$ Then $$ AA^t= \left(\begin{matrix} a^2+b^2+c^2+d^2 & 0 & 0 & 0\\ 0 & a^2+b^2+c^2+d^2 & 0 & 0 \\ 0 & 0 & a^2+b^2+c^2+d^2 & 0 \\ 0 & 0 & 0 &a^2+b^2+c^2+d^2 \end{matrix}\right). $$ Note that $$ \det(A)^2=\det(A)\det(A^t). $$

3

You do not need any references or any remotely complicated calculations to show what is required. The matrix is indeed related to quaternion representation, but it is not the discussed representation.

First, note that if you factor out $-1$ from 3 last rows of your matrix you will get the skew-symmetric matrix $$ (-1)\begin{vmatrix} a & b & c & d\\ -b & a & -d &c \\ -c & d & a &-b \\ -d & -c & b &a \end{vmatrix}. $$
Now this is the representation two other answers refer to. Take this matrix and multiply by its transpose. You will find the diagonal matrix with all four elements on the main diagonal $(a^2+b^2+c^2+d^2)$. Now conclude...

Artem
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I like the Artem methode but this is a direct methode :

We can assume that $a\neq0$ : \begin{eqnarray} \begin{vmatrix} a & b & c & d\\ b & -a & d &-c \\ c & -d & -a &b \\ d & c & -b &-a \end{vmatrix} &=&\frac{1}{a}\begin{vmatrix} a^2 & b & c & d\\ ab & -a & d &-c \\ ac & -d & -a &b \\ ad & c & -b &-a \end{vmatrix} \\ &=& \frac{1}{a}\begin{vmatrix} a^2 +b^2+c^2+d^2 & b & c & d\\ (ab-ab+dc-cd) & -a & d &-c \\ (ac-bd-ac+bd) & -d & -a &b \\ (ad+cb-bc-ad) & c & -b &-a \end{vmatrix}((C_1)+b(C_2)+c(C_3)+d(C_4)\to (C_1) ) \\ &=& \frac{1}{a}\begin{vmatrix} a^2 +b^2+c^2+d^2 & b & c & d\\ 0 & -a & d &-c \\ 0 & -d & -a &b \\ 0 & c & -b &-a \end{vmatrix} \\ &=& \frac{a^2 +b^2+c^2+d^2}{a}\begin{vmatrix} -a & d &-c \\ -d & -a &b \\ c & -b &-a \end{vmatrix} \\ &=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2)-d(ad-bc)-c(db+ac) \right)\\ &=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2)-ad^2+dbc-cdb-ac^2 \right)\\ &=&\frac{a^2 +b^2+c^2+d^2}{a}\left(-a(a^2+b^2+b^2+c^2) \right)\\ &=& -(a^2+b^2+c^2+d^2)^2 \end{eqnarray}

Hamza
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