1. identity transformation
Now let's find this particular boring transform \ (T (\ boldsymbol v) = \ boldsymbol v \) corresponding to the matrix. The identity transformation have nothing to do, the corresponding matrix is the identity matrix, and the input-yl group if the output of the same word.
If \ (T (\ boldsymbol v_J) = \ boldsymbol v_J = \ boldsymbol w_j \) , then the transformation matrix is \ (the I \) .
However, if the substrate is not the same, then \ (T (\ boldsymbol v_1) = \ boldsymbol v_1 \) will be \ (\ boldsymbol w \) a combination of \ (m_ {11} \ boldsymbol w_1 + \ cdots + m_ {n1 } \ boldsymbol w_n \) , i.e. the combination coefficient matrix \ (M \) of the first column.
Before and after the transform base vectors themselves changed but not changed, the input and output when the group is not the same when the transformation matrix is not an identity matrix.
2. wavelet transform to the wavelet base change =
Wavelets of different lengths and in different places, is not in fact a first wavelet basis vector, which is a useful constant vector. The following is an example of a wavelet:
These vectors are orthogonal, very good. You can see, \ (\ boldsymbol w_3 \) is positioned in the first half, while \ (\ boldsymbol w_4 \) positioned in the second half. Is intended to find a set of wavelet transform coefficients \ (c_1, c_2, c_3, c_4 \) to an input signal with a wavelet basis vector \ (V = (V_1, V_2, V_3, V_4) \) .
Coefficient \ (c_1 \) represents the mean and coefficient \ (c_3 \) and \ (c_4 \) , respectively, to tell us more about the current half and half. Why should we change the basis vectors it? May be considered \ (v_1, v_2, v_3, v_4 \) is the intensity of the signal, of course, a very small 4 figures may actually have \ (n-10,000 = \) . We want to compress the signal, leaving only 5% of the maximum coefficient, so that we can achieve 20: 1 compression ratio.
If we are to retain 5% of the standard base under coefficient, we will lose out 95% of the signal. However, if we select a better base set of 5%, and the basis vectors can be restored to the original signal very close. In the field of image processing and audio coding, you do not see not hear the difference, we do not need the other 95%.
在线性代数里，一切都是完美的，我们省略压缩的步骤。输出 \(\hat {\boldsymbol v}\) 和输入 \(\boldsymbol v\) 一模一样，变换得到 \(c=W^{-1}\boldsymbol v\)，重建过程则将我们带回到原点 \(\boldsymbol v=Wc\)。在真正的信号处理领域，没有什么是完美的但一切都很快，无损失的变换和只丢失不必要信息的压缩过程是成功的关键，我们有 \(\hat {\boldsymbol v}=W\hat c\)。
3. 傅里叶变换=改变到傅里叶基底
一个电气工程师对一个信号做的第一件事就是求它的傅里叶变换。针对有限的向量，我们要讨论的是离散傅里叶变换。离散傅里叶变换涉及到复数，但如果我们选择 \(n=4\)，矩阵非常小并且仅有的复数是 \(i\) 和 \(i^3=-i\)。
第一列仍然是常向量，代表信号均值或者直流分量。是一个频率为零的波，第三列则以最高的频率改变。傅里叶变换将信号分解成等间隔频率的波。傅里叶矩阵绝对是数学、科学和工程学领域最重要的复数矩阵，快速傅里叶变化通过加速傅里叶变化的过程，彻底改变了工业界。漂亮的是傅里叶矩阵和其逆矩阵非常像，改变一下符号即可。
4. 一点备忘
假设第一组基为 \(\boldsymbol v_1, \cdots, \boldsymbol v_n\)，第二组基为 \(\boldsymbol w_1, \cdots, \boldsymbol w_n\)，由 \(V\to W\) 的基变换矩阵为 \(M\)，那么我们有：
\[V=WM\]
一个向量 \(\boldsymbol s\) 在 \(V\) 和 \(W\) 下的坐标分别为 \(\boldsymbol x\) 和 \(\boldsymbol y\)，那么我们有：
\[\boldsymbol s = V\boldsymbol x=W\boldsymbol y \to M\boldsymbol x=\boldsymbol y\]
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