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<article class="md-content__inner md-typeset">
<h1 id="_1">矩阵运算<a class="headerlink" href="#_1" title="Permanent link">&para;</a></h1>
<p><em>矩阵运算是深度学习的计算引擎。本文涵盖矩阵加法、标量乘法、矩阵-向量积、矩阵乘法、逐元素运算、Kronecker积和广播——支撑每一次前向传播和梯度更新的运算。</em></p>
<ul>
<li>
<p>矩阵可以像向量一样进行加法和缩放。</p>
</li>
<li>
<p>加法要求两个矩阵维度相同，然后逐元素相加：</p>
</li>
</ul>
<div class="arithmatex">\[
\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 6 & 8 \\ 10 & 12 \end{bmatrix}
\]</div>
<ul>
<li>标量乘法将每个元素乘以标量：</li>
</ul>
<div class="arithmatex">\[
3 \times \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix}
\]</div>
<ul>
<li>矩阵能做的最简单的事情是乘以一个向量。<strong>矩阵-向量乘法</strong> <span class="arithmatex">\(A\mathbf{x}\)</span> 使用 <span class="arithmatex">\(\mathbf{x}\)</span> 的分量作为权重来组合 <span class="arithmatex">\(A\)</span> 的列：</li>
</ul>
<div class="arithmatex">\[
\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 \\ 6 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 3 \end{bmatrix} + 6 \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 17 \\ 39 \end{bmatrix}
\]</div>
<ul>
<li>
<p>这是机器学习中的核心运算。每个神经网络层都计算 <span class="arithmatex">\(A\mathbf{x} + \mathbf{b}\)</span>：矩阵乘以输入向量，再加上偏置。</p>
</li>
<li>
<p>一般情况是<strong>矩阵乘法</strong>。给定 <span class="arithmatex">\(A\)</span>（<span class="arithmatex">\(m \times n\)</span>）和 <span class="arithmatex">\(B\)</span>（<span class="arithmatex">\(n \times p\)</span>），乘积 <span class="arithmatex">\(C = AB\)</span> 是一个 <span class="arithmatex">\(m \times p\)</span> 矩阵，每个元素都是一个点积：</p>
</li>
</ul>
<div class="arithmatex">\[C_{ij} = \sum_{k=1}^{n} A_{ik} B_{kj}\]</div>
<ul>
<li>
<p>结果中的每个条目都是 <span class="arithmatex">\(A\)</span> 的一行与 <span class="arithmatex">\(B\)</span> 的一列的点积。内部维度必须匹配（<span class="arithmatex">\(n\)</span>），结果取外部维度（<span class="arithmatex">\(m \times p\)</span>）。</p>
</li>
<li>
<p>另一种理解方式：结果的每一列都是 <span class="arithmatex">\(A\)</span> 的列的<strong>加权和</strong>，其中权重来自 <span class="arithmatex">\(B\)</span> 的对应列。</p>
</li>
<li>
<p>如果 <span class="arithmatex">\(B\)</span> 的某一列为 <span class="arithmatex">\([2, 3]^T\)</span>，则结果列就是 <span class="arithmatex">\(2 \times (\text{A的第1列}) + 3 \times (\text{A的第2列})\)</span>。</p>
</li>
<li>
<p>一个有用的特例：矩阵与其转置相乘总是得到一个方阵。<span class="arithmatex">\(AA^T\)</span> 是 <span class="arithmatex">\(m \times m\)</span>，<span class="arithmatex">\(A^TA\)</span> 是 <span class="arithmatex">\(n \times n\)</span>：</p>
</li>
</ul>
<div class="arithmatex">\[
\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 14 & 32 \\ 32 & 77 \end{bmatrix}
\]</div>
<ul>
<li>
<p>矩阵乘法有重要的运算规则：</p>
<ul>
<li>
<p><strong>不满足交换律</strong>：通常 <span class="arithmatex">\(AB \neq BA\)</span>。顺序很重要。</p>
</li>
<li>
<p><strong>满足结合律</strong>：<span class="arithmatex">\((AB)C = A(BC)\)</span>。你可以任意分组乘法。</p>
</li>
<li>
<p><strong>满足分配律</strong>：<span class="arithmatex">\(A(B + C) = AB + AC\)</span>。</p>
</li>
<li>
<p><strong>单位矩阵</strong>：<span class="arithmatex">\(AI = IA = A\)</span>。</p>
</li>
</ul>
</li>
<li>
<p><strong>Hadamard积</strong>（逐元素乘积）将两个相同大小的矩阵逐项相乘，记作 <span class="arithmatex">\(A \odot B\)</span>：</p>
</li>
</ul>
<div class="arithmatex">\[
\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \odot \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 12 \\ 21 & 32 \end{bmatrix}
\]</div>
<ul>
<li>
<p>与标准矩阵乘法不同，Hadamard积满足交换律（<span class="arithmatex">\(A \odot B = B \odot A\)</span>），且要求两个矩阵维度相同。它在机器学习中广泛用于门控机制：通过与一个取值在0到1之间的掩码逐元素相乘，控制每个条目"通过"多少。</p>
</li>
<li>
<p>两个向量 <span class="arithmatex">\(\mathbf{u}\)</span> 和 <span class="arithmatex">\(\mathbf{v}\)</span> 的<strong>外积</strong>产生一个矩阵：<span class="arithmatex">\(\mathbf{u}\mathbf{v}^T\)</span>。每个条目是 <span class="arithmatex">\(\mathbf{u}\)</span> 的一个元素与 <span class="arithmatex">\(\mathbf{v}\)</span> 的一个元素的乘积：</p>
</li>
</ul>
<div class="arithmatex">\[
\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 4 & 5 \end{bmatrix} = \begin{bmatrix} 4 & 5 \\ 8 & 10 \\ 12 & 15 \end{bmatrix}
\]</div>
<ul>
<li>
<p>结果总是秩为1，因为每一行都是 <span class="arithmatex">\(\mathbf{v}^T\)</span> 的缩放版本。任何矩阵都可以写成秩-1外积之和，这正是SVD所做的事情（见分解章节）。</p>
</li>
<li>
<p>矩阵乘法的计算开销很大。两个 <span class="arithmatex">\(n \times n\)</span> 矩阵相乘需要 <span class="arithmatex">\(O(n^3)\)</span> 次运算。对于一个 <span class="arithmatex">\(1000 \times 1000\)</span> 的矩阵，那就是十亿次乘法。</p>
</li>
<li>
<p>当矩阵是<strong>稀疏的</strong>（大部分为零）时，朴素的乘法会浪费时间乘以零。<strong>压缩稀疏行（CSR）</strong>格式只存储非零元素及其位置：</p>
<ul>
<li><strong>值</strong>：按行顺序排列的非零条目</li>
<li><strong>列索引</strong>：每个值属于哪一列</li>
<li><strong>行偏移</strong>：每一行在值列表中的起始位置</li>
</ul>
</li>
<li>
<p>例如，矩阵：</p>
</li>
</ul>
<div class="arithmatex">\[
A = \begin{bmatrix} 5 & 0 & 0 & 2 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}
\]</div>
<ul>
<li>
<p>存储为：values = [5, 2, 3, -1], columns = [0, 3, 2, 3], row offsets = [0, 2, 3, 4]。这跳过了所有零，使稀疏运算快得多。</p>
</li>
<li>
<p>矩阵的一个核心用途是求解<strong>线性方程组</strong>。方程组 <span class="arithmatex">\(A\mathbf{x} = \mathbf{b}\)</span> 问的是："什么向量 <span class="arithmatex">\(\mathbf{x}\)</span> 被 <span class="arithmatex">\(A\)</span> 变换后，会得到 <span class="arithmatex">\(\mathbf{b}\)</span>？"</p>
</li>
<li>
<p>例如，假设你在买水果。苹果每个 <span class="arithmatex">\(x_1\)</span> 元，香蕉每个 <span class="arithmatex">\(x_2\)</span> 元。已知2个苹果和1个香蕉共5元，1个苹果和3个香蕉共10元。用矩阵形式表示：</p>
</li>
</ul>
<div class="arithmatex">\[
\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 5 \\ 10 \end{bmatrix}
\]</div>
<ul>
<li>矩阵逐行乘以向量（每一行与 <span class="arithmatex">\([x_1, x_2]^T\)</span> 点积）得到两个方程：</li>
</ul>
<div class="arithmatex">\[2x_1 + 1x_2 = 5 \qquad \text{(第1行)} \qquad \qquad x_1 + 3x_2 = 10 \qquad \text{(第2行)}\]</div>
<ul>
<li>
<p>从第1行得 <span class="arithmatex">\(x_2 = 5 - 2x_1\)</span>。代入第2行：<span class="arithmatex">\(x_1 + 3(5 - 2x_1) = 10\)</span>，解得 <span class="arithmatex">\(x_1 = 1\)</span>，则 <span class="arithmatex">\(x_2 = 3\)</span>。苹果每个1元，香蕉每个3元。</p>
</li>
<li>
<p>验证——结果正确：</p>
</li>
</ul>
<div class="arithmatex">\[
\begin{bmatrix} 2 & 1 \\ 1 & 3 \end{bmatrix} \begin{bmatrix} 1 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 + 3 \\ 1 + 9 \end{bmatrix} = \begin{bmatrix} 5 \\ 10 \end{bmatrix}
\]</div>
<ul>
<li>
<p>如果 <span class="arithmatex">\(A\)</span> 有逆矩阵，解就是简单的 <span class="arithmatex">\(\mathbf{x} = A^{-1}\mathbf{b}\)</span>。但直接计算逆矩阵代价高昂且数值不稳定。实践中我们使用分解方法。</p>
</li>
<li>
<p>并非所有矩阵都是方阵，也不是所有方阵都可逆。<strong>伪逆</strong> <span class="arithmatex">\(A^+\)</span> 将逆推广到任意矩阵。它总是存在，并提供"尽可能好的"逆：</p>
</li>
</ul>
<div class="arithmatex">\[A^+ = (A^TA)^{-1}A^T\]</div>
<ul>
<li>
<p>当 <span class="arithmatex">\(A\)</span> 是下三角矩阵时，通过<strong>前向代入</strong>求解 <span class="arithmatex">\(L\mathbf{x} = \mathbf{b}\)</span> 很容易：先解出 <span class="arithmatex">\(x_1\)</span>，然后用它求出 <span class="arithmatex">\(x_2\)</span>，依此类推。</p>
</li>
<li>
<p>当 <span class="arithmatex">\(A\)</span> 是上三角矩阵时，通过<strong>回代</strong>求解 <span class="arithmatex">\(U\mathbf{x} = \mathbf{b}\)</span>：先解出最后一个变量，然后向上求解。</p>
</li>
<li>
<p>这就是为什么将矩阵分解为三角因子（如分解章节所述）如此有用——它将一个难题转化为两个简单问题。</p>
</li>
</ul>
<h2 id="colabjupyter-notebook">编程练习（使用CoLab或Jupyter Notebook）<a class="headerlink" href="#colabjupyter-notebook" title="Permanent link">&para;</a></h2>
<ol>
<li>将两个矩阵相乘并验证维度。然后交换顺序，观察结果如何变化（或者，如果维度不匹配，运算失败）。</li>
</ol>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">jax.numpy</span><span class="w"> </span><span class="k">as</span><span class="w"> </span><span class="nn">jnp</span>
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a>
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a><span class="n">A</span> <span class="o">=</span> <span class="n">jnp</span><span class="o">.</span><span class="n">array</span><span class="p">([[</span><span class="mf">1.0</span><span class="p">,</span> <span class="mf">2.0</span><span class="p">],</span>
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a> <span class="p">[</span><span class="mf">3.0</span><span class="p">,</span> <span class="mf">4.0</span><span class="p">]])</span>
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a><span class="n">B</span> <span class="o">=</span> <span class="n">jnp</span><span class="o">.</span><span class="n">array</span><span class="p">([[</span><span class="mf">5.0</span><span class="p">,</span> <span class="mf">6.0</span><span class="p">],</span>
<a id="__codelineno-0-6" name="__codelineno-0-6" href="#__codelineno-0-6"></a> <span class="p">[</span><span class="mf">7.0</span><span class="p">,</span> <span class="mf">8.0</span><span class="p">]])</span>
<a id="__codelineno-0-7" name="__codelineno-0-7" href="#__codelineno-0-7"></a>
<a id="__codelineno-0-8" name="__codelineno-0-8" href="#__codelineno-0-8"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;A @ B:</span><span class="se">\n</span><span class="si">{</span><span class="n">A</span><span class="w"> </span><span class="o">@</span><span class="w"> </span><span class="n">B</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-0-9" name="__codelineno-0-9" href="#__codelineno-0-9"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;B @ A:</span><span class="se">\n</span><span class="si">{</span><span class="n">B</span><span class="w"> </span><span class="o">@</span><span class="w"> </span><span class="n">A</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-0-10" name="__codelineno-0-10" href="#__codelineno-0-10"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;Equal: </span><span class="si">{</span><span class="n">jnp</span><span class="o">.</span><span class="n">allclose</span><span class="p">(</span><span class="n">A</span><span class="w"> </span><span class="o">@</span><span class="w"> </span><span class="n">B</span><span class="p">,</span><span class="w"> </span><span class="n">B</span><span class="w"> </span><span class="o">@</span><span class="w"> </span><span class="n">A</span><span class="p">)</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
</code></pre></div>
<ol>
<li>求解线性方程组 <span class="arithmatex">\(A\mathbf{x} = \mathbf{b}\)</span>，并通过回代乘法验证解。尝试改变 <span class="arithmatex">\(\mathbf{b}\)</span>，观察解如何变化。</li>
</ol>
<div class="highlight"><pre><span></span><code><a id="__codelineno-1-1" name="__codelineno-1-1" href="#__codelineno-1-1"></a><span class="kn">import</span><span class="w"> </span><span class="nn">jax.numpy</span><span class="w"> </span><span class="k">as</span><span class="w"> </span><span class="nn">jnp</span>
<a id="__codelineno-1-2" name="__codelineno-1-2" href="#__codelineno-1-2"></a>
<a id="__codelineno-1-3" name="__codelineno-1-3" href="#__codelineno-1-3"></a><span class="n">A</span> <span class="o">=</span> <span class="n">jnp</span><span class="o">.</span><span class="n">array</span><span class="p">([[</span><span class="mf">2.0</span><span class="p">,</span> <span class="mf">1.0</span><span class="p">],</span>
<a id="__codelineno-1-4" name="__codelineno-1-4" href="#__codelineno-1-4"></a> <span class="p">[</span><span class="mf">5.0</span><span class="p">,</span> <span class="mf">3.0</span><span class="p">]])</span>
<a id="__codelineno-1-5" name="__codelineno-1-5" href="#__codelineno-1-5"></a><span class="n">b</span> <span class="o">=</span> <span class="n">jnp</span><span class="o">.</span><span class="n">array</span><span class="p">([</span><span class="mf">4.0</span><span class="p">,</span> <span class="mf">7.0</span><span class="p">])</span>
<a id="__codelineno-1-6" name="__codelineno-1-6" href="#__codelineno-1-6"></a>
<a id="__codelineno-1-7" name="__codelineno-1-7" href="#__codelineno-1-7"></a><span class="n">x</span> <span class="o">=</span> <span class="n">jnp</span><span class="o">.</span><span class="n">linalg</span><span class="o">.</span><span class="n">solve</span><span class="p">(</span><span class="n">A</span><span class="p">,</span> <span class="n">b</span><span class="p">)</span>
<a id="__codelineno-1-8" name="__codelineno-1-8" href="#__codelineno-1-8"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;Solution x: </span><span class="si">{</span><span class="n">x</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
<a id="__codelineno-1-9" name="__codelineno-1-9" href="#__codelineno-1-9"></a><span class="nb">print</span><span class="p">(</span><span class="sa">f</span><span class="s2">&quot;A @ x: </span><span class="si">{</span><span class="n">A</span><span class="w"> </span><span class="o">@</span><span class="w"> </span><span class="n">x</span><span class="si">}</span><span class="s2">&quot;</span><span class="p">)</span>
</code></pre></div>
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