# Linear Maps, Basis of Domain, and Matrix

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?

• 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. – Arturo Magidin Feb 19 '19 at 6:48
• 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. – 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 $$T(v) = T(c_1 e_1 + ... + c_n e_n) = c_1 T(e_1) + ... + c_n T(e_n)$$.
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).