Here is the proof from the book I was reading
Now, I know about linear transformations and how in its matrix representation , the columns are coordinates of the image of basis vectors in the transformation. I don't know how and why but today while reading this proof I thought of working this proof out and it turned out that matrix multiplication doesn't make sense to me anymore.It is written that $T \alpha_1 = A_{11} \alpha_1$ . Now this makes sense since first column is the coordinate of image of first basis vector. But if I go by explicit matrix multiplication, it turns out be something else and I got confused .Basically how to prove that this is correct by explicit matrix-multiplication?? This might too stupid to ask but I am badly stuck.
Here is what I mean by explicit multiplication:
Best Answer
You are making several category errors, all based on confusing abstract vectors with their coordinate representations given a certain basis.
First, $T\alpha_1$ is a vector in the abstract vector space $V$. It is not necessarily a column vector in $\mathbb{R}^n$. Thus your first equation $T\alpha_1=A\alpha_1$ is already an error. You cannot multiply the $n\times n$ matrix $A$ and the abstract vector $\alpha_1$; the right side of your equation is simply nonsense. You are confusing the vector itself with its coordinate representation.
You can, however, multiply $A$ and the coordinate representation of $\alpha_1$ in the given basis, which is just the column vector $(1, 0, \ldots, 0)^T$. This is your second error: you represented $\alpha_1$ using arbitrary coordinates $x_i$ when in fact the coordinates should be $x_1=1$ and $x_i=0$ otherwise, because $\alpha_1$ is part of the basis. When you represent $\alpha_1$ correctly in coordinates, the product is the vector $(A_{11}, 0, \ldots, 0)^T$, which is indeed $A_{11}$ multiplied by the (correct) coordinate representation of $\alpha_1$.
But all of this is overkill. What you need to recall is that the numbers in the first column of $A$ tell you the coordinates of the abstract vector $T\alpha_1$ with respect to the given basis. (There is a theorem in the book that says this.) Since $A$ is upper triangular, the only nonzero coordinate is $A_{11}$, which tells you that the abstract vector $T\alpha_1$ is just $A_{11}\alpha_1$, as desired.