Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 18 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 3 / 2 / a / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 34 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 3 / 5 / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 45 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 55 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 71 / 4 / 1 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 71 / 6 / Solution

Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 71 / 7 / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 2 / 2B / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / ia / Paper 2 / 5B / Solution

Past exam of the mathematics course of the University of Cambridge / 2005 / ib / Paper 1 / 9C / Solution

{wiki=Basis_(linear_algebra)}

= Vector space basis
{synonym}

= Bases
{synonym}

= Hamel basis
{c}
{synonym}

A basis is a <linearly independent> <spanning set> of <vectors> in a <vector space>. Every <vector> is a unique finite <linear combination> of <basis vectors>. Independence supplies uniqueness, and spanning supplies existence. An <ordered basis> turns the <coefficients> into a <coordinate vector>.

For $\mathcal B=(b_1,b_2)=((2,1),(1,2))$, the <matrix> with these <vectors> as columns is
$$
B=\begin{pmatrix}2&1\\1&2\end{pmatrix},\qquad
B^{-1}=\frac13\begin{pmatrix}2&-1\\-1&2\end{pmatrix}.
$$
Hence the <coordinate vector> of $v=(5,4)$ is $[v]_{\mathcal B}=B^{-1}v=(2,1)$, and $B[v]_{\mathcal B}=v$ recovers the usual coordinates. The <determinant> $3$ is nonzero, so this recovery works for every <vector> in $\mathbb R^2$. A <basis> need not have perpendicular or unit-length <vectors>; an <orthonormal basis> adds those geometric conveniences when an <inner product> is available.

The point of changing <basis> is to simplify the operation being studied. <Diagonalization> uses an <eigenbasis> to turn a coupled <linear map> into independent scalings. In a space of <polynomials>, the <basis> $1,x,x^2$ makes <differentiation> easy to encode, while <orthogonal polynomials> can make approximation errors easier to measure. Gilbert Strang's MIT lecture on independence, basis and dimension develops this connection between attainable columns, redundant columns and coordinates.https://ocw.mit.edu/courses/18-06sc-linear-algebra-fall-2011/[Strang, MIT 18.06SC, Independence, Basis and Dimension]{ref}

Basis

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