The example is given below:

enter image description here

But I do not understand the details of calculating $\phi_{BB}(\alpha_{v})$, could anyone explain this for me please?

The definition of $\phi_{BB}(\alpha_{v})$ is given below:

enter image description here

EDIT: I mean how the definition of the linear transformation given affect the matrix?


Part of the problem is that Proposition 8.1 is not a definition. It doesn't tell you what $\Phi_{BD}$ is, or how to compute it. It simply asserts existence.

It's also not particularly well-stated as a proposition, since it asserts the existence of a family of isomorphisms based on pairs of bases $(B, D)$ on $V$ and $W$ respectively, but doesn't specify any way in which said isomorphisms differ. If you could find just one (out of the infinitely many) isomorphisms between $\operatorname{Hom}(V, W)$ and $M_{k \times n}(F)$ (call it $\phi$), then letting $\Phi_{BD} = \phi$ would technically satisfy the proposition, and constitute a proof!

Fortunately, I do know what the proposition is getting at. There is a very natural map $\Phi_{BD}$, taking a linear map $\alpha : V \to W$, to a $k \times n$ matrix.

The fundamental, intuitive idea behind this map is the idea that linear maps are entirely determined by their action on a basis. Let's say you have a linear map $\alpha : V \to W$, and a basis $B = (v_1, \ldots, v_n)$ of $V$. That is, every vector $v \in V$ can be expressed uniquely as a linear combination of the vectors $v_1, \ldots, v_n$. If we know the values of $\alpha(v_1), \ldots, \alpha(v_n)$, then we essentially know the value of $\alpha(v)$ for any $v$, through linearity. The process involves first finding the unique $a_1, \ldots, a_n \in F$ such that $$v = a_1 v_1 + \ldots + a_n v_n.$$ Then, using linearity, $$\alpha(v) = \alpha(a_1 v_1 + \ldots + a_n v_n) = a_1 \alpha(v_1) + \ldots + a_n \alpha(v_n).$$

As an example of this principle in action, let's say that you had a linear map $\alpha : \Bbb{R}^2 \to \Bbb{R}^3$, and all you knew about $\alpha$ was that $\alpha(1, 1) = (2, -1, 1)$ and $\alpha(1, -1) = (0, 0, 4)$. What would be the value of $\alpha(2, 4)$?

To solve this, first express $$(2, 4) = 3(1, 1) + 1(1, -1)$$ (note that this linear combination is unique, since $((1, 1), (1, -1))$ is a basis for $\Bbb{R}^2$, and we could have done something similar for any vector, not just $(2, 4)$). Then, $$\alpha(2, 4) = 3\alpha(1, 1) + 1 \alpha(1, -1) = 3(2, -1, 1) + 1(0, 0, 4) = (6, -3, 7).$$ There is a converse to this principle too: if you start with a basis $(v_1, \ldots, v_n)$ for $V$, and pick an arbitrary list of vectors $(w_1, \ldots, w_n)$ from $W$ (not necessarily a basis), then there exists a unique linear transformation $\alpha : V \to W$ such that $\alpha(v_i) = w_i$. So, you don't even need to assume an underlying linear transformation exists! Just map the basis vectors wherever you want in $W$, without restriction, and there will be a (unique) linear map that maps the basis in this way.

That is, if we fix a basis $B = (v_1, \ldots, v_n)$ of $V$, then we can make a bijective correspondence between the linear maps from $V$ to $W$, and lists of $n$ vectors in $W$. The map $$\operatorname{Hom}(V, W) \to W^n : \alpha \mapsto (\alpha(v_1), \ldots, \alpha(v_n))$$ is bijective. This is related to the $\Phi$ maps, but we still need to go one step further.

Now, let's take a basis $D = (w_1, \ldots, w_m)$ of $W$. That is, each vector in $W$ can be uniquely written as a linear combination of $w_1, \ldots, w_m$. So, we have a natural map taking a vector $$w = b_1 w_1 + \ldots + b_n w_n$$ to its coordinate column vector $$[w]_D = \begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix}.$$ This map is an isomorphism between $W$ and $F^m$; we lose no information if we choose to express vectors in $W$ this way.

So, if we can express linear maps $\alpha : V \to W$ as a list of vectors in $W$, we could just as easily write this list of vectors in $W$ as a list of coordinate column vectors in $F^m$. Instead of thinking about $(\alpha(v_1), \ldots, \alpha(v_n))$, think about $$([\alpha(v_1)]_D, \ldots, [\alpha(v_n)]_D).$$ Equivalently, this list of $n$ column vectors could be thought of as a matrix: $$\left[\begin{array}{c|c|c} & & \\ [\alpha(v_1)]_D & \cdots & [\alpha(v_n)]_D \\ & & \end{array}\right].$$ This matrix is $\Phi_{BD}$! The procedure can be summed up as follows:

  1. Compute $\alpha$ applied to each basis vector in $B$ (i.e. compute $\alpha(v_1), \ldots, \alpha(v_n)$), then
  2. Compute the coordinate column vector of each of these transformed vectors with respect to the basis $D$ (i.e. $[\alpha(v_1)]_D, \ldots, [\alpha(v_n)]_D$), and finally,
  3. Put these column vectors into a single matrix.

Note that step 2 typically takes the longest. For each $\alpha(v_i)$, you need to find (somehow) the scalars $b_{i1}, \ldots, b_{im}$ such that $$\alpha(v_i) = b_{i1} w_1 + \ldots + b_{im} w_m$$ where $D = (w_1, \ldots, w_m)$ is the basis for $W$. How to solve this will depend on what $W$ consists of (e.g. $k$-tuples of real numbers, polynomials, matrices, functions, etc), but it will almost always reduce to solving a system of linear equations in the field $F$.

As for why we represent linear maps this way, I think you'd better read further in your textbook. It essentially comes down to the fact that, given any $v \in V$, $$[\alpha(v)]_D = \Phi_{BD}(\alpha) \cdot [v]_B,$$ which reduces the (potentially complex) process of applying an abstract linear transformation on an abstract vector $v \in V$ down to simple matrix multiplication in $F$. I discuss this (with different notation) in this answer, but I suggest looking through your book first. Also, this answer has a nice diagram, but different notation again.

So, let's get into your example. In this case, $B = D = ((1, 0, 0), (0, 1, 0), (0, 0, 1))$, a basis for $V = W = \Bbb{R}^3$. We have a fixed vector $w = (w_1, w_2, w_3)$ (which is $v$ in the question, but I've chosen to change it to $w$ and keep $v$ as our dummy variable). Our linear map is $\alpha_w : \Bbb{R}^3 \to \Bbb{R}^3$ such that $\alpha_w(v) = w \times v$. Let's follow the steps.

First, we compute $\alpha_w(1, 0, 0), \alpha_w(0, 1, 0), \alpha_w(0, 0, 1)$: \begin{align*} \alpha_w(1, 0, 0) &= (w_1, w_2, w_3) \times (1, 0, 0) = (0, w_3, -w_2) \\ \alpha_w(0, 1, 0) &= (w_1, w_2, w_3) \times (0, 1, 0) = (-w_3, 0, w_1) \\ \alpha_w(0, 0, 1) &= (w_1, w_2, w_3) \times (0, 0, 1) = (w_2, -w_1, 0). \end{align*}

Second, we need to write these vectors as coordinate column vectors with respect to $B$. Fortunately, $B$ is the standard basis; we always have, for any $v = (a, b, c) \in \Bbb{R}^3$, $$(a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) \implies [(a, b, c)]_B = \begin{bmatrix} a \\ b \\ c\end{bmatrix}.$$ In other words, we essentially just transpose these vectors to columns, giving us, $$\begin{bmatrix} 0 \\ w_3 \\ -w_2\end{bmatrix}, \begin{bmatrix} -w_3 \\ 0 \\ w_1\end{bmatrix}, \begin{bmatrix} w_2 \\ -w_1 \\ 0\end{bmatrix}.$$

Last step: put these in a matrix:

$$\Phi_{BB}(\alpha_w) = \begin{bmatrix} 0 & -w_3 & w_2 \\ w_3 & 0 & -w_1 \\ -w_2 & w_1 & 0 \end{bmatrix}.$$

  • $\begingroup$ what about if we have 4 $2 \times 2$ matrices what will be the second step and what will be the dimension of $\phi_ (B, B)$ in this case? $\endgroup$ – Secretly Sep 19 at 3:17
  • $\begingroup$ @hopefully Well, the second step really depends on the elements of the codomain $W$, not so much the dimensions of $V$ and $W$. It really comes from the fact that $D$ is a basis for $W$; in order for this to be true, there must be a proof that $D$ spans $W$, and in that proof must be instructions for how to express $\alpha(v_1), \ldots, \alpha(v_n)$ as linear combinations in terms of $D$. But, the details of this proof will depend on the specific vector space (and perhaps, the basis as well). I can't really say anything more specifically without a specific problem. $\endgroup$ – Theo Bendit Sep 19 at 4:04
  • 1
    $\begingroup$ @hopefully Now that I've seen (and answered) your latest question, I think I see what you mean. $\endgroup$ – Theo Bendit Sep 19 at 4:21

With the equations of $\alpha_v$:

Let $\:w={}^{\mathrm t\mkern-1.5mu}(x, y,z)$. The coordinates of $v\times w$ are obtained as the cofactors of the determinant (along the first row):

$$\begin{vmatrix} \vec i&\vec j&\vec k \\ a_1&a_2 & a_3 \\ x&y&z \end{vmatrix} \rightsquigarrow \begin{pmatrix} a_2z-a_3y\\a_3x-a_1z \\a_1y-a_2x \end{pmatrix}=\begin{pmatrix} 0&-a_3&a_2\\a_3& 0 &-a_1 \\ -a_2 &a_1&0 \end{pmatrix}\begin{pmatrix} x \\y\\z \end{pmatrix}$$

  • $\begingroup$ what about if we have 4 $2 \times 2$ matrices ..... I will insert the link of the question in a comment. $\endgroup$ – Secretly Sep 19 at 3:42

The details probably come in the proof of Theorem 8.1 (which you should read).

Let $B = (v_1,\dots,v_n)$ and $D = (w_1,\dots,w_k)$ be the given bases. Suppose that $\alpha\in\operatorname{Hom}(V,W)$. For each $i$ in $1,\dots,n$ there exist scalars $\phi_{ij} \in F$ such that $$ \alpha(v_i) = \phi_{1i}w_1 + \phi_{2i}w_2 + \dots + \phi_{ki} w_k $$ Set $\Phi_{BD}(\alpha)$ to be the $k\times n$ matrix whose $(i,j)$-th entry is $\phi_{ij}$.

Now we come to angryavian's suggestion. Here $V = W = \mathbb{R}^3$, and $B = D = (e_1,e_2,e_3)$. Moreover, $\alpha(w) = v \times w$ for a fixed $v = \begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$. So you need to find the coefficients of $\alpha(e_1)$, $\alpha(e_2)$ and $\alpha(e_3)$ in the basis $(e_1,e_2,e_3)$.


The first column of the matrix is $v \times \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$, the second column is $v \times \begin{bmatrix}0 \\ 1 \\ 0\end{bmatrix}$, and the third is $v \times \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$.

  • $\begingroup$ I mean how the definition of the linear transformation given affect the matrix? $\endgroup$ – Secretly Sep 18 at 18:58

If $B = \{e_1,\dots,e_n\}$ and $D = \{f_1,\dots,f_m\}$ and $T$ is a linear transformation, then $\Phi_{BD}(T)$ is obtained by applying $T$ to each element of $B$ and witting the result in terms of $f_1,\dots,f_m$. That is, if

$$ T(e_j) = \sum_{i=1}^m a_{i,j}f_i, $$

then the $j$-th column of $\Phi_{BD}(T)$ is

$$ \begin{bmatrix} a_{1,j} \\ a_{2,j} \\ \vdots \\ a_{m,j} \end{bmatrix}. $$

For example, $\alpha_v(e_1) = v \times e_1 = [0,a_3,-a_2]^T = 0e_1 + a_3e_2 -a_2e_3$ so the first column of $\Phi_{BB}(\alpha_v)$ is $[0,a_3,-a_2]^T$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.