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Permutations and Determinants ... Walschap, Theorem 1.3.1 ... ... Another question ...

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,887
Hobart, Tasmania
I am reading Gerard Walschap's book: "Multivariable Calculus and Differential Geometry" and am focused on Chapter 1: Euclidean Space ... ...

I have a further question concerning an aspect of the proof of Theorem 1.3.1 ...

Theorem 1.3.1 and its proof read as follows:


Walschap - 1 -  Theorem 1.3.1 & Proof ... PART 1 ... .png
Walschap - 2 -  Theorem 1.3.1 & Proof ... PART 2 ... .png


Equation 1.3.1 in the proof implies that


\(\displaystyle \text{det} ( v_1, \cdot \cdot \cdot , v_n ) = \text{det} \left( \ \begin{bmatrix} a_{ 11} \\ \ \\ . \\ . \\ . \\ \ \\ a_{n1} \end{bmatrix} , \cdot \cdot \cdot , \begin{bmatrix} a_{ 1n} \\ \ \\ . \\ . \\ . \\ \ \\ a_{nn} \end{bmatrix} \ \right) \)


so ... in other words ...

\(\displaystyle v_j = \begin{bmatrix} a_{ 1j} \\ \ \\ . \\ . \\ . \\ \ \\ a_{nj} \end{bmatrix}\)


Now I can see that \(\displaystyle v_j = a_{ 1j} e_1 + a_{ 2j} e_2 + \cdot \cdot \cdot + a_{nj} e_n \)


... and believe that to represent \(\displaystyle v_j\) as \(\displaystyle \begin{bmatrix} a_{ 1j} \\ \ \\ . \\ . \\ . \\ \ \\ a_{nj} \end{bmatrix}\)


... is just a convention to represent \(\displaystyle v_j\) by its coordinates with respect to the current basis ... ...


Can someone please confirm that this is the correct interpretation of the left hand side of equation 1.3.1 in the proof ...


Hope someone can help ...

Peter
 
Last edited:

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,485
I think you are basically right. However, det is announced as a function from $(\mathbb{R}^n)^n$ to $\mathbb{R}$. That is, det takes an $n\times n$ matrix of real numbers, and each of the $n$ arguments is an element of $\mathbb{R}^n$ and can be viewed as a column of that matrix. So, $v_i$'s are columns of numbers to begin with, and not abstract vectors. However, the book does not seem to require that $(e_1,\ldots,e_n)$ is the standard basis; it can be an arbitrary basis of $\mathbb{R}^n$. That's how I see it.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,887
Hobart, Tasmania
I think you are basically right. However, det is announced as a function from $(\mathbb{R}^n)^n$ to $\mathbb{R}$. That is, det takes an $n\times n$ matrix of real numbers, and each of the $n$ arguments is an element of $\mathbb{R}^n$ and can be viewed as a column of that matrix. So, $v_i$'s are columns of numbers to begin with, and not abstract vectors. However, the book does not seem to require that $(e_1,\ldots,e_n)$ is the standard basis; it can be an arbitrary basis of $\mathbb{R}^n$. That's how I see it.

Thanks Evgeny ...

Just for interest ... after proving Theorem 1.3.1 Walschap defines a determinant as follows:


Walschap - Defn of Determinant ... Page 15 ... .png



Peter
 
Last edited: