Invertible Matrices and Rank 1 Matrices: Understanding Linear Transpose

[bananasplit]

Homework Statement

I have an idea on how to part 1, but I have no clue on how to do part 2 and 3.

1.Suppose A is invertible. Check that (A^-1)^TA^T=I and A^T(A-1)T=I, and deduce that AT is likewise invertible with inverse (A^-1)^T.

2. Suppose A is an mxn matrix with rank 1. Prove that there are nonzero vectors u element in Rm and v element in Rn such that A=uv^T.

3.Suppose A is an mxn matrix and x is an element of Rn satisfies (A^TA)x=0. Prove that AX=0.

Homework Equations

For part one I'm guessing (AB)^T=B^TA^T and (A^-1A^-1)=I

The Attempt at a Solution

Part 1. I know that some kind of way that it is due to the relationship of (A^-1A^-1)=I

 

[lanedance]

Homework Statement

I have an idea on how to part 1, but I have no clue on how to do part 2 and 3.

1.Suppose A is invertible. Check that (A^-1)^TA^T=I and A^T(A-1)T=I, and deduce that AT is likewise invertible with inverse (A^-1)^T.

2. Suppose A is an mxn matrix with rank 1. Prove that there are nonzero vectors u element in Rm and v element in Rn such that A=uv^T.

3.Suppose A is an mxn matrix and x is an element of Rn satisfies (A^TA)x=0. Prove that AX=0.

Homework Equations

For part one I'm guessing (AB)^T=B^TA^T and (A^-1A^-1)=I

The Attempt at a Solution

Part 1. I know that some kind of way that it is due to the relationship of (A^-1A^-1)=I

(A^-1A^-1) does not equal I


(A^-1.A) = (A.A^-1) = I

what is

(A^-1.A)^T ?

 

[bananasplit]

(A.A^-1) = I I am sorry I typed that wrong

 

[lanedance]

ok so ideas for 1) ?

for 2), what does it mean to be a matrix of rank 1? and uv^T looks like an outer product, do you know how this is defined? Thinking about row reduction may help make the connection...

for 3) think about multiplying both sides of your equation by something...