# The inner product of column vectors is the same as matrix multiplication

I am very much new on the topic of inner product:

Definition. The inner product of vectors $$x, y \in \mathbb{R}^n$$ is $$\langle x, y\rangle =\sum_{i=1}^{n} x_ky_k=x_1y_1+x_2y_2+\dots+x_ny_n$$

I can't understand 3 properties of inner product:

(1) The inner product of column vectors is the same as matrix multiplication: $$\langle x, y\rangle=x^Ty=\begin{bmatrix} x_1 & x_2\dots x_n\\ \end{bmatrix}\begin{bmatrix} y_1\\ y_2\\ \dots\\ y_n \end{bmatrix}$$

What is the column vector here?

(2) The inner product satisfies the usual distributive rules of multiplication: $$\langle x,cy+dz\rangle =c\langle x, y\rangle + d\langle x, z\rangle$$ for all $$c, d \in \mathbb{R}$$ and $$x, y, z \in \mathbb{R}^n$$.

How is this distributive rule true? Is $$c+d =1?$$

(3) Let $$A$$ be a $$m\times n$$ matrix, let $$u \in \mathbb{R}^n$$ and let $$v \in \mathbb{R}^m$$. Then

$$\langle Au, v \rangle=\langle u, A^Tv \rangle$$.

I don't understand this point specially, would you elaborate this point by taking any matrix(A) and vectors(u, v) example .

• $\langle Au, v \rangle = (Au)^Tv$ by definition. Can you solve it from here? Commented Apr 23 at 15:45
• @CyclotomicField I need the example, by taking any matrix(A) and vectors(u, v) as an example
– user1290851
Commented Apr 23 at 15:48
• @CyclotomicField would you explain point 1?
– user1290851
Commented Apr 23 at 16:02
• Column vectors are matrices with a single column. The transpose of a column vector is a row vector, which is a matrix with a single row. So $v^Tu$ is a $1 \times n$ matrix times a $n \times 1$ matrix which gives you a $1 \times 1$ matrix. You can identify $1 \times 1$ matrices with scalars from the base field. Commented Apr 23 at 16:07
• @CyclotomicField in my question column vector is y(which is form of matrix) ?
– user1290851
Commented Apr 23 at 16:11

Both $$x$$ and $$y$$ are column vectors which are matrices of size $$n \times 1$$. The inner product is given by $$x^Ty$$ because $$x^T$$ is a $$1 \times n$$ row vector so multiplying $$x^Ty$$ gives a $$1 \times 1$$ matrix which we can identify as a scalar from the base field. Writing the inner product in this way allows you to use basic properties of matrices to reveal properties of the inner product. For example we can see that for some scalar $$a$$ that $$\langle x,ay \rangle = x^T(ay)=a(x^Ty) = (ax^T)y = \langle ax,y\rangle$$ and you can do a similar computation for $$\langle x,y+z \rangle = \langle x,y\rangle + \langle x,z\rangle$$ to prove (2) and for $$\langle Ax,y \rangle = \langle x,A^Ty\rangle$$ to prove (3). There form also motivates the name of the outer product given by $$xy^T$$ which is also highly useful. For example if $$u$$ is a unit vector then $$uu^T$$ is an orthogonal projection matrix onto the subspace spanned by $$u$$.