1. ## linear algebra

Let A =
[ 3 0 1 ]
[-4 1 -2 ]
[ -4 0 -1 ] which is a 3x3 matrix. and let I be a 3x3 identity matrix.
Find a basis for the intersection of R(A-I) and N(A-I), where R(A-I) is the range of A-I and N(A-I) is the nullspace of A-I.

the solution says this basis is [1 -2 -2]^T (this T means transpose).
and the procedure is the following.

1. Find a basis {x1,x2,..,} for R(A-I)
2. Let X be the matrix that consists of the vectors in this basis for R(A-I).
3. Find a basis {v1,v2,....,}for N((A-I)X)
4. B={Xv1, Xv2,...,..} is the basis of the intersection.

i ve been trying to do this problem by following the procedure but i cant get the solution and i dont know what i am doing wrong. please help me.can please anyone explain and show me how to find this basis?

2. Looks to me like it is just a matter of doing the calculation.

$A- I= \left[\begin{array}{ccc}2 & 0 & 1 \\ -4 & 0 & -2 \\ -4 & 0 & -2\end{array}\right]$

The null space is the set of all (x, y, z) such that
$A- I= \left[\begin{array}{ccc}2 & 0 & 1 \\ -4 & 0 & -2 \\ -4 & 0 & -2\end{array}\right]\left[\begin{array}{c} x \\ y \\ z \end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]$
That means we must have 2x+ z= 0, -4x- 2z= 0, and -4x- 2z= 0, all of which reduce to the single equation z= -2x. Of course, y does not appear in that equation and can be anything. That tells us that any vector in the nullity is of the form <x, y, -2x>= x<1, 0, -2>+ y<0, 1, 0> so {<1, 0, -2>, <0, 1, 0>} is a basis for the null space.

Of course, any vector in the range is of the form
$A- I= \left[\begin{array}{ccc}2 & 0 & 1 \\ -4 & 0 & -2 \\ -4 & 0 & -2\end{array}\right]\left[\begin{array}{c} x \\ y // z \end{array}\right]= \left[\begin{array}{c}2x+ z \\ -4x- 2z \\ -4x- 2z\end{array}\right]$
or <2x+ z, -4x- 2z, -4x- 2z>= (2x+z)<1, -2, -2> for any numbers x, z so <1, -2, -2> is a basis for the range, as you say.

But <1, -2, -2>= <1, 0, -2>- 2<0, 1, 0> and so is itself in the null space. That means that the range is contained in the null space: the intersection of range and null space is just the range itself. A basis for the intersection is a basis for the range, <1, -2, -2>.

And, Since Ax is in the range which is in the null space of A, $A(Ax)= A^2x= 0$ for all x: $A^2= 0$.