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I need to calculate the following determinant in order to prove the following equality:

$$\det\begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{pmatrix} = (a^2+b^2+c^2+d^2)^2.$$

I tried using Gauss-algorithm to get an easier matrix, but I'm not sure if I did it correctly.

Calling the $4$ lines $I$, $II$, $III$ and $IV$, I did:

(1) $II \cdot a$

(2) $III \cdot a$

(3) $IV \cdot a$

After this I did:

(4) $II' + I \cdot b$

(5) $III' + I \cdot c$

(6) $IV' + I \cdot d$

So finally I got the following matrix:

$$\begin{pmatrix} a & b & c & d \\ 0 & a^2+b^2 & bc-ad & ac+bd \\ 0 & ad+bc & a^2+c^2 & cd-ab \\ 0 & bd-ac & ab-cd & a^2+d^2 \end{pmatrix}.$$

I thought this would make the determinant a bit easier, unfortunately I must have done something wrong. Is multiplication with single lines allowed as I have done it?

Thank you very much.

TMM
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Vazrael
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1 Answers1

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Notice that the rows are perpendicular to each other and the length of the rows are all the same.If you call the matrix $M$ then $\det(M^tM)=\det(M^t)\det(M)=\det(M)^2$ But $$ M^tM= (a^2+b^2+c^2+d^2)\left( \begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array} \right)$$ Therefor $\det(M^tM)=(a^2+b^2+c^2+d^2)^4\Rightarrow \det(M)=\pm(a^2+b^2+c^2+d^2)^2$

Set $N=\frac{1}{\sqrt{a^2+b^2+c^2+d^2}}M$ then all the rows of $N$ have length $1$. Further $N$ is completely determined by its first row. Now we have $\det(N)=\pm 1$ . Notice that the determinant gives a smooth function, $$\det:S^3\to \{-1,1\} $$ By sending $(a,b,c,d)$ to our matrix and then to the determinant of this matrix. (This is smooth because it is a polynomial function). But the point $(1,0,0,0)\in S^3$ is mapped to $1$. Now we can conclude that because $S^3$ is connected, $\{-1,1\}$ is discrete, and our function is continuous, that the entire $S^3$ is mapped to $1$. Thus $\det(N)=1$ and also $\det(M)=(a^2+b^2+c^2+d^2)^2$

Kaladin
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  • thank you very much, this looks great, however there're still things I don't understand yet. Why is $M^tM = (a^2+b^2+c^2+d^2) \begin{pmatrix} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 1 \end{pmatrix}$ ? I thought for orthogonal matrices (haven't heard of them yet, so they'll appear soon in lesson I guess) $M^t M = I$ ? – Vazrael Mar 10 '14 at 11:39
  • I think you've got it a bit wrong because $\det(M^tM) = (\det M)^2 = x^2 \Longrightarrow \det (M) = \pm x$. So why is $\det(M)$ positive here? – ElThor Mar 10 '14 at 11:42
  • I've have made an addition to show it has to be positive – Kaladin Mar 10 '14 at 12:29
  • Hello once again. What you've written above contains things that I don't have learned so far. Is there a different way, or can you explain me the things you wrote above (before editing your post!)? – Vazrael Mar 10 '14 at 13:34
  • It is probably a bit much to explain all. If this is indeed the beginning of the course they likely want you to do the brute calculation. Then mind the comment of Gerry Myerson – Kaladin Mar 10 '14 at 13:43
  • Okay, then I'll have to try my best ;-) I hope I will soon get an answer to why $M^t m = (a^2+b^2+c^2+d^2)...$ (see above). Thank you for your help and time that you spent :) – Vazrael Mar 10 '14 at 21:39