15

We have been given matrix: $$A = \begin{pmatrix} a& b& c &d \\ b &−a& d& −c\\ c& −d &−a& b \\ d &c& −b& −a\\ \end{pmatrix} $$
...and have been asked to calculate $\det(A)$ using $AA^T$.

We see that:

$$AA^T=\begin{pmatrix} a^2+ b^2+ c^2+ d^2& 0& 0&0\\ 0 &a^2+ b^2+ c^2+ d^2& 0& 0\\ 0& 0 &a^2+ b^2+ c^2+ d^2& b \\ 0 &0& 0& a^2+ b^2+ c^2+ d^2\\ \end{pmatrix} $$

So, $\det(AA^T)= (a^2+ b^2+ c^2+ d^2)^4$

Now which should I choose: $(a^2+ b^2+ c^2+ d^2)^2 $ or $ -(a^2+ b^2+ c^2+ d^2)^2$ ?

Please explain me which one and why.

apnorton
  • 17,929
  • 6
  • 55
  • 115
xarix
  • 183

1 Answers1

17

This might be an overkill but it is a useful observation nonetheless. You have a matrix $A = A(a,b,c,d)$ which depends on four real parameters and you computed $\det(A)^2$. From your computation, it is clear that $\det(A) = 0$ iff $a = b = c = d = 0$. Set $U = \mathbb{R}^4 \setminus \{ (0,0,0,0) \}$. If $(a,b,c,d) \in U$ then $\det(A) \neq 0$ so either $\det(A) > 0$ or $\det(A) < 0$. However, the set $U$ is connected and the function $U \rightarrow \mathbb{R}$ given by $(a,b,c,d) \mapsto \det(A(a,b,c,d))$ is continuous so we must have $\det(A) > 0$ for all $(a,b,c,d) \in U$ or $\det(A) < 0$ for all $(a,b,c,d) \in U$ and we can check which by plugging in specific values for $a,b,c,d$. For example, when $b = c = d = 0$ we have

$$ \det(A) = a(-a)^3 = -a^4 $$

so $\det(A) < 0$ if $a \neq 0$ and hence

$$ \det(A(a,b,c,d)) = -(a^2 + b^2 + c^2 + d^2)^2 $$

for all $(a,b,c,d) \in U$.

levap
  • 67,610
  • but can we do some observation using $AA^T$ ? – xarix Apr 04 '17 at 18:29
  • 2
    @xarix: Not really. If instead of your matrix, I would have given you the matrix $A = \begin{pmatrix} -a & b & c & d \ -b & -a & d & -c \ -c & -d & -a & b \ -d & c & -b & -a \end{pmatrix}$ and you would have calculated $AA^T$, you'd get the same result but $\det(A) = (a^2 + b^2 + c^2 + d^2)^2$. – levap Apr 04 '17 at 18:36
  • why with + when you showed that should be with - ? – xarix Apr 04 '17 at 18:41
  • @xarix: Because if you plug in $a = 1, b = 0, c = 0, d = 0$ you get a minus sign and the continuity argument shows that is must hold for all $a,b,c,d$. For the matrix I wrote in the comment above, you'd get a plus sign and the same continuity argument would show it continues to hold for all $a,b,c,d$. – levap Apr 04 '17 at 18:43
  • The result is true if the entries of the matrix are in an arbitrary field (not necessarily $\Bbb R$), but unfortunately I don't recall the general argument – Ivo Terek Apr 05 '17 at 08:39
  • @levap very nice answer actually I don't understand how the function is continuous and suppose it is continuous then how can we say that either $det(A)<0$ or $det(A) > 0$ can only happen ? – BAYMAX Apr 05 '17 at 12:29
  • 1
    The function is polynomial in $a,b,c,d$ so it is continuous. And since it is a continuous function on a connected set that is never zero, it must be always positive or negative. Think about the one-dimensional case where you have a continuous function $f$ on an interval $[a,b]$ and you know that $f(x) \neq 0$ for all $x \in [a,b]$. By the mean value theorem, we must have $f(x) > 0$ for all $x \in [a,b]$ or $f(x) < 0$ for all $x \in [a,b]$. – levap Apr 05 '17 at 17:33