- #1
"Don't panic!"
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Hi all,
Just doing a bit of personal study on vector spaces and wanted to clear up my understanding on the following. This is my description of what I'm trying to understand, is it along the right lines? (apologies in advance, I am a physicist, not a pure mathematician, so there are most probably errors in the mathematical formalism I've used below):Given an n-dimensional vector space V and an ordered basis E= \left[\mathbf{v}_{i}\right]_{i=1,2,\ldots ,n}, one can represent any vector \mathbf{v}\in V with respect to this basis as a linear combination \qquad\qquad\qquad\qquad\qquad\qquad \mathbf{v}=\sum_{i=1}^{n}a_{i}\mathbf{v}_{i} where a_{i} are the components of \mathbf{v} with respect to the basis vectors \mathbf{v}_{i}.We can define an isomorphism between V and \mathbb{R}^{n}, f:V\rightarrow\mathbb{R}^{n} such that, with respect to the basis E
\qquad\qquad\qquad\qquad\qquad\qquad \sum_{i=1}^{n}a_{i}\mathbf{v}_{i} \;\; \longmapsto \;\; \left(\begin{matrix}a_{1} \\ a_{2} \\ \vdots \\ a_{n}\end{matrix}\right) Denoting this isomorphism as
\qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v}\right]_{E} = \left(\begin{matrix}a_{1} \\ a_{2} \\ \vdots \\ a_{n}\end{matrix}\right) we observe that \qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v} +\mathbf{w} \right]_{E} = \left[\mathbf{v} \right]_{E} +\left[\mathbf{w} \right]_{E} \;\;\forall \,\mathbf{v},\mathbf{w} \in V and
\qquad\qquad\qquad\qquad\qquad\qquad \left[c\mathbf{v} \right]_{E} = c\left[\mathbf{v} \right]_{E} \;\;\forall \,\mathbf{v}\in V\;\;\text{and} \;\; c\in F where F is the underlying scalar field of V.
Given this, we have that \qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v}\right]_{E} = \left[\sum_{i=1}^{n}a_{i}\mathbf{v}_{i}\right]_{E} = \sum_{i=1}^{n}a_{i} \left[\mathbf{v}_{i}\right]_{E} = \left(\begin{matrix}a_{1} \\ a_{2} \\ \vdots \\ a_{n}\end{matrix}\right) which implies that \qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v}_{i}\right]_{E} = \left(\begin{matrix}0 \\ \vdots \\ 1 \\ \vdots \\ 0\end{matrix}\right) where the only non-zero component is the i^{th} component with a value of unity. Hence, it can be seen that, using such an isomorphism, one can make the basis vectors of a given n-dimensional ordered basis, resemble the standard basis in \mathbb{R}^{n}. This is true for all such n-dimensional ordered bases for a given vector space V.
I am currently working my way through Nadir Jeevanjee's book "An Introduction to Tensors & Group Theory for Physicists" and in it he mentions that "given an ordered basis, it is always possible to represent the basis vectors, in the basis that they define, such that they resemble the standard basis" (in words close to that effect), and I'm just trying to make sense of the notion. (Please ignore the "change of basis" part of the title. I initially got a little over enthusiastic and subsequently realized discussing both in one thread might be a little too long-winded).
Just doing a bit of personal study on vector spaces and wanted to clear up my understanding on the following. This is my description of what I'm trying to understand, is it along the right lines? (apologies in advance, I am a physicist, not a pure mathematician, so there are most probably errors in the mathematical formalism I've used below):Given an n-dimensional vector space V and an ordered basis E= \left[\mathbf{v}_{i}\right]_{i=1,2,\ldots ,n}, one can represent any vector \mathbf{v}\in V with respect to this basis as a linear combination \qquad\qquad\qquad\qquad\qquad\qquad \mathbf{v}=\sum_{i=1}^{n}a_{i}\mathbf{v}_{i} where a_{i} are the components of \mathbf{v} with respect to the basis vectors \mathbf{v}_{i}.We can define an isomorphism between V and \mathbb{R}^{n}, f:V\rightarrow\mathbb{R}^{n} such that, with respect to the basis E
\qquad\qquad\qquad\qquad\qquad\qquad \sum_{i=1}^{n}a_{i}\mathbf{v}_{i} \;\; \longmapsto \;\; \left(\begin{matrix}a_{1} \\ a_{2} \\ \vdots \\ a_{n}\end{matrix}\right) Denoting this isomorphism as
\qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v}\right]_{E} = \left(\begin{matrix}a_{1} \\ a_{2} \\ \vdots \\ a_{n}\end{matrix}\right) we observe that \qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v} +\mathbf{w} \right]_{E} = \left[\mathbf{v} \right]_{E} +\left[\mathbf{w} \right]_{E} \;\;\forall \,\mathbf{v},\mathbf{w} \in V and
\qquad\qquad\qquad\qquad\qquad\qquad \left[c\mathbf{v} \right]_{E} = c\left[\mathbf{v} \right]_{E} \;\;\forall \,\mathbf{v}\in V\;\;\text{and} \;\; c\in F where F is the underlying scalar field of V.
Given this, we have that \qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v}\right]_{E} = \left[\sum_{i=1}^{n}a_{i}\mathbf{v}_{i}\right]_{E} = \sum_{i=1}^{n}a_{i} \left[\mathbf{v}_{i}\right]_{E} = \left(\begin{matrix}a_{1} \\ a_{2} \\ \vdots \\ a_{n}\end{matrix}\right) which implies that \qquad\qquad\qquad\qquad\qquad\qquad \left[\mathbf{v}_{i}\right]_{E} = \left(\begin{matrix}0 \\ \vdots \\ 1 \\ \vdots \\ 0\end{matrix}\right) where the only non-zero component is the i^{th} component with a value of unity. Hence, it can be seen that, using such an isomorphism, one can make the basis vectors of a given n-dimensional ordered basis, resemble the standard basis in \mathbb{R}^{n}. This is true for all such n-dimensional ordered bases for a given vector space V.
I am currently working my way through Nadir Jeevanjee's book "An Introduction to Tensors & Group Theory for Physicists" and in it he mentions that "given an ordered basis, it is always possible to represent the basis vectors, in the basis that they define, such that they resemble the standard basis" (in words close to that effect), and I'm just trying to make sense of the notion. (Please ignore the "change of basis" part of the title. I initially got a little over enthusiastic and subsequently realized discussing both in one thread might be a little too long-winded).
