TITLE: If T is a linear operator, has dim V distinct eigen values, then T has a diagonal matrix with respect to some basis of V.
QUESTION [0 upvotes]: My question is "Why did they write T has a diag matrix w.r.t some basis of V".
Give T belongs to L(v), dosent that say we already fixed the basis ?

REPLY [1 votes]: You're mixing the idea of a linear operator with that of its matrix representation in a given basis. Suppose $V$ is a $K$-vector space with $\text{dim} \; V=n$, a linear operator $T$ on $V$ is just a map $T:V \longrightarrow V$ which takes a vector $v \in V$ into a new vector $w =T(v)$. Of course, $T$ has to be linear:
$$ T(\alpha v + \beta w) = \alpha T(v) + \beta T(w) \hspace{1cm} \forall v,w \in V, \forall\alpha, \beta \in K$$
but this isn't a really important point here. Until now, we haven't made any mention to basis, matrices... So, keep in mind, a linear operator in $V$ is just a map: it takes vectors in $V$ into new vectors in $V$.
Now, representing this linear operator by a $n \times n$ matrix is a really useful trick when dealing with concrete calculations, for example. But the matrix representation of a linear map is not unique, because we have to fix a basis in order to get that matrix. I'll show you how it goes through an example. Suppose we have the usual vector space $\mathbb{R}^2$, and the linear transformation $T: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$ acting on the canonical basis $C=\{(1,0),\,(0,1)\}$ in the following way:
$$T(1,0)=(0,1) \hspace{1cm} T(0,1)=(-1,0)$$
(You can specify the effect of a linear transformation giving its effect on a basis). Now, write any vector $v \in \mathbb{R}^2$ in the canonical basis as $v = x (1,0) + y(0,1)$ : when you apply $T$ to this vector you get a new vector $w=T(v)=\bar{x}(1,0) + \bar{y}(0,1)$ with coordinates $\bar{x}, \bar{y}$ related to the old ones by
$$\begin{pmatrix}
\bar{x} \\
\bar{y}
\end{pmatrix} = \begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix} \begin{pmatrix}
x \\
y
\end{pmatrix}$$
Check it out for yourself if you don't believe this is the true relation between $x,y$ and $\bar{x},\bar{y}$ (you only have to apply the fact that $T$ is linear and the definition of $T$ in terms of its action on the canonical basis). Now, take a new basis of $\mathbb{R}^2$, for example $B=\{ (1,1),\, (0,1) \}$. You can write $v$ as:
$$ v = x(1,0) + y(0,1) = x'(1,1)+y'(0,1) \hspace{1cm}x'=x ,\,y'=y-x $$
Again, I'm sure you can verify this is true by yourself. So $x'$ and $y'$ are now the coordinates of $v$ in the new basis $B$. When you apply $T$, getting $w=T(v)$, you can now write $w$ in the basis $B$: $w=\bar{x}'(1,1)+\bar{y}'(0,1)$. In matrix form, the relation between the coordinates of $v$ and $w$ takes the form:
$$\begin{pmatrix}
\bar{x}' \\
\bar{y}'
\end{pmatrix} = \begin{pmatrix}
-1 & -1 \\
2 & 1
\end{pmatrix} \begin{pmatrix}
x' \\
y'
\end{pmatrix}$$
As you see, we have obtained a new matrix, which we could call the matrix of $T$ in the basis $B$. The old one would be the matrix of $T$ in the canonical basis:
$$ (T)_C = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \hspace{1cm} (T)_B = \begin{pmatrix} -1 & -1 \\ 2 & 1 \end{pmatrix} $$
The subscripts here represent the basis in which we are working. Now you can see that the matrix representation of a linear map is not unique, so when you diagonalize a linear operator what you really do is look for a basis in which the matrix associated to $T$ is diagonal. Never forget this: you don't have a matrix form for a linear map until you specify a basis of the vector space!
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