Construct the linear functional $\phi$ Let 
$A$=$\begin{bmatrix}
1 &1  &1 \\ 
0 & 2 & 1\\ 
0 &0  & 2
\end{bmatrix}$ 

let  $W=\{x \in \mathbb{R^3} \mid Ax=2x\}$

Construct a linear functional $\phi$ on $\mathbb{R^3}$ such that $\phi(x_0)=1$ where $x^t_0=(1,2,3),$   and $\phi(x)=0$ $\forall x \in W$
The solution I tried - Here the eigenvalues of given matrix are $1,2$ and the set $W$ is set of all the eigenvalues corresponding to eigenvalue $2$. Here I have to find the functional, I don't have that much idea about functionals.
Please help
Thank you
 A: Hint: A functional in this context is a linear map $\phi:\Bbb R^3 \to \Bbb R$.  Every functional on $\Bbb R^3$ can be written in the form $\phi((x_1,x_2,x_3)^T) = a_1x_2 + a_2 x_2 + a_3 x_3$ for some $a_1,a_2,a_3$.  In other words, you can think of $\phi$ as being a row-vector $(a_1,a_2,a_3)$.
$W$ is one-dimensional; find a vector $v = (v_1,v_2,v_3)^T$ which forms a basis of $W$.  Once you have done that, you want a that satisfies $\phi(v) = 0$ and $\phi(x) = 1$.  In other words, you need a solution $(a_1,a_2,a_3)$ to the system of equations
$$
\phi(v) = a_1 v_1 + a_2 v_2 + a_3 v_3 = 0\\
\phi(x) = a_1(1) + a_2(2) + a_3(3) = 0.
$$
A: It is easy to check that if $Ax=2x$ then $x$ is a multiple of $(1,1,0)$. Any linear functional has the form $\phi(a,b,c)=\lambda a+\mu b+\nu c$. We have $\phi(1,1,0)=0$ and $\phi(1,2,3)=1$, so we have $$\lambda+\mu=0\text{ and }\lambda+2\mu+3\nu=1$$ So we find $$\mu=-\lambda,\nu=\frac{1+\lambda}{3}$$ In particular, we can take $\lambda=2,\mu=-2,\nu=1$. So $\phi(a,b,c)=2a-2b+c$.
Check: $\phi(1,1,0)=2\cdot1-2\cdot1+0=0$ and $\phi(1,2,3)=2\cdot1-2\cdot2+1\cdot3=1$.
A: A linear functional is just a linear function that takes a vector and outputs a scalar. The Riesz representation theorem tells us that if we have an inner product $\langle\cdot,\cdot\rangle:\mathbb R^n\times\mathbb R^n\to\mathbb R$, then for any linear functional $\phi$ on $\mathbb R^n$, there is a unique element $y\in\mathbb R^n$ such that for all $x\in\mathbb R^n$, $\phi(x)=\langle x,y\rangle$. 
Now, you can find either by inspection or other methods that the eigenspace of $2$ is spanned by $(1,1,0)$, so, taking the standard Euclidean inner product (dot product) for our inner product on $\mathbb R^3$, this problem is equivalent to finding some vector $y$ such that $y\cdot(1,1,0)=0$ and $y\cdot(1,2,3)=1$. The first equation tells us that $y$ is orthogonal to $(1,1,0)$, so we can write $y=\lambda(1,-1,0)+\mu(0,0,1)$. Substituting this into the second equation and simplifying produces $3\mu-\lambda=1$. Any pair of real numbers that satisfy this equation produce a linear functional that satisfies the given conditions.
