Linear Algebra pre-requisites

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We say that \(\mathbf{b}_{\mathbf{1}}, \ldots, \mathbf{b}_{\mathbf{n}}\) are linearly independent if \(
c_1 \mathbf{b}_{\mathbf{1}}+c_2 \mathbf{b}_{\mathbf{2}}+\ldots+c_n \mathbf{b}_{\mathbf{n}}=d_1 \mathbf{b}_{\mathbf{1}}+d_2 \mathbf{b}_{\mathbf{2}}+\ldots+d_n \mathbf{b}_{\mathbf{n}} \text { implies } c_1=d_1, c_2=d_2 \ldots, c_n=d_n\) or, equivalently,\(\mathbf{b}_{\mathbf{1}}, \ldots, \mathbf{b}_{\mathbf{n}}\) are linearly independent if \(
c_1 \mathbf{b}_{\mathbf{1}}+c_2 \mathbf{b}_{\mathbf{2}}+\ldots+c_n \mathbf{b}_{\mathbf{n}}=0 \text { implies } c_1=c_2=\ldots =c_n=0 \text {. }
\)

Example 1: Lets look at the vectors \(\mathbf{b}_{\mathbf{1}}=(1,0,0), \mathbf{b}_{\mathbf{2}}=(0,1,0), \mathbf{b}_{\mathbf{3}}=(0,0,1)\) . Observe that if \( a\mathbf{b}_{\mathbf{1}} + b\mathbf{b}_{\mathbf{2}} + c\mathbf{b}_{\mathbf{3}} = (1,2,3), \text{ then } a=1, b=2, \text{ and } c=3 \) and there is no other values for \(a,b, \text{ and } c \) to give the vector \((1,2,3)\). Next let's for the same values of \(\mathbf{b}_{\mathbf{1}}, \mathbf{b}_{\mathbf{2}}, \text{ and }\mathbf{b}_{\mathbf{3}}\), the only values of \(a,b, \text{ and } c \) such that \( a\mathbf{b}_{\mathbf{1}} + b\mathbf{b}_{\mathbf{2}} + c\mathbf{b}_{\mathbf{3}} = (0,0,0)\) is \(a=0, b=0, \text{ and } c=0. \) Therefore vectors \(\mathbf{b}_{\mathbf{1}}=(1,0,0), \mathbf{b}_{\mathbf{2}}=(0,1,0), \mathbf{b}_{\mathbf{3}}=(0,0,1)\) are linearly independent.

Example 2: \(\mathbf{b}_{\mathbf{1}}=1, \mathbf{b}_{\mathbf{2}}=t, \mathbf{b}_{\mathbf{3}}=t^2\).
\[
1+2 t+3 t^2 \neq 1+2 t+4 t^2 .
\]

If \(a+b t+c t^2=1+2 t+3 t^2\) then \(a=1, b=2, c=3\).

Application: Partial Fractions
\[
\frac{4}{\left(x^2+1\right)(x-3)}=\frac{A x+B}{x^2+1}+\frac{C}{x-3}
\]

If you don't like denominators, get rid of them:
\[
\begin{aligned}
& 4=(A x+B)(x-3)+C\left(x^2+1\right) \\
4= & A x^2+B x-3 A x-3 B+C x^2+C \\
4= & (A+C) x^2+(B-3 A) x-3 B+C
\end{aligned}
\]
I.e., \(0 x^2+0 x+4=(A+C) x^2+(B-3 A) x-3 B+C\)

Thus \(0=A+C, \quad 0=B-3 A, \quad 4=-3 B+C\).
\[
C=-A, \quad B=3 A, 4=-3(3 A)+-A \Rightarrow 4=-10 A \text {. }
\]

Hence \(A=-\frac{2}{5}, B=3\left(-\frac{2}{5}\right)=-\frac{6}{5}, C=\frac{2}{5}\).

Thus,
\[
\begin{aligned}
\frac{4}{\left(x^2+1\right)(x-3)} & =\frac{-\frac{2}{5} x-\frac{6}{5}}{x^2+1}+\frac{\frac{2}{5}}{x-3} \\
& =\frac{-2 x-6}{5\left(x^2+1\right)}+\frac{2}{5(x-3)}
\end{aligned}
\]

Alternatively, can plug in \(x=3\) to quickly find \(C\) and then solve for \(A\) and \(B\). We can also use matrices to solve linear equations.

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