Algebra question??

[uther4117]

Hello,

I've been trying to solve the 2 problems i'm about to ask for hours. I've make search on internet, read all the related topics on the book but still. As today is holiday here(china) i decided to ask for help


1) Let A be an nxn matrix. Show that if A²=0, then I-A is nonsingular and (I-A)¯1 = I+A. ((I-A) over minus one but couldn't found the 1 on over )

In this one nonsingular means there is multiplicative inverse for it, cool but i'm new to these all proving thing and there is really not much in book about it. I need to make use of rules of multiplicative inverse but how is the question here

2)Let A be an mxn matrix. Show that A(over T).A and A.A(over T) are both symetric.

In this one A.AT should be mxm matrix and the other is nxn but how to prove they are symetric? It dimension needs to be square for sure but...

Thanks...

 

[HallsofIvy]

First, do NOT think of "\(\displaystyle (I+ A)^{-1}\)" as \(\displaystyle \frac{1}{I+ A}\)! Since multiplication of matrices is not commutative, \(\displaystyle A^{-1}B\) and \(\displaystyle BA^{-1}\) are not, in general, the same though, if we were using \(\displaystyle \frac{1}{A}\) for \(\displaystyle A^{-1}\) we would be tempted to write both as \(\displaystyle \frac{B}{A}\). Do NOT think of "multiply by the inverse matrix" as "divide by the matrix". In fact, in working with matrices, we do not talk about "dividing" at all.

Now, you want to show that "if \(\displaystyle A^2= 0\) then I- A is non-singular". Look at \(\displaystyle (I- A)(I+ A)= I- AI+ IA- A^2= I\).

(Notice that because matrix multiplication is NOT commutative, (B- A)(B+ A)= B^2-AB+ BA- A^2 is NOT in general equal to "\(\displaystyle A^2- B^2\)" but here, with B= I, IA= AI= A so IA- AI= 0.)

Now, what is the multiplicative inverse of I- A?