MATH 240

Mon. September 9th, 2019

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Why work in the abstract?

Many different types of problems can be represented as systems of linear equations, and many operations can be represented as linear transformations. By working in the abstract, we only have to learn a small number of techniques to work with this wide variety of equivalent problems.

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Ex. Every quadratic equation can be represented as $c_0+c_1x+c_2x^2$. The only thing that changes between them is the values of the coefficients, which can be represented as a vector $\begin{pmatrix} c_1\\c_2\\c_3 \end{pmatrix}.$ This implies that we can use linear algebra to answer questions about quadratic equations.

This idea can also be extended to polynomials of any degree. When working with cubics like $c_0+c_1x+c_2x^2+c_3x^3$, they can be represented by the vector $\begin{pmatrix} c_0\\c_1\\c_2\\c_3 \end{pmatrix}.$ If you wanted to represent only the quadratics within this vector space, you could use $\text{Span}\{\begin{pmatrix} 1\\0\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\1\\0\\0 \end{pmatrix},\begin{pmatrix} 0\\0\\1\\0 \end{pmatrix}\}.$
This set will only contain points where $c_3=0$, i.e. cubics where the coefficient of $x^3$ is 0, i.e. quadratics. This demonstrates how looking at span can be useful in a concrete sense.

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The question of whether $Ax=b$ has a solution is equivalent to a question regarding the span of the columns that make up $A$.

Key theorem from Friday:

Let $A$ be an $m \times n$ matrix. The following are equivalent:

• For each $b$ in $\R^n$, $Ax=b$ has a solution.
• Each $b$ in $\R^n$ is a linear combination of the columns of $A$.
• The columns of $A$ span $\R^n$.
• $A$ has a pivot position in every row.

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A homogeneous system of linear equations has the form $Ax=0.$ They are always consistent (always have at least one solution), since the trivial solution (where every variable equals 0) will always be a solution.

Ex.
$\begin{cases} 3x_1+5x_2-4x_3=0\\ -3x_1-2x_2+4x_3=0\\ 6x_1+x_2-8x_3=0 \end{cases}$
$\begin{aligned} &\begin{pmatrix} 3&5&-4&0\\ -3&-2&4&0\\ 6&1&-8&0 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 3&5&-4&0\\ 0&3&0&0\\ 0&-9&0&0 \end{pmatrix}\\ \rightarrow&\begin{pmatrix} 3&5&-4&0\\ 0&3&0&0\\ 0&0&0&0 \end{pmatrix}\\ \end{aligned}$
$\begin{aligned} 3x_2=0\rightarrow x_2&=0\\ 3x_1+5(0)-4x_3=0\rightarrow x_1&=\frac{4}{3}x_3 \end{aligned}$
$=\begin{pmatrix} \frac{4}{3}x_3\\0\\x_3 \end{pmatrix}=x_3\begin{pmatrix} \frac{4}{3}\\0\\1 \end{pmatrix}=\text{Span}\{\begin{pmatrix} \frac{4}{3}\\0\\1 \end{pmatrix}\}.$