Linear functional on Banach algebra Let $A$ be the space of all matrices of the form $\begin{pmatrix} a & b \\0 & a\end{pmatrix}$, $2\times2$ over complex field.
Then the spectrum of any element of $A$ comes out to be $\{a\}$. I then have to prove that $A$ has only one multiplicative linear functional. I tried to relate to the result giving correspondence between the set of all maximal ideals of $A$ and the set of all non zero homomorphisms on $A$. But could not get it. Any help on this one.
 A: Here is how you can do that using a few fundamental facts about the maximal ideal space, which is the approach you wanted. Note nevertheless that this can be done in a completely elementary fashion. I do it at the end.
First observe that $A$ is a commutative Banach algebra with unit $I_2$, for any operator norm we put on it. Say we work with the operator norm induced by the Euclidean norm on $\mathbb{C}^2$.
Recall that $\Phi_A$, the set of all characters (nonzero multiplicative bounded functionals on $A$) of $A$ is locally compact and Hausdorff when equipped with the induced weak* topology inherited from the continuous dual $A*$ (here we are in finite dimension, so it is equal to the algebraic dual). Note that $\Phi_A$ is in bijection with the set of maximal ideals of $A$ (via $I=\ker \phi$). So it is nonempty. 
Since $A$ has a unit, $\Phi_A$ is actually compact (this is equivalent to the existence of unit). And we have
$$
\mbox{spectrum}\;( x)=\{\phi(x)\;;\;\phi\in \Phi_A\}\qquad\forall x\in A.
$$
That is, the spectrum of $x$ is the range of $\hat{x}$ over $\Phi_A$, where 
$$
\hat{x}(\phi):=\phi(x)
$$
and $x\longmapsto\hat{x}$ is the Gelfand representation.
In our case, the spectrum of every element of $A$ is a singleton. Given your computation for the spectrum of an arbitrary element, we get
$$
\phi\left( \left(\matrix{a&b\\0&a} \right)\right)=a \qquad\forall\phi\in\Phi_A\;\;\forall a,b\in\mathbb{C}.
$$
So there is a a unique character on $A$, and it is given by the above formula.
Elementary approach: take a character $\phi$. Since it is linear and since we must have $\phi(I_2)=1$, we have
$$
\phi\left( \left(\matrix{a&b\\0&a} \right)\right)= a+ \mu b.
$$
Now commutativity yields
$$
aa'+\mu(ab'+a'b)+\mu^2bb'=aa'+\mu(ab'+a'b)\quad\forall a,a',b,b'\in\mathbb{C}.
$$
For $a=a'=0$ and $b=b'=1$, we get $\mu=0$ and we recover the character we found above. Note that we could also have directly taken 
$$
x=\left(\matrix{0&1\\0&0} \right)
$$
and observed that $x^2=0$, hence $\mu^2=\phi(x)^2=\phi(x^2)=0$.
