[Mathematical knowledge] Coordinate base representation of vectors, Matlab code verification

serial number	content
1	[Mathematical knowledge] Coordinate base representation of vectors, Matlab code verification
2	[Mathematical knowledge] Inner product of vector and base, Matlab code verification

1. Coordinate base representation of vector

Suppose there is a vector $\vec{a}$, in different coordinate systems (or coordinate bases), vector $\vec{a}$Represented by different coordinate values.

When the coordinate base is uniquely determined, the corresponding coordinate values ​​are also uniquely determined. At the same time, vectors can also be represented by a linear combination of coordinate values ​​and coordinate lines.

• When the vector is a two-dimensional plane vector, it can be expressed as
  $$\vec{a}=a_x{\vec{e}}_1+a_y{\vec{e}}_2=\left[\begin{array}{cc}{\vec{e}}_1& {\vec{e}}_2\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]$$

• When the vector is a vector in a three-dimensional space, it can be expressed as
  $$\vec{a}=a_x{\vec{e}}_1+a_y{\vec{e}}_2+a_z{\vec{e}}_3=\left[\begin{array}{ccc}{\vec{e}}_1& {\vec{e}}_2& {\vec{e}}_3\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\\ a_z\end{array}\right]$$

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2. Example of two-dimensional plane vector

The following is an example based on a vector in a two-dimensional plane, but the case in three-dimensional space has the same properties and conclusions.

Suppose there is a two-dimensional plane vector $\vec{a}$, in the standard coordinate base ${\vec{e}}_1=\left[\begin{array}{c}1\\ 0\end{array}\right],{\vec{e}}_2=\left[\begin{array}{c}0\\ 1\end{array}\right]$ The coordinate value under is$\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$ . Then this vector can be expressed as

$$\begin{array}{rl}\vec{a}& =a_x{\vec{e}}_1+a_y{\vec{e}}_2=\left[\begin{array}{cc}{\vec{e}}_1& {\vec{e}}_2\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\\ & =3\left[\begin{array}{c}1\\ 0\end{array}\right]+4\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}3\\ 4\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]\end{array}$$

Now, we change the coordinate base to ${\vec{e}}_{1^{\prime }}=\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right],{\vec{e}}_{2^{\prime }}=\left[\begin{array}{c}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]$ , the coordinate value under this new basis is$\left[\begin{array}{c}{a}_{x^{\prime }}\\ a_{y^{\prime }}\end{array}\right]=\left[\begin{array}{c}\frac{7}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]$ . Then this vector can be expressed as

$$\begin{array}{rl}\vec{a}& =\frac{7}{\sqrt{2}}\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]+\frac{1}{\sqrt{2}}\left[\begin{array}{c}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{c}\frac{7}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]\end{array}$$

From the above example, we can see that no matter which coordinate base is used, there will always be the following equation

$$\begin{array}{rl}\vec{a}& =a_x{\vec{e}}_1+a_y{\vec{e}}_2=\left[\begin{array}{cc}{\vec{e}}_1& {\vec{e}}_2\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\\ & =a_{x^{\prime }}{\vec{e}}_{1^{\prime }}+a_{y^{\prime }}{\vec{e}}_{2^{\prime }}=\left[\begin{array}{cc}{\vec{e}}_{1^{\prime }}& {\vec{e}}_{2^{\prime }}\end{array}\right]\left[\begin{array}{c}{a}_{x^{\prime }}\\ a_{y^{\prime }}\end{array}\right]\end{array}$$

Similar conclusions can be drawn for vectors in three-dimensional space.

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As for how to obtain the coordinate values ​​​​under the new basis, we can achieve it through the following steps

$$\begin{array}{rl}\left[\begin{array}{cc}{\vec{e}}_1& {\vec{e}}_2\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]& =\left[\begin{array}{cc}{\vec{e}}_{1^{\prime }}& {\vec{e}}_{2^{\prime }}\end{array}\right]\left[\begin{array}{c}{a}_{x^{\prime }}\\ a_{y^{\prime }}\end{array}\right]\\ {\left[\begin{array}{cc}{\vec{e}}_{1^{\prime }}& {\vec{e}}_{2^{\prime }}\end{array}\right]}^{-1}\left[\begin{array}{cc}{\vec{e}}_1& {\vec{e}}_2\end{array}\right]\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]& =\left[\begin{array}{c}{a}_{x^{\prime }}\\ a_{y^{\prime }}\end{array}\right]\\ {\left[\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right]}^{-1}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}3\\ 4\end{array}\right]& =\left[\begin{array}{c}\frac{7}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]\end{array}$$

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3. Matlab code verification

a_x = 3;
a_y = 4;

e_1 = [ 1
        0];
e_2 = [ 0
        1];

a_x_prime = 7/sqrt(2);
a_y_prime = 1/sqrt(2);

e_1_prime = [ sqrt(2)/2
              sqrt(2)/2];
e_2_prime = [-sqrt(2)/2
              sqrt(2)/2];

>> pinv([e_1_prime  e_2_prime]) * [e_1  e_2] * [a_x; a_y]
ans =
    4.9497
    0.7071

>> a_x_prime
ans =
    4.9497

>> a_y_prime
ans =
    0.7071

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Ref

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