How to find $(Ker(A^{*}))^{\perp}$ Let $$A = \begin{pmatrix} 1 & 1 & 1 & 1 \\ 1 & 0 & -2 & -1 \\ 1 & 2 & 4 & 3 \end{pmatrix}$$ Find a basis for $(Ker(A^{*}))^{\perp}$.
Find vectors $b_i$ such that $ y \perp b$ implies $Ax = y$ is solvable.
I know that $A^{*}$ is simply $A^T$ since $A$ is a real matrix.  I can find a basis for $Ker(A^{*})$ easily.  How does the $\perp$ operator change things?  Also, how do I find vectors $b_i$ such that $ y \perp b$ implies $Ax = y$ is solvable?
 A: Firstly, I don't know if for a matrix $B$ one usually defines $Ker(B):=\{v\in V\mid Bv=0\}$,
I have seen this only for linear operators (of course one can consider
a matrix as an operator defined by $v\mapsto Bv$ ), but I assume
this is the case here. 
You are right that since all entries are real then 
$$
A^{*}=A^{T}
$$
now, I assume you know how to find a basis for $Ker(B)$ when $B$
is given - in this case take $B=A^{*}$, say it is $\{v_{1},...v_{k}\}$
Now, we can complete this basis to a basis for all of $V$ and use
the Graham-Schmidt process on this basis - since when we apply this
process to 
$$
\{v_{1},...,v_{k}\}
$$
and get 
$$
\{u_{1},...,u_{k}\}
$$
with the same span it follows that $\{u_{1},...,u_{k}\}$ is a basis
for $Ker(A^{*})$ as well. 
Now - the other vectors in the complete basis for $V$ are all orthogonal
to the $u_{i}$ - hence orthogonal to any linear combination of them,
thus to all of $Ker(A^{*})$.
It is a theorem that 
$$
V=W\oplus W^{\perp}
$$
for every subspace $W$ and thus we conclude that the remaining vectors
are in fact a basis for $Ker(A^{*})^{\perp}$
I leave it to you to apply it to this example.
A: The "four subspace" theorem tells us that the orthogonal complement of the null space of $A^*$ is the range (column space) of $A$.  So for the first part, you need only find a basis for the column space of $A$.
If $y$ is orthogonal to every vector in $N(A^*)$, then $y \in R(A)$, so $Ax = y$ has a solution.  You can pick the vectors $b_i$ to form a basis for $N(A^*)$.
