# Prove that $(A + B) \mathbf{v} = A\mathbf{v} + B\mathbf{v}$ Using the Matrix Components

## Problem 635

Let $A$ and $B$ be $n \times n$ matrices, and $\mathbf{v}$ an $n \times 1$ column vector.

Use the matrix components to prove that $(A + B) \mathbf{v} = A\mathbf{v} + B\mathbf{v}$.

## Solution.

We will use the matrix components $A = (a_{i j})_{1 \leq i, j \leq n}$, $B = (b_{i j})_{1 \leq i , j \leq n }$, and $\mathbf{v} = (v_i)_{1 \leq i \leq n}$.

Then
\begin{align*}
(A + B) \mathbf{v} = \begin{bmatrix} \sum_{j=1}^n (a_{1 j} + b_{1 j} ) v_j \3pt] \sum_{j=1}^n (a_{2 j} + b_{2 j} ) v_j \\[3pt]\sum_{j=1}^n (a_{3 j} + b_{3 j} ) v_j \\ \vdots \\ \sum_{j=1}^n (a_{n j} + b_{n j} ) v_j \end{bmatrix}. \end{align*} On the other hand, \[A\mathbf{v} + B\mathbf{v} = \begin{bmatrix} \sum_{j=1}^n a_{1 j} v_j \\[3pt] \sum_{j=1}^n a_{2 j} v_j \\[3pt] \sum_{j=1}^n a_{3 j} v_j \\ \vdots \\ \sum_{j=1}^n a_{n j} v_j \end{bmatrix} + \begin{bmatrix} \sum_{j=1}^n b_{1 j} v_j \\[3pt] \sum_{j=1}^n b_{2 j} v_j \\[3pt] \sum_{j=1}^n b_{3 j} v_j \\ \vdots \\ \sum_{j=1}^n b_{n j} v_j \end{bmatrix} = \begin{bmatrix} \sum_{j=1}^n (a_{1 j} + b_{1 j} ) v_j \\[3pt] \sum_{j=1}^n (a_{2 j} + b_{2 j} ) v_j \\[3pt] \sum_{j=1}^n (a_{3 j} + b_{3 j} ) v_j \\ \vdots \\ \sum_{j=1}^n (a_{n j} + b_{n j} ) v_j \end{bmatrix}.

We can see that they are the same.

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