# Is there a relationship between a matrix of vectors and a matrix of equations?

I’m just getting a more in depth intuition for Linear Algebra, and I’ve come across a couple uses of matrices that seem like there should be a relationship, but I’m not sure.

Thinking about matrices as columns of vectors that determine a transformation from the basis, we can insert vectors $$\begin{bmatrix}1 \\ 3 \end{bmatrix}$$ and $$\begin{bmatrix}5 \\ 7 \end{bmatrix}$$ into a matrix:

$$\begin{bmatrix} 1 & 5 \\ 3 & 7 \end{bmatrix}$$

If $$\hat{i}$$ is the $$x$$-axis unit vector and $$\hat{j}$$ is the $$y$$-axis unit vector then $$\begin{bmatrix}1 \\ 3 \end{bmatrix} = \ \hat{i} + 3\hat{j} \quad \text{and} \quad \begin{bmatrix}5 \\ 7 \end{bmatrix} = \ 5\hat{i} + 7\hat{j}$$ so any vector in the space can have its coordinates determined after the transformation.

But then we can also use matrices as a way to express systems of linear equations. So if we have:

\begin{align} x + 5y &= 0 \\ 3x + 7y &= 0 \end{align}

And they can be expressed as:

$$\begin{bmatrix}1 & 5 \\ 3 & 7 \end{bmatrix} \begin{bmatrix}x \\ y \end{bmatrix} = \begin{bmatrix}0 \\ 0 \end{bmatrix}$$

Is there some relationship between the first transformation matrix and this system of linear equations with the same coefficients values?

• Yes, and you've just written it down: you can turn a bunch of vectors into a matrix $A$, and that matrix can be used to write down the system of linear equations $Ax = 0$. Sep 26, 2017 at 22:56
• @QiaochuYuan but how do those new equations relate to the original vectors? You end up merging the $x$ coefficients into one equation and the $y$ coefficients into another. What does that mean graphically? Sep 26, 2017 at 23:06
• The columns of the matrix $A$ are the vectors $A e_1, A e_2$, etc. where $e_i$ is the vector which is $0$ everywhere except for the $i^{th}$ entry, where it is $1$. The more general question of whether $Ax = b$ is the question of whether some linear combination of these columns is equal to $b$. Sep 26, 2017 at 23:11

There are two fundamental and complementary ways to specify a subspace $$V$$ of $$\mathbb R^n$$. The first is as the span of some set of vectors $$\{v_1,\dots,v_k\}$$, i.e., as the set of all linear combinations $$\sum_{i=1}^kc_iv_i$$. In matrix form, this is $$\begin{bmatrix}v_1&\cdots&v_k\end{bmatrix}\begin{bmatrix}c_1\\\vdots\\c_k\end{bmatrix}$$ i.e., $$V$$ is the column space of the matrix with the $$v_i$$ as its columns. We can view the product $$Av$$ as specifying a particular element of $$A$$’s column space, namely, the linear combination of the columns of $$A$$ with the corresponding components of $$v$$ as coefficients.

The other way is to specify a set of vectors $$\{w_1,\dots,w_l\}$$ that span its orthogonal complement $$W=V^\perp$$. $$V$$ is then the subspace that consists of all the vectors that are orthogonal to every $$w_i$$. This condition can be expressed as the set of homogeneous linear equations \begin{align} w_1\cdot(x_1,\dots,x_n)&=0\\ w_2\cdot(x_1,\dots,x_n)&=0 \\ &\ \ \vdots \\ w_l\cdot(x_1,\dots,x_n)&=0. \end{align} In matrix form, this is $$\begin{bmatrix}w_1^T\\\vdots\\w_l^T\end{bmatrix}\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}.$$ This gives us another view of the product $$Av$$ as the vector of dot products of $$v$$ with the rows of $$A$$. The subspace $$V$$ is then the null space of the matrix that has the $$w_i$$ for its rows, and we can see that the row space and null space of a matrix are orthogonal complements.

We can connect the two views through the equation $$Ax=b$$. The row view shows us that this equation is equivalent to a system of linear equations involving dot products of the rows of $$A$$, which the column view tells us that the matrix equation, and so the system of linear equations, has a solution iff $$b$$ can be expressed as a linear combination of the columns of $$A$$, in other words, iff $$b$$ lies in the column space of $$A$$.

I’ve described the second method in terms of the standard basis of $$\mathbb R^n$$ and the Euclidean inner (dot) product, but it’s possible to express the same idea in a way that requires neither an inner product nor choice of basis. Instead of elements of $$V$$’s orthogonal complement, we can use elements of the dual space that annihilate $$V$$, i.e., linear functionals $$\varphi:\mathbb R^n\to\mathbb R$$ such that $$\varphi[v]=0$$ for $$v\in V$$.

\begin{align} x+ 5y &= 0 \\ 3x + 7y &= 0 \end{align}

Can be written as

$$x\begin{bmatrix} 1\\ 3 \end{bmatrix} + y\begin{bmatrix} 5\\ 7 \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$

The system of equations has a solution if and only if $$\begin{bmatrix} 0\\ 0 \end{bmatrix}$$ is in the span of $$\begin{bmatrix} 1\\ 3 \end{bmatrix}$$ and $$\begin{bmatrix} 5\\ 7 \end{bmatrix}$$.

Also note that $$\begin{bmatrix} 1 & 5 \end{bmatrix}$$ and $$\begin{bmatrix} 3 & 7 \end{bmatrix}$$ are elements of the row space of the matrix while $$\begin{bmatrix} 1 \\ 3 \end{bmatrix}$$ and $$\begin{bmatrix} 5 \\ 7 \end{bmatrix}$$ are elements of the column space of the matrix.