Correspondences between linear transformations and matrices

Question

EDIT: I value all the answers given to me in this question, and upon reading them and thinking about the question for some time, I have come up with my own answer and explanation for this question. I have typed this up and submitted it as an answer here, so that people may read it and can hopefully let me know if my current understanding is fully correct and useful. And furthermore if it is incorrect I would very much appreciate corrections and amendments to my answer. Thank you.

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I am a bit confused about representing linear transformations as matrices, I will try to explain my confusion here, and hopefully someone can clear it up.

Note: Let $V$ and $V'$ be finite dimensional vector spaces over an arbitrary field $F$ such that $\operatorname{dim}(V) = n$ and $\operatorname{dim}(V') = m$, $V$ has basis $B$, and $V'$ has basis $B'$.

Now we want to create an isomorphism:

$f: \operatorname{Hom}(V,V') \rightarrow M_{m\times n}(F)$

According to notes we map $$ h \mapsto [h]^{B'}_{B} = [P^{-1}_{B'}hP_{B}] = \left( \begin{array}{cc} P^{-1}_{B'}\circ h\circ P_B \begin{pmatrix} 1\\ 0 \\ \vdots \\ 0\end{pmatrix} &\cdots &P^{-1}_{B'}\circ h\circ P_B\begin{pmatrix} 0\\ 0 \\ \vdots\\ 1\end{pmatrix} \end{array} \right)$$

Where $P_B : F^n \rightarrow V$ is defined by $$\begin{pmatrix} a_1 \\ \vdots \\ a_n \end{pmatrix} \mapsto a_1v_1 + ... + a_nv_n$$

Now, I'm quite confused as to where this mapping came from, and furthermore I'm confused as to exactly what matrix we are mapping $h$ to, and why is this matrix called the matrix of $h$ with respect to $B$ and $B'$ ($[h]^{B'}_{B}$).

I apologize if the question isn't quite clear and thank you in advance for any answers.

Answer 1

Often I feel that to truly understand something you must be able to explain it, or teach it, which is the reason I have put thought into devising an answer to my own question. I have considered all the answers (and will select one as best soon), and have come up with my own explanation of where the matrix comes from, how we get it, and how it acts. This explanation follows.

EDIT: NOTE: I realize after reading this that in this answer I have defined the function $P_B$ slightly differently and the result is that $[T]_B^{B'}$ appears slightly different aesthetically than how it did in the original question.

EDIT: I have undeleted this answer incase it is helpful to someone else but I will accept one of the other answers. Thank you all for your help

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Let $T: V\rightarrow W$ be a linear transformation, and suppose that

$V$ has basis $B = \{v_1, ..., v_n\}$

$W$ has basis $B' = \{w_1, ..., w_m\}$

We have an isomorphism (I):
$$P_B : V\rightarrow F^n$$ $$v = \sum\limits_{i=1}^na_iv_i\mapsto (a_1, a_2, ..., a_n)$$

Proof. First we show that this is a homomorphism. Let $v = \sum\limits_{i=1}^{n}a_iv_i$ and $w = \sum\limits_{i=1}^nb_iv_i$ then $f(v + w) = (a_1 + b_1, ..., a_n + b_n)$ and $f(v) + f(w) = (a_1, a_2, ..., a_n) + (b_1, b_2, ..., b_n) = (a_1 + b_1, ..., a_n + b_n)$ thus $f(v + w) = f(v) + f(w)$ and $f(cv) = (cv_1, ..., cv_2) = c(v_1, ..., v_2) = cf(v)$. Now we know that this is a homomorphism, so we must show that it is a bijection. First we show that it is one-to-one. Suppose $f(w) = f(v)$ where $v$ and $w$ are as they were earlier. Then $f(v) - f(w) = f(v-w) = 0$. Thus $(a_1 - b_1, ..., a_n - b_n) = (0, 0, ..., 0)$, which implies that $a_1 = b_1, ..., a_n = b_n$, and thus $v = w$. Next we show that the map is onto. Consider $x\in F^{n} = (r_1, ..., r_n)$. But since $\{v_1, ..., v_n\}$ spans $V$ we can write $f(\sum\limits_{i=1}^nr_iv_i) = (r_1, ..., r_n)$ and thus the map is onto. $\Box$

Similarly we have the isomorphism (II):

$$P_{B'} : W\rightarrow F^m$$

$$w = \sum\limits_{i=1}^mb_iw_i\mapsto (b_1, b_2, ..., b_n) $$

Thus, it seems that once we choose bases, $B$ and $B'$, every vector $v\in V$ has a unique representation in $F^n$ and every vector $w\in W$ has a unique representation in $F^m$.

So we wonder, how can we represent $T: V\rightarrow W$ in terms of a map from $F^n \rightarrow F^m$?

Said another way, can we find a mapping which will satisfy the following diagram:

$$F^n \rightarrow V \xrightarrow{T} W\rightarrow F^m?$$

Using isomorphisms (I) and (II) we can piece together a mapping from $F^n\rightarrow F^m$ which is equivalent to $T: V\rightarrow W$:

$$P_B^{-1} : F^n\rightarrow V$$ $$T: V\rightarrow W$$ $$P_{B'} : W \rightarrow F^m$$

Hence, we have found the equivalent transformation from $F^n\rightarrow F^m$ that we were looking for:

$$P_{B'}\circ T\circ P_B^{-1} : F^n \rightarrow F^m$$

Finally we have the isomorphism (III):

$$\{\text{Linear Transformations } F^n\rightarrow F^m\}\rightarrow M_{m\times n}(F)$$

$$T\mapsto [T] =\left(\begin{smallmatrix} T\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & T\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... & T\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right)$$

Proof. First we show that this is a homomorphism.

$$[c_1f_1 + c_2f_2] $$

$$= \left(\begin{smallmatrix} c_1f_1\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) + c_2f_2\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & c_1f_1\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) + c_2f_2\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right)& ... &c_1f_1\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right) + c_2f_2\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right)$$

$$= \left(\begin{smallmatrix} c_1f_1\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & c_1f_1\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... & c_1f_1\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right) + \left(\begin{smallmatrix} c_2f_2\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & c_2f_2\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... &c_2f_2\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right)$$

$$= c_1\left(\begin{smallmatrix} f_1\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & f_1\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... &f_1\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right) + c_2\left(\begin{smallmatrix} f_2\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & f_2\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... &f_2\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right)$$

$$= c_1[f_1] + c_2[f_2]$$

Now we show that this is onto. Consider an arbitrary matrix $A$. Since a linear transformation is completely determined by its action on the basis vectors, we can define a linear transformation that gives $A$, and thus the function is onto. Next we show that the map is one-to-one. Suppose $S(f) = S(g)$. Then $\left(\begin{smallmatrix} f\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & f\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... &f\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right) = \left(\begin{smallmatrix} g\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & g\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... &g\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right)$, however, since again a linear transformation is completely determined by how it acts on the basis vectors, these two functions must be equivalent, and hence $f = g$. $\Box$

Therefore, we know that we can write a linear transformation $F^n \rightarrow F^m$ as an $m\times n$ matrix

This tells us that we can now write our transformation $T:V\rightarrow W$ as an $m\times n$ matrix. Or, more precisely, we can write $P_{B'}\circ T\circ P_B^{-1} : F^n \rightarrow F^m$ equivalently as:

$[T]_{B}^{B'} = [P_{B'}\circ T\circ P_B^{-1}] = \left(\begin{smallmatrix} P_{B'}\circ T\circ P_B^{-1}\left(\begin{smallmatrix} 1\\ 0 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & P_{B'}\circ T\circ P_B^{-1}\left(\begin{smallmatrix} 0\\ 1 \\ . \\ . \\ . \\ 0\end{smallmatrix}\right) & ... & P_{B'}\circ T\circ P_B^{-1}\left(\begin{smallmatrix} 0\\ 0 \\ . \\ . \\ . \\ 1\end{smallmatrix}\right)\end{smallmatrix}\right)$

Therefore, once we chooses bases, we can create an isomorphism (IV)

$$H: \operatorname{Hom}(V,W) = \{\text{Linear transformations } V\rightarrow W\}\rightarrow M_{m\times n}(F)$$

$$T\mapsto [T]_{B}^{B'}$$

And hence, after a choice of bases, we can write any linear transformation with a matrix.

Answer 2

The map $P_B$ converts $\left(\begin{matrix}a_1\\\vdots\\a_n\end{matrix}\right)\in F^n$ to the corresponding linear combination of the basis vectors in $B$, namely, $a_1v_1+\dots+a_nx_n$; its inverse converts each vector in $V$ to the $n$-tuple of coefficients needed to write it as a linear combination of the $B$-basis vectors. The map $P_{B'}$ does the same thing between $F^m$ and $V'$: it converts an $m$-tuple of coefficents into the corresponding linear combination of the vectors in $B'$, and its inverse converts each vector in $V'$ to the $m$-tuple of coefficients in its unique representation as a linear combination of the $B'$-vectors.

The operation of $[H]_B^{B'}$ on a member of $F^n$ is then as follows. First it applies $P_B$, converting the $n$-tuple to an actual element of $V$. Then it applies the linear transformation $h$ to get a vector in $V'$. And finally it applies $P_{B'}^{-1}$ to convert that to the $m$-tuple in $F^m$ that represents it with respect to the basis $B'$.

To see just how the operation of $h$ itself is incorporated in the matrix, consider the special case in which $V=\mathbb{R}^n$, $V'=\mathbb{R}^m$, and $B$ and $B'$ are the standard bases. Then $H$ is simply the matrix $[\begin{matrix}h(e_1)&h(e_2)&\dots&h(e_n)\end{matrix}]$. When you form a product $$[\begin{matrix}h(e_1)&h(e_2)&\dots&h(e_n)\end{matrix}]\left[\begin{matrix}a_1\\a_2\\\vdots\\a_n\end{matrix}\right],$$ you get $a_ah(e_1)+\dots+a_nh(e_n) = h(a_1v_1+\dots+a_nv_n)$, exactly as you want.

Answer 3

The thing is that we want a matrix such that the next calculation holds $$(\text{Column vector of the coordinates of the image in the basis B'})=$$ $$=[h_B^{B'}](\text{Column vector of the coordinates of the element of V in the basis B})$$ The point fact is that columns in the matrix represent the coordinates of the images of each of the elements of the basis $B$ in the basis $B'$. Since given $$x=\sum x^iv_i$$ in coordinates of the basis $B=\{v_i\}$, we have by linearity that $$h(x)=\sum x^ih(v_i)$$ which explains why the matrix notation works.

To sum up, the thing is that the matrix takes coordinates in one basis of the domain vector space and returns coordinates of the image in a basis of the codomain vector space, since coordinates changes if we changes the basis, it is important to see that matrixes are dependant on the basis that we choose -many linear algebra is searching a good basis for a nice form of the matrix in that basis (see Jordan triangularization)-.

Answer 4

I am not sure what you are saying, but perhaps I can explain it like this:

Consider the vector space $V$ as you said of dimension $n$. Let $B$ be its basis consisting of vectors $v_1, v_2 \ldots v_n$. Hence any vector little $v$ in big $V$ will be a linear combination of these $v_i's$, say $a_1v_1 + a_2v_2 + \ldots a_nv_n$. Now what we do is to this little $v$, we attach to it its coordinate vector $[v]_B$ in $\mathbb{F}^n$, which is just the $n-$tuple of numbers

$$(a_1, a_2 \ldots a_n).$$

Prove that the mapping $P_B : V \rightarrow \mathbb{R}^n$ that sends $v \mapsto [v]_B$ is an isomorphism of vector spaces.

Now suppose you have another vector space $V'$ with basis $B'$ consisting of vectors $w_1, w_2 \ldots w_m$. Suppose too that you have a linear transformation $T$ from $V$ to $V'$ such that:

$$\begin{array}{ccc} T(v_1) &=& a_{11}w_1 + a_{21}w_2 + \ldots + a_{m1}w_m, \\ T(v_2) &=& a_{21}w_1 + a_{22}w_2 + \ldots + a_{m2}w_m,\\ & \vdots& \vdots \\ & \vdots& \vdots \\ T(v_n) &=& a_{1n}w_1 + a_{2n}w_2 + \ldots + a_{mn}w_m.\end{array}$$ Such a linear transformation can then be represented by the following matrix (why?):

$$\left[ \begin{array}{cccc} a_{11} & a_{12} & \ldots & a_{1n} \\ a_{21} & a_{22} & \ldots & a_{2n} \\ && \vdots & \\ a_{m1} & a_{m2} & \ldots & a_{mn} \end{array}\right]$$

Now at this point we bring together all these facts, including those about coordinate vectors above.

To apply this linear transformation to a vector $v \in V$, you first need to find its coordinate vector $[v]_B$ in $\mathbb{F}^n$.

Now notice that the columns of this matrix are the coordinate vectors of the $T(v_i)$ in the basis of B'. This is the crucial bit - which is why we call the matrix associated to this linear map not just say $M$, but $M^{B}_{B'}(T)$. In fact this entire explanation can be summed up in the following elegant way:

$$[T(v)]_{B'} = M^{B}_{B'}\big([v]_B\big).$$

Now it is your turn to do the explanation: Why is it that the association $T \mapsto M^{B}_{B'}(T)$ gives rise to an isomorphism between $\operatorname{Hom}(V, V')$ and $\operatorname{Mat}_{m\times n} (F)$ once we have fixed bases $B$ and $B'$?

What happens if we assume that two matrices in some given bases give rise to the same linear map?

Answer 5

Ok so lets have a look at what a linear map does to an arbitrary vector in $W$.

$T(b_1 w_1 + b_2 w_2 + ... + b_n w_n) = b_1 T(w_1) + ... + b_n T(w_n)$

Now each $T(w_i)$ lies in $V$ and so is expressible in terms of the basis of $V$.

Thus for each $i$ between $1$ and $n$ we have:

$T(w_i) = a_{i,1} v_1 + a_{i,2} v_2 ... + a_{i,m} v_m$

Plugging these into the above gives you:

$T(b_1 w_1 + b_2 w_2 + ... + b_n w_n) = \left(\sum_{j=1}^n\sum_{i=1}^m a_{i,j} b_j\right) v_i$

You can see from the RHS that the components of the corresponding vector in $V$ just come from a matrix multiplication of the matrix $A = (a_{i,j})$ by the vector $b = (b_1, ..., b_n)^T$