Understanding an example in Golan's “Linear Algebra”

The example is given below:

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:

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:

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

• 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? – Secretly Sep 19 at 3:17
• @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. – Theo Bendit Sep 19 at 4:04
• @hopefully Now that I've seen (and answered) your latest question, I think I see what you mean. – 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):

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

• what about if we have 4 $2 \times 2$ matrices ..... I will insert the link of the question in a comment. – 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 $$v \times \begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}$$.

• I mean how the definition of the linear transformation given affect the matrix? – 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 $$[0,a_3,-a_2]^T$$.