[Mathematical knowledge] Inner product of vector and base, 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. Inner product of vector and basis

Suppose there is a vector a ⃗ \vec{a} in a two-dimensional plane$\vec{a}$, which is in the coordinate base ${\vec{e}}_1,{\vec{e}}_2$The coordinate value below is $\left[\begin{array}{c}x\\ y\end{array}\right]$。

Let’s first look at the vector $\vec{a}$Self and coordinate base ${\vec{e}}_1$The inner product of. For the principle of inner product, please refer to the article [Mathematical Knowledge] Vector multiplication, inner product, outer product, matlab code implementation . Here we directly use its conclusion, that is, the inner product of a vector is, the projected length of one vector in the direction of another vector, multiplied by the length of the projected vector, as shown in the figure below

Described by the formula as

$$\vec{a}\cdot {\vec{e}}_1=\Vert \vec{a}\Vert \Vert {\vec{e}}_1\Vert \cos \left(\theta \right)$$

In our case, the projected vector is the basis vector ${\vec{e}}_1$, and the basis vector ${\vec{e}}_1$Its module length $\Vert {\vec{e}}_1\Vert$ is 1 1again$1$ , therefore

$$\begin{array}{rl}\vec{a}\cdot {\vec{e}}_1& =\Vert \vec{a}\Vert \Vert {\vec{e}}_1\Vert \cos \left(\theta \right)\\ & =\Vert \vec{a}\Vert \cos (\theta \end{array}$$

Numerically $\Vert \vec{a}\Vert \cos \left(\theta \right)$ is equal to the vector$\vec{a}$In the coordinate base ${\vec{e}}_1$coordinate values ​​on. If the coordinate base ${\vec{e}}_1$We think of it as the abscissa, then $\vec{a}\cdot {\vec{e}}_1$Numerically it is equal to the value of the abscissa, that is

$$\begin{array}{rl}{a}_x& =\vec{a}\cdot {\vec{e}}_1\end{array}$$

In the same way, we can also get $\vec{a}\cdot {\vec{e}}_2$Numerically equal to the value of the ordinate.

$$\begin{array}{rl}{a}_y& =\vec{a}\cdot {\vec{e}}_2\end{array}$$

Finally, the formulaic description conclusion is

$$\begin{array}{rl}{a}_x& =\vec{a}\cdot {\vec{e}}_1=\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\cdot \left[\begin{array}{c}{e}_{11}\\ e_{12}\end{array}\right]=a_x{e}_{11}+a_y{e}_{12}\\ a_y& =\vec{a}\cdot {\vec{e}}_2=\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\cdot \left[\begin{array}{c}{e}_{21}\\ e_{22}\end{array}\right]=a_x{e}_{21}+a_y{e}_{22}\end{array},\Vert {\vec{e}}_1\Vert =\Vert {\vec{e}}_2\Vert =1$$

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2. Example of two-dimensional plane vector

The following is an example based on a vector on a two-dimensional plane.

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$\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]=\left[\begin{array}{c}3\\ 4\end{array}\right]$。

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$\left[\begin{array}{c}{a}_{x^{\prime }}\\ a_{y^{\prime }}\end{array}\right]=\left[\begin{array}{c}\frac{7}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]$。

Verify the conclusion first

$$\begin{array}{rl}{a}_x& =\vec{a}\cdot {\vec{e}}_1=\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\cdot \left[\begin{array}{c}{e}_{11}\\ e_{12}\end{array}\right]=a_x{e}_{11}+a_y{e}_{12}\\ & =\left[\begin{array}{c}3\\ 4\end{array}\right]\cdot \left[\begin{array}{c}1\\ 0\end{array}\right]=3\times 1+4\times 0=3\end{array}$$

$$\begin{array}{rl}{a}_{x^{\prime }}& =\vec{a}\cdot {\vec{e}}_{1^{\prime }}=\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\cdot \left[\begin{array}{c}{e}_{11^{\prime }}\\ e_{12^{\prime }}\end{array}\right]=a_x{e}_{11^{\prime }}+a_y{e}_{12^{\prime }}\\ & =\left[\begin{array}{c}3\\ 4\end{array}\right]\cdot \left[\begin{array}{c}\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]=3\times \frac{1}{\sqrt{2}}+4\times \frac{1}{\sqrt{2}}=\frac{7}{\sqrt{2}}\end{array}$$

By observing the figure below, we can also roughly see the vector $\vec{a}$In the new basis ${\vec{e}}_{1^{\prime }}$The projection length on is $7/\sqrt{2}$。

This is also consistent with the effect in the coordinate chart.

Continue below to verify the conclusion

$$\begin{array}{rl}{a}_y& =\vec{a}\cdot {\vec{e}}_2=\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\cdot \left[\begin{array}{c}{e}_{21}\\ e_{22}\end{array}\right]=a_x{e}_{21}+a_y{e}_{22}\\ & =\left[\begin{array}{c}3\\ 4\end{array}\right]\cdot \left[\begin{array}{c}0\\ 1\end{array}\right]=3\times 0+4\times 1=4\end{array}$$

$$\begin{array}{rl}{a}_{y^{\prime }}& =\vec{a}\cdot {\vec{e}}_{2^{\prime }}=\left[\begin{array}{c}{a}_x\\ a_y\end{array}\right]\cdot \left[\begin{array}{c}{e}_{11^{\prime }}\\ e_{12^{\prime }}\end{array}\right]=a_x{e}_{11^{\prime }}+a_y{e}_{12^{\prime }}\\ & =\left[\begin{array}{c}3\\ 4\end{array}\right]\cdot \left[\begin{array}{c}-\frac{1}{\sqrt{2}}\\ \frac{1}{\sqrt{2}}\end{array}\right]=3\times (-\frac{1}{\sqrt{2}})+4\times \frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\end{array}$$

The second conclusion also means that the vector $\vec{a}$In the new basis ${\vec{e}}_{2^{\prime }}$The projection length on is $1/\sqrt{2}$。

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3. Code verification

a_x = 3;
a_y = 4;
a = [a_x
     a_y];

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

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

>> dot(a, e_1)
ans =
     3

>> dot(a, e_2)
ans =
     4

>> dot(a, e_1_prime)
ans =
    4.9497

>> dot(a, e_2_prime)
ans =
    0.7071

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Ref

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