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I tried to learn dual vectors online but failed to exactly understand it, I know that it could be understood using change of basis. Below is a example for change of basis, kindly help me with this. Let V be a space over $\Bbb{R}^3$ and the basis be $\begin{Bmatrix} \begin{bmatrix}1\\1\\0\\ \end{bmatrix},& \begin{bmatrix}0\\2\\0\\ \end{bmatrix},& \begin{bmatrix}1\\0\\1\\ \end{bmatrix} \end{Bmatrix}$ and U be another space over $\Bbb{R}^3$ with basis $\begin{Bmatrix} \begin{bmatrix}5\\0\\3\\ \end{bmatrix},& \begin{bmatrix}2\\3\\4\\ \end{bmatrix},& \begin{bmatrix}1\\6\\2\\ \end{bmatrix} \end{Bmatrix}$, $\begin{bmatrix}3\\5\\7\\ \end{bmatrix}$ is a vector in U and we want this vector's coefficient in V. The approach would be:

$$c_1\left[\begin{matrix}1\\1\\0\\ \end{matrix}\right]+ c_2\left[\begin{matrix}0\\2\\0\\ \end{matrix}\right]+ c_3\left[\begin{matrix}1\\0\\1\\ \end{matrix}\right] = 3\left[\begin{matrix}5\\0\\3\\ \end{matrix}\right]+ 5\left[\begin{matrix}2\\3\\4\\ \end{matrix}\right]+ 7\left[\begin{matrix}1\\6\\2\\ \end{matrix}\right] $$

Where $\mathbf{c}$ are coefficients of $V$'s basis.

$$\left[\begin{matrix}1 & 0 &1 \\ 1&2&0 \\0& 0 & 1\\ \end{matrix}\right] \left[ \begin{matrix}c_1\\c_2\\c_3\\ \end{matrix}\right]= \left[\begin{matrix}5 &2&1\\ 0&3&6 \\3& 4 & 2\\ \end{matrix}\right] \left[ \begin{matrix}3\\5\\7\\ \end{matrix}\right] $$

$$ \left[ \begin{matrix}c_1\\c_2\\c_3\\ \end{matrix}\right]= \left[\begin{matrix}1 & 0 &1 \\ 1&2&0 \\0& 0 & 1\\ \end{matrix}\right]^{-1}\left[\begin{matrix}5 &2&1\\ 0&3&6 \\3& 4 & 2\\ \end{matrix}\right] \left[ \begin{matrix}3\\5\\7\\ \end{matrix}\right]= \left[ \begin{matrix}-11\\34\\43\\ \end{matrix}\right] $$

Now I want to know from above example:

  1. What is/are dual vector?
  2. What is/are covector?
  3. What is dual space?

Let me know if I have understood it completely wrong. Thanks

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    $\begingroup$ The dual space of a vector space $V$ is the spaceof linear functional on $V$ and has nothing to do with a change of basis in $V$. $\endgroup$ – Emilio Novati Sep 22 '15 at 12:29
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    $\begingroup$ dual vectors and covectors are the same. $\endgroup$ – tired Sep 22 '15 at 12:31
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    $\begingroup$ Covectors are just linear functions whose input is a vector and whose output is a number. If your vectors are the space of $3\times 1$ column matrices, then the dual space (space of covectors) is the $1\times 3$ column matrices. Then when you left multiply a vector (column matrix) by a covector (row matrix), you get a $1\times 1$ matrix. I.e. a number. $\endgroup$ – got it--thanks Sep 22 '15 at 12:32
  • $\begingroup$ @EmilioNovati i'll have a closer look at linear functional and then i'll return. $\endgroup$ – pkj Sep 22 '15 at 12:37
  • $\begingroup$ For what reason are you asking about dual vectors in the context of the change of basis problem you've written? Are you being asked something else (for instance, "Find the coefficients $c_1, c_2, c_3$ without using matrix inversion, but by using the dual basis of $V$ instead")? $\endgroup$ – Muphrid Sep 23 '15 at 2:42
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Given a vector space $V$ over field $K,$ the dual space of $V,$ denoted $V^\ast,$ is the set of all linear maps $\varphi:V\to K.$ For example, consider $V=\mathbb{R}\times\mathbb{R}$ with addition $+:V\times V\to V$ defined as: $$\begin{bmatrix}a \\ b\end{bmatrix}+\begin{bmatrix}c \\ d\end{bmatrix}=\begin{bmatrix}a+c \\ b+d\end{bmatrix}$$ and vector multiplication $\cdot:K\times V\to V$ defined as $$k\cdot\begin{bmatrix}a \\ b\end{bmatrix}=\begin{bmatrix}k\cdot a \\ k\cdot b\end{bmatrix}$$ The $+$ and $\cdot$ inside represent field addition and multiplication, but I'll just use the same symbol because I'm lazy. Define a linear mapping $\varphi:V\to K$ as $\varphi\!\left(\begin{bmatrix}a\\b\end{bmatrix}\right)=a+b.$ One can show that this function adds linearly: $$\varphi\!\left(\begin{bmatrix}a\\b\end{bmatrix}\right)+\varphi\!\left(\begin{bmatrix}c\\d\end{bmatrix}\right)=a+b+c+d=a+c+b+d=\varphi\!\left(\begin{bmatrix}a+c\\b+d\end{bmatrix}\right)$$ and that it scales linearly: $$k\cdot\varphi\!\left(\begin{bmatrix}a\\b\end{bmatrix}\right)=k\cdot\left(a+b\right)=k\cdot a+k\cdot b=\varphi\!\left(\begin{bmatrix}k\cdot a\\k\cdot b\end{bmatrix}\right).$$ Since it's linear, it's said to be a co-vector. Thus, one can show that this is equivalent to the "row vector" $\begin{bmatrix}1 & 1\end{bmatrix}$ using matrix multiplication.$\begin{bmatrix}1 & 1\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}=a+b.$ When this collection of maps comes equipped with addition and multiplication that behaves linearly, the set itself also becomes a vector space over $K$ called the vector dual space that contains dual vectors.

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