Quick Linear Algebra Proof: Why is A^T nonsingular if A is nonsingular?

squaremeplz
Messages
114
Reaction score
0

Homework Statement



Prove that if A is nonsingular then A^T is nonsingular and

(A^T)^(-1) = (A^(-1))^T

Homework Equations



(AB)^T = (B^T)(A^T)



The Attempt at a Solution



Step 1: Multiply both sides by B^T

B^T * (A^T)^(-1) = B^T * (A^(-1))^T

B^T * (A^T)^(-1) = (A^(-1)*B)^T

(A^(-1)*B)^T = (A^(-1)*B)^T

I feel that my last step is wrong. Any suggestions would be helpful.

I was also noticing that if I take just the right side of the equation and do this

A^T*(A^(-1))^T = (A^(-1)*A)^T

= I^T = I

which suggests that the left side has to equal I if I multiply it by A^T as well so

A^T*(A^T)^(-1) = I

which makes sense just looking at it.

Thanks!
 
Last edited:
Physics news on Phys.org
Start from A*A^(-1)=I. Now take the transpose of both sides and use your relevant equation. Stare at it for a while and figure out what it means.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top