2
$\begingroup$

P and Q are non-zero $3$X$3$ matrices and satisfy the equation

$(PQ)^T+Q^{-1}P=0$

(i) Prove that if Q is orthogonal, then, P is skew symmetric.

(ii) Without assuming Q is orthogonal, prove that P is singular.

Part (i) is correct for me. Using cyclic properties of Transpose matrix, $(PQ)^T=Q^TP^T$.

Then, since Q is orthogonal,$Q^T=Q^{-1}$.

Using associative law for matrix operation,

$Q^{-1}[P^T+P]=0$

Pre- multiplying by Q on both sides ends up giving $P=-P^T$

For part (ii), we need to show $det(P)=0$ for singular matrices

$|Q^TP^T|+|Q^{-1}P|=|0|$

Using properties that $|Q^{-1}|=\frac{1}{|Q|}$

We end up with $|Q||P^T|+\frac{1}{|Q|}|P|=0$

How to show $|P|=0$ from here.

$\endgroup$
0
$\begingroup$

$$Q^tP^t+Q^{-1}P=0\Rightarrow Q^tP^t=-Q^{-1}P\Rightarrow QQ^tP^t=-P$$

Witch means $$\tag 1 det(QQ^t)det(P^t)=-det(P)$$

Now $det(QQ^t)$ is always positive (since $QQ^t$ is positive definite, because Q is invertible). So we must have $det(P)=0$ for equation (1) to hold.

PS: I think the equation $|Q^TP^T|+|Q^{-1}P|=|0|$ you used in your question is incorrect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.