Proof relating to Orthogonal Complement 
a.) Show that if $A=A^T$ is a symmetric matrix, then $A\mathbf{x}=\mathbf{b}$ has a
  solution iff b is orthogonal to $\ker A$.
b.) Prove that if $K$ is a positive semi-definite matrix and $\mathbf{f}\notin
 \operatorname{rng}K$, then the quadratic function $$p(\mathbf{x}) = \mathbf{x}^\mathrm{T}K\mathbf{x} -2\mathbf{x}^\mathrm{T}\mathbf{f} + c$$ has no
  minimum value.
c.) Suppose $\{\mathbf{v_1},\ \cdots,\ \mathbf{v_n}\}$ span a subspace $V \subset \mathbb{R}^m$.
  Prove that $\mathbf{w}$ is orthogonal to $V$ iff $\mathbf{w}\in \operatorname{coker}A$ where
  $A=\begin{pmatrix}\mathbf{v_1} & \mathbf{v_2} & \cdots & \mathbf{v_n}\end{pmatrix}$ is the matrix with the indicated columns.

My attempt:
a.) We know that $A=A^\mathrm{T}$, then a vector $\mathbf{x}\in \mathbb{R}^n$ lies in $\ker A$ iff $A\mathbf{x} = \mathbf{0}$. 
By matrix multiplication we know that the $i^{\text{th}}$ entry of $A\mathbf{x}$ equals the vector product of the $i^{\text{th}}$ row $\mathbf{r_i}^T$ of $A$ and $\mathbf{x}$, hence $\mathbf{r_i}^{T}\cdot \mathbf{x} = \mathbf{r_i} \cdot  \mathbf{x} = 0$ iff $\mathbf{x}$ is orthogonal to $\mathbf{r_i}$. 
Therefore $\mathbf{x}\in \ker A$ iff $\mathbf{x}$ is orthogonal to all the rows of $A$.  Thus $A\mathbf{x} = \mathbf{b}$ has a solution iff $\mathbf{b}$ is orthogonal to $\ker A$. Is this correct?
b.) I do not know how to do this
c.) In this do I have to prove that $\mathbf{w} \in \operatorname{coker}A$ is orthogonal to the range of $A$? I am not exactly sure what they are asking. 
 A: For a, your argument is showing that the rowspace is the orthogonal complement of the nullspace. That is not what the question is asking (although it is related). Notice that $\mathbf{b}$ is an element in the columnspace of $A$. The columnspace, rowspace and nullspace of a symmetric matrix are related in a very specific manner. See if you can use these relations.
For part b, consider the critical points of the quadratic form. Where do they occur? Does your function have any?
For c, what can you say about the left-nullspace and the columnspace of a matrix?
A: I'll assume $A$ has real entries.
For part a),
\begin{align*}
& u \in R(A)^{\perp} \\
\iff & \langle u, y \rangle = 0 \, \forall \, y \in R(A) \\
\iff & \langle u, Ax \rangle = 0 \, \forall \, x \\
\iff & \langle A^T u, x \rangle = 0 \, \forall \, x \\
\iff & A^T u = 0 \\
\iff & u \in N(A^T).
\end{align*}
This shows that $N(A^T)$ is the orthogonal complement of $R(A)$.  Because $A$
is symmetric, $N(A)$ is the orthogonal complement of $R(A)$.  Hence $b \in R(A)$
if and only if $b$ is orthogonal to $N(A)$.
For part b), let $v$ be the projection of $f$ onto $N(A)$.  Note that $v^T f = \|v\|^2 \neq 0$
and the function
\begin{align*}
g(\alpha) &= p(\alpha v) \\
&= -2\alpha v^T f + c
\end{align*}
is unbounded below.
