Some basic matrix proofs

**paupsers** · June 16th 2009, 03:46 PM

Can anyone help me solve these? I'm totally stuck...

1. Let A be invertible. Prove that the transpose of A (denoted $A^t$ )
is invertible and $(A^t)^{-1}=(A^{-1})^t$

2. Prove that if A is invertible and $AB=0$ , then $B=0$ .

Any help on these?

**Krizalid** · June 16th 2009, 04:06 PM

$\left( A^{-1} \right)^{t}\cdot A^{t}=\left( A\cdot A^{-1} \right)^{t}=I_{n}.$

Thus $\left( A^{-1} \right)^{t}\cdot A^{t}\cdot \left( A^{t} \right)^{-1}=I_{n}\cdot \left( A^{t} \right)^{-1}\implies \left( A^{-1} \right)^{t}=\left( A^{t} \right)^{-1}.\quad\blacksquare$

**Random Variable** · June 16th 2009, 04:13 PM

$AB = 0$

$A^{-1} AB = A^{-1} 0 = 0$

$(A^{-1}A)B = 0$

$B = 0$

I have no idea why it's making everything lowercase.

EDIT: I guess it was an issue with Firefox.

**putnam120** · June 16th 2009, 07:00 PM

You could also use the fact that $\det(A)=\det(A^t)$ .

**Krizalid** · June 16th 2009, 07:08 PM

as far as paupsers knows, since these are basic proofs then the use of determinants wouldn't be valid, though he(she) may clarify this.

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