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 March 3rd, 2012, 05:49 AM #1 Member   Joined: Jan 2012 Posts: 63 Thanks: 0 matrix invertible problem Let A be n*n matrix such that A^k=0(n,n) (the n*n zero matrix) for some natural integer k. show that In+A is invertible. attempt solutions: for A^k=0(n,n) then we have A=0 or A is a matrix not equal to 0 ,but for some k s.t. A^k=0 for case 1, if A=0 then In+A =In ,and det(In+A) not zero then it is invertible for case 2. if A is not 0 then i dont know how to argue this one.cause for some A is not zero , maybe det(In+A)=0 or maybe A is not zero but A^k is not zero as well can someone give me some helps
 March 3rd, 2012, 07:37 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: matrix invertible problem Not sure if this works, but I believe if $A^k= 0(n, n)$ then A must be of the form $\begin{bmatrix} 0 & 0 & 0 \\ a & 0 & 0 \\ b & c & 0 \end{bmatrix}$ (or the transpose), in which case $A^n= 0(n, n)$. Now, if you add $nI$ to this matrix, you get $\begin{bmatrix} n & 0 & 0 \\ a & n & 0 \\ b & c & n \end{bmatrix}$, the determinant of which is $n^n$, hence it is invertible.
 March 3rd, 2012, 12:55 PM #3 Member   Joined: Jul 2010 Posts: 44 Thanks: 0 Re: matrix invertible problem how do i delete this post?
 March 6th, 2012, 11:45 AM #4 Newbie   Joined: Dec 2011 Posts: 15 Thanks: 0 Re: matrix invertible problem Suppose $(I_n+A)x=0$ for some vector $x\neq 0$. So $Ax=-x$. This means that $\lambda=-1$ is an eigenvalue of A which is a contradiction since zero is the only eigenvalue of A.

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