 ### Hongyi Li

I am a PhD student in the University of Macau, supervised by Prof Hongcai Zhang and Prof Yonghua Song. As a participant of UM-Imperial College London 1+3 PhD program, I am currently studying at Imperial as an MSc student. At Imperial, I work closely with Dr Papadaskalopoulos Dimitrios. My research interest includes smart grid technologies, consensus-based coordination and distributed optimization.

# Lecture 01：The Geometry of Linear Equations

The fundamental problem of linear algebra is to solve n linear equations in n unknowns, for example:

\begin{aligned} 2x-y\,=\,0\\ -x+2y\,=\,3 \end{aligned}

In the first lecture, Dr. Strang show us three ways to view this problem.

## Row Picture

Plot the points that satisfied each equation. The intersection of the plots (if they do intersect) is the solution of the system of above equations, which is $x\,=\,1, y\,=\,2$. Figure 1: The plots of the above equations intersect at the point (1,2)

We substitute (1,2) into the original system of equations to check it’s validity:

\begin{aligned} 2\cdot1-2\,=\,0\\ -1+2\cdot2\,=\,3 \end{aligned}

Similarly, the solution of a three dimensional system is the common intersection of those three planes (if there does exist one).

## Column Picture

In the column picture, we rewrite the system as a single equation by turning the coefficients in the column of the system into vectors:

\begin{aligned} x \left[ \begin{array}{c} 2\\ 1 \end{array} \right] +y \left[ \begin{array}{c} -1\\ 2 \end{array} \right] = \left[ \begin{array}{c} 0\\ 3 \end{array} \right] \end{aligned}

Given two vectors c and d and scalars x and y, the sum xc+yd is called a linear combination of c and d, which is an important concept throughout Linear Algebra.

Geometrically, we are looking for a pair of x and y which satisfies that x copies of vector $\left[\begin{array}{c}2\\-1\end{array}\right]$ added to y copies of vector $\left[\begin{array}{c}-1\\2\end{array}\right]$ equals the vector $\left[\begin{array}{c}0\\3\end{array}\right]$. As shown in Figure 2, $x=1 \;\text{and}\; y=2$ agreeing with the result we got from row picture. Figure 2: A linear combination of the column vectors.

In the three dimensions, the column picture requires to find a linear combination of 3-dimensional vectors that equals to the vector b.

## Matrix Picture

Rewrite the system of equations as a single equation by using matrices and vectors:

\begin{aligned} \left[ \begin{array}{cc} 2 & -1\\ -1 & 2\\ \end{array} \right] \left[ \begin{array}{c} x\\ y \end{array} \right] =\left[ \begin{array}{c} 0\\ 3 \end{array} \right] \end{aligned}

The matrix $\left[\begin{array}{cc}2&-1\\-1&2\end{array}\right]$ is called the coefficient matrix. The vector $\bold{x}=\left[\begin{array}{c}x\\y\end{array}\right]$ is the vector of unknowns. The value on the right hand side of the equations form the vector b:

$A\bold{x}=\bold{b}$

The three dimensional matrix picture is similar to the two dimensional one, except that the vectors and matrices increase in size.

## Matrix Multiplication

How do we multiply a matrix A by a vector x?

$\left[ \begin{array}{cc} 2&5\\ 1&3 \end{array} \right] \left[ \begin{array}{c} 1\\2 \end{array} \right] =\;?$

The method Dr. Strang suggests is to think of the entries of x as the coefficients of a linear combination of the column vectors of the matrix:

$\left[ \begin{array}{cc} 2&5\\1&3 \end{array} \right] \left[ \begin{array}{c} 1\\2 \end{array} \right] =1\left[ \begin{array}{c} 2\\1 \end{array} \right] +2\left[ \begin{array}{c} 5\\3 \end{array} \right] =\left[ \begin{array}{c} 12\\7 \end{array} \right]$

The technique shows that Ax is a linear combination of the columns of A.

Also, you can calculate the product Ax by taking dot product of each row of A with the vector x:

$\left[ \begin{array}{cc} 2&5\\1&3 \end{array} \right] \left[ \begin{array}{c} 1\\2 \end{array} \right] =\left[ \begin{array}{c} 2\cdot1+5\cdot2\\1\cdot1+3\cdot2 \end{array} \right] =\left[ \begin{array}{c} 12\\7 \end{array} \right]$

## Linear Independence

In the column and matrix pictures, the right hand side of the equation is a vector b. Given a matrix A, if we can solve:

$A\bold{x}=\bold{b}$

for every possible b, we say that A is an inversible matrix, which means that the linear combinations of the column vectors fill the xy-plane (in two dimensional case). Otherwise, we say that A is a singular matrix, whose column vectors are linear dependent, or in other words, all linear combinations of those vectors lie on a point or line (in two dimensional case). In such case, the combinations don’t fill the whole space.