Does the basis of column space of a square matrix remains the same after squaring i.e is c(A)=c(AxA)?

Well I'm working on matrices, and this thing is bugging me , though according to me it should be true because: let there be a nxn matrix:(aij ; i=row number, j=column number)

a11 a12 a13.....a1n
a21 a22 a23......
a31 a32 a33......
.   .   .  ......
.   .   .  ......
an1 an2 an3......


then A^2 will be:

1             column 1                          column 2
a11*a11+a12*a21+a13*a31....+a1n*an1|a11*a12+a12*a22+a13*a32....+a1n*an2|
a21*a11+a22*a21+a23*a31....+an2*an1|a21*a12+a22*a22+a23*a32....+an2*an2|

.                                  |.                                  |
.                                  |.                                  |
.                                  |.                                  |
an1*a11+an2*a21+an3*a31....+ann*an1|an1*a12+an2*a22+an3*a32....+ann*an2|


so if we take the row elements to be constants x,y,z,..... then all the columns are essentially x*column1+y*column2............ What do you all think?

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Your argument only shows the inclusion $c(A^2) \subset c(A)$. The opposite inclusion doesn't hold in general (as the answers you've already received show). – Hans Lundmark May 1 '11 at 16:45

If you have a nilpotent matrix where $N^2 = 0$ and $N$ is not the zero matrix,then the column space of $N$ is not the same as the column space of $N^2$.

For example, the matrix Jiangwei has written is a nilpotent matrix of size $2\times 2$.

Edit: check out http://en.wikipedia.org/wiki/Nilpotent_matrix for more details on Nilpotent matrices.

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Well then I think C(AxA)is a subset of c(A) – manik May 1 '11 at 16:44
column space of $ab$ is a subspace of the column space of $a$ for any compatible matrix $b$ because $ax$ is a linear combination of the columns of $a$ for a vector $x.$ – abel May 1 '11 at 20:38

In general, no. Take $A=\left(\begin{smallmatrix} 0 & 1 \\ 0 & 0 \end{smallmatrix}\right)$ for example.

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