# Vector Component Transformation Matrix in Shilov's Linear Algebra

I am working through Georgi Shilov's Linear Algebra, and I am having trouble understanding the vector component transformation matrix definition he gives in section 5.31. I will describe this definition below.

Let $$e_1, e_2, \dots, e_n$$ and $$f_1, f_2, \dots, f_n$$ be two bases in a vector field of dimension $$n$$ such that for some quantities $$p^{(j)}_i$$ we have $$f_j = p^{(j)}_1 e_1 + p^{(j)}_2 e_2 + \dots + p^{(j)}_n e_n$$ Shilov now defines the matrix of the transformation from basis $$\{e\}$$ to basis $$\{f\}$$ as $$P = \begin{bmatrix} p^{(1)}_1 & p^{(2)}_1 & \dots & p^{(n)}_1 \\ p^{(1)}_2 & p^{(2)}_2 & \dots & p^{(n)}_2 \\ \vdots & \vdots & \ddots & \vdots \\ p^{(1)}_n & p^{(2)}_n & \dots & p^{(n)}_b \end{bmatrix}$$ So, $$P$$ is the matrix with components of $$f_i$$ in terms of the basis $${e}$$ as columns.

Now, suppose we have some vector $$x = \xi_1 e_1 + \xi_2 e_2 + \dots + \xi_n e_n = \eta_1 f_1 + \dots + \eta_n f_n$$. Then, the author claims that the matrix describing the transformation from the components $$\xi_1, \dots, \xi_n$$ to the components $$\eta_1, \dots , \eta_n$$ is $$S = (P^{-1})^T$$ In my understanding, the "matrix describing the transformation from the components $$\xi_1, \dots, \xi_n$$ to the components $$\eta_1, \dots , \eta_n$$" means that $$S\begin{bmatrix} \xi_1 \\ \xi_2 \\ \vdots \\ \xi_n \end{bmatrix} = \begin{bmatrix} \eta_1 \\ \eta_2 \\ \vdots \\ \eta_n \end{bmatrix}$$ However, that doesn't seem to be the case. Consider an example where $$e_1 = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, e_2 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, e_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$ and $$f_1 = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, f_2 = \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, f_3 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$. Then, the matrix of the transformation from $$\{e\}$$ to $$\{f\}$$ and the respective component transformation matrix is $$P = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{bmatrix} \quad S = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}^T = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}$$ However, if we try to use it to transform the vector $$[5, 1, 2]_{\{e\}} = [5, -4, 1]_{\{f\}}$$ we get $$\begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ -1 \\ 2 \end{bmatrix}$$ Which is not what I expected. However, if we instead take $$S = P^{-1}$$, it seems to work out: $$\begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 5 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 5 \\ -4 \\ 1 \end{bmatrix}$$ So, am I misunderstanding what the component transformation matrix is supposed to do, or did I construct $$P$$ incorrectly? I did my best to write the definitions as they are written in the text, however, it is very possible I just mixed up my indices.

• P.S.: A pdf of the text is available here. The relevant section is chapter 5, sections 5.1-3, pages 118-123. Commented Aug 5, 2021 at 21:15

Let's set $$\mathcal{E} = (e_1, \dots, e_n)$$ and $$\mathcal{F} = (f_1, \dots, f_n)$$. Also, let's denote the coordinates of a vector $$v$$ with respect to a basis $$\mathcal{E}$$ (represented as a column vector) by $$[v]_{\mathcal{E}}$$. Then

$$v = \begin{bmatrix} 5 \\ 4 \\ 2 \end{bmatrix}, \, [v]_{\mathcal{E}} = \begin{bmatrix} \xi_1 \\ \xi_2 \\ \xi_3 \end{bmatrix} = \begin{bmatrix} 5 \\ 4 \\ 2 \end{bmatrix}, \, [v]_{\mathcal{F}} = \begin{bmatrix} \eta_1 \\ \eta_2 \\ \eta_3 \end{bmatrix} = \begin{bmatrix} 5 \\ -4 \\ 1 \end{bmatrix}.$$

First of all, it seems you have the role of $$S$$ and $$P^{-1}$$ in your question reversed as you actually have $$P^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{bmatrix}, \,\,\, S = \left( P^{-1} \right)^T = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{bmatrix}.$$ The multiplications you perform later are also incorrect. However, your actual conclusion is true: In order to compute the coordinates of $$[v]_{\mathcal{F}}$$ from $$[v]_{\mathcal{E}}$$, you need to multiply $$[v]_{\mathcal{E}}$$ by $$P^{-1}$$, not $$\left( P^{-1} \right)^T$$. This is the meaning of equation $$(12)$$ Shilov writes on page 122 (where $$Q = P^{-1}$$).

However, Shilov is also not entirely wrong in claiming that the matrix "describing the transformation from the components $$\xi_1,\dots,\xi_n$$ to the components $$\eta_1,\dots,\eta_n$$ is $$\left( P^{-1} \right)^T$$". Why? If you write explicitly $$Q [v]_{\mathcal{E}} = [v]_{\mathcal{F}}$$ (equaton $$(12)$$) you get $$\eta_1 = q_1^{(1)} \xi_1 + \dots + q^{(n)}_1 \xi_n, \\ \dots, \\ \eta_n = q_n^{(1)} \xi_1 + \dots + q^{(n)}_n \xi_n.$$

If you think of the components $$\xi_1, \dots, \xi_n$$ and $$\eta_1, \dots, \eta_n$$ as two "bases" and compare this equation to the equation you wrote in the beginning of the question, you will see that the "matrix of the transformation from the basis $$\{ \xi \}$$ to the basis $$\{ \eta \}$$" is actually $$Q^T$$ and not $$Q$$. To make this statement precise, one needs to discuss dual spaces and then the "components" of a vector with respect to a basis become the dual basis to the original basis and the matrix $$\left( P^{-1} \right)^T$$ becomes the change of basis matrix between the dual bases.

• You are right about the multiplication being incorrect, I think I copied it down incorrectly from my scratchwork. I also think Shilov's statements make sense now, what he was saying was just not something that I expected he would mean. Thank you for the detailed answer! Commented Aug 6, 2021 at 12:31

Let $$B_1 = \{e_1, \dots, e_n\}$$, $$B_2 = \{f_1, \dots, f_n\}$$. Then $$P = M_{B_2}^{B_1}(I)$$, meaning $$P$$ represents the identity map in the bases $$B_2$$, $$B_1$$. It seems that $$S = M_{B_1}^{B_2}(I)$$. So indeed $$S = P^{-1}$$ because of the general identity $$M_{B_2}^{B_3}(T)M_{B_1}^{B_2}(S) = M_{B_1}^{B_3}(TS)$$, essentially by definition of matrix multiplication.

Edit: Let $$V$$ and $$W$$ be finite dimensional vector spaces over field $$F$$ and $$T \colon V \to W$$ be a linear map. Suppose $$B_1 = \{v_1, \dots, v_n\}$$ is a basis of $$V$$ and $$B_2 = \{w_1, \dots, w_m\}$$ is a basis of $$W$$. Then $$A = M_{B_1}^{B_2}(T)$$ is a $$m \times n$$ matrix with entries in $$F$$ where $$Tv_j = \sum_{i = 1}^{m}a_{ij}w_j$$. That is, the $$j$$th column of $$A$$ is the coordinates of $$Tv_j$$ in the basis $$\{w_1, \dots, w_m\}$$.

• What is the notation $M^{B_1}_{B_2}$ that you use? Commented Aug 5, 2021 at 21:38
• @RamanAliakseyeu I defined the notation. The notation might not be standard, but the matrix itself is standard; $A$ is usually called the matrix representation of $T$. Commented Aug 5, 2021 at 21:47