I come back to study Linear Algebra Done Right from reading How to Prove It, and I am currently on 3.C Matrices.

It said:

We know that if $v_1,...,v_n$ is a basis of $V$ and $T:V \to W$ is linear, then the values of $Tv_1,...,Tv_n$ determine the values of $T$ on arbitrary vectors in $V$ (See 3.5.) As we will soon see, matrices are used as an efficient method of recording the values of the $Tv_j's$ in terms of a basis of $W$.

So, the first sentence, does it mean that the values of $W$ is determined by the basis of $V$? Intuitively, since every vector in $V$ can be represented by some combinations of its basis, then the transformation should determine by the basis. Am I correct?

For the (See 3.5.), the theorem said that

Suppose $v_1,...,v_n$ is a basis of $V$ and $w_1,...w_n \in W$. Then there exists a unique linear map $T:V \to W$ such that $$Tv_j = w_j$$ for each $j=1,...,n$

What is the significance of this to understand Matrix? The book explained that the existence of the result means that we can find a linear map that takes on whatever values we wish on the vectors in a basis. The uniqueness part of the next result means that a linear map is completely determined bu its value on a basis. I don't quite understand.

And lastly, For the last sentence

matrices are used as an efficient method of recording the values of the $Tv_j's$ in terms of a basis of $W$.

I kind of get the pattern from the book that the columns are the basis of domain, and the rows are the basis of the codomain. Am I correct?

  • $\begingroup$ I think your first question, the $W$ should be a $T$. Yes. A linear transformation is completely determined by its values on a basis. As to your second question, it’s not about matrices, it’s about linear transformations. The previous result tells you: if you know what happens to the basis, you know exactly what the linear transformation does at every vector. The second result tells you: you can make a linear transformation do anything particular you want done to a basis: that is, you specify what you want to happen to a basis, you get a (unique) linear transformation that does exactly that. $\endgroup$ – Arturo Magidin Feb 19 '19 at 6:48
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    $\begingroup$ No, the columns of a matrix ar enot a basis for the domain, and no, the rows are not a basis of a codomain. The columns of the matrix are a spanning set for range (which is a subspace of the codomain). They need not be a basis (they may not be linearly independent, after all). The rows do span a space, but it is not easily related to either the range or the other important subspace associated to a matrix, which is the nullspace. $\endgroup$ – Arturo Magidin Feb 19 '19 at 6:50

First question: $W$ is some arbitrary space that YOU CHOSE as the codomain in your map $T$. But $ T $ IS defined by its action on a basis of its domain. Since $ T $ is a linear transformation, we know that for any vectors $a, b \in V , c \in F$, where $ F $ is the field over which your vector spaces lie, $T(ca + b) = cT(a) + T(b)$. So, if you have a basis $ \{ e_1, ..., e_n \} $ and you know the outputs $ T(e_1), ..., T(e_n) $, given a linear combination of those vectors, say $v = c_1 e_1 + ... + c_n e_n $, now you know exactly what happens to $v $ under $T$, since you already have the values $ T(e_1), ..., T(e_n) $ and you just split up $ T(v) = T(c_1 e_1 + ... + c_n e_n) = c_1 T(e_1) + ... + c_n T(e_n) $.

Also, I would like to point out that the matrices are useful for recording values if you have a basis for the domain and one for the codomain. The matrix will tell you what happens to COORDINATES in those bases. If you change your bases, you get a different matrix.

For the second question: This just means there is only one linear map $T $ for which $T(v_1) = w_1, T(v_2) = w_2, ... $. This is again for the same aforementioned reason.

Last question: $ V , W $ are ABSTRACT vector spaces. So a vector could be a row vector, column vector, polynomial, linear transformation, etc. The COORDINATES are column vectors. If you have an $n$ dimensional vector space, then in any basis, the coordinates will be of the form $ \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix},\begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, ..., \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \end{pmatrix} $.

Second thing: the columns of the matrix may not be linearly independent, so they may not be a basis, but they SPAN the image up to isomorphism. If you put in coordinates of a basis vector into a linear transformation, then the matrix tells you the coordinates to which that vector is sent by the transformation and the coordinates of the $j$th ordered basis vector will be sent to the $j$th column of the matrix. Try working an example yourself to see this. To see the power of this idea in the abstract, try the space of polynomials up to degree n and note that differentiation is a linear transformation. Find a reasonable basis for the polynomials (you can take the easy way out, or try something fun like the Lagrange interpolation polynomials), find a matrix for the differentiation transformation, and test out this matrix on some polynomials (which you first have to convert into coordinates in the bases you chose).


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