# I I don't quite understand what i read

1. Aug 31, 2016

### Terrell

i am having a really tough time following what this is saying with all the abstractions. i am assuming that since it's a transformation from V to W. then why does the equation below have y vectors instead of x? and since linear transformation follow traditional matrix multiplication, then why is a column vector of A multiplied to the column vectors of W???

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2. Aug 31, 2016

### Staff: Mentor

I think what its saying is that you have some basis vector xj from the V space that you would like to represent in the W space so you apply the transformation T to it.

T is in fact the matrix A.

So you are doing a matrix multiplication of A and xj to get the xjvector represented by the yi basis vectors in the W space.

Does that make sense?

Hopefully @Mark44 correct me if I'm wrong.

3. Aug 31, 2016

### Krylov

No.
No, it does not.

The appearance of $Ax_j$ in (6) in the quoted text is misleading and incorrect. Try to read (6) without it. The scalars $a_{1j},\ldots,a_{mj}$ appearing in the linear combination are the matrix elements of the representation $A$ of $T$ w.r.t. the particular bases in $V$ and $W$.

Last edited: Aug 31, 2016
4. Aug 31, 2016

### Staff: Mentor

Thanks Krylov. I'll remember to add you to the post when I need verification.

I guess I looked at it in a more operational sense where you have the linear transformation T implemented as the matrix A and you then applied a vector in the V space to get its representation in the W space via matrix multiplication. Its been many years since I've played with it and when I'm not sure I call in an advisor.

5. Aug 31, 2016

### Staff: Mentor

A is a matrix representation of the transformation T. A transformation can have multiple matrix representations, depending on the bases for the domain and codomain.

I've been busy with some other things this morning and early afternoon -- I'll try to get back to this a little later.

6. Aug 31, 2016

### Terrell

so the equation is incorrect? i think it is

7. Sep 1, 2016

### Krylov

Yes, it is incorrect. The vectors $T(x_j)$ and $a_{1j}y_1 + a_{2j}y_2 + \ldots + a_{mj}y_m$ are in W, which could be any $m$-dimensional vector space.
$Ax_j$ is not even defined, because $A$ maps from $\mathbb{R}^n$ to $\mathbb{R}^m$ but $x_j$ is a vector in $V$, which could be any $n$-dimensional vector space.

When you are learning linear algebra, it is important that your book distinguishes carefully between a linear transformation $T$ and its matrix representation $A$ w.r.t. a certain pair of bases. Afterwards, people often become sloppy about it. (There is a good reason for this sloppiness, but I do not consider it helpful at the beginning.)

(I remember that a tutorial was written about this topic a while ago: Matrix Representations of Linear Transformations. Its notation is different from your book, though.)

Last edited: Sep 1, 2016
8. Sep 1, 2016

### Terrell

exactly the cause of my confusion. will check the article. thanks!

9. Sep 28, 2016

### Stephen Tashi

Are you referring to the right hand side of the equation? The image of the transform is in W, so it would be a vector expressed in the basis for W and that basis is the set of the $y_i$.

The vectors of W aren't necessarily row vectors or column vectors. The usual way to represent a linear transformation is to represent it as multiplication by a matrix A on the left hand side of a column vector, which produces another column vector. ( However, I suppose there are a few books where the representation is done by multiplying a row vector on the right hand side by a matrix A to produce another row vector. Let's assume your text uses the conventional approach.)

The representation of the linear transformation $T$ by a matrix involves assuming we will represent the vectors in V and W in a certain manner. For example, if $X$ is a vector in $V$ expressed in the (orthonormal) x-basis as $X = a1 x_1 + a2 x_2 + a3 x_3$ then a "natural" way to represent $X$ by the ordered triple $(a1,a2,a3)$. Notice this is not "naturally" a column vector. In fact, it looks like a row vector. But, after reading a lot of algebra texts, one begins to accept that $X$ will be represent by a column vector - i.e. as the transpose of $(a1,a2,a3) = (a1,a2,a3)^T$.

Using the convention of representing the vectors involved as column vectors, the representation of $x_2$ in the x-basis is $(0,1,0)^T$. The image of $x_2$ under a linear transformation $T$ will be some vector $T(x_2)$ in W. Suppose W has dimension two and $T(x_2) = b1 y_1 + b2 y_2$ , which is represented as $(b1, b2)^T$. Compare $(b1,b2)^T$ with the result of multiplying $(0,1,0)$ on the left by a matrix $A$ of the appropriate dimensions.

$\begin{pmatrix} a_{11}&a_{12}& a_{13} \\ a_{21}&a_{22}&a_{23} \end{pmatrix} \begin{pmatrix}0\\1\\0\end{pmatrix} = \begin{pmatrix}a_{12}\\a_{22}\end{pmatrix}$

So $a_{12} = b_1$ and $a_{22} = b_2$. As your text indicated, the second column of $A$ is determined by the coefficients of $T(x_2)$ when it is represented as a vector in the y-basis.