MATH 240

Mon. October 21st, 2019


Unique Representation Theorem

Theorem. Let B={b1,...,bn}B=\{b_1,...,b_n\} be a basis for a vector space VV. Then for each xx in VV, there exists unique c1,...,cnc_1,...,c_n such that x=c1b1+...+cnbnx=c_1b_1+...+c_nb_n.

Coordinates

Definition. Suppose B={b1,...,bn}B=\{b_1,...,b_n\} is a basis for VV and xx is in VV.
The coordinates of xx relative to the basis BB (or the BB-coordinates of xx) are the weights c1,...,cnc_1,...,c_n such that c1b1+...+cnbn=xc_1b_1+...+c_nb_n=x.
[x]B=(c1...cn)[x]_B=\begin{pmatrix} c_1\\...\\c_n \end{pmatrix}


From BB-Coordinates to Standard Basis Coordinates

Ex. If b1=(10), b2=(12), B={b1,b2},b_1=\begin{pmatrix} 1\\0 \end{pmatrix},\ b_2=\begin{pmatrix} 1\\2 \end{pmatrix},\ B=\{b_1,b_2\}, and [x]B=(23)[x]_B=\begin{pmatrix} -2\\3 \end{pmatrix}, then find xx. x=c1b1+c2b2=2(10)+3(12)=(16).\begin{aligned} x&=c_1b_1+c_2b_2\\ &=-2\begin{pmatrix} 1\\0 \end{pmatrix}+3\begin{pmatrix} 1\\2 \end{pmatrix}\\ &=\begin{pmatrix} 1\\6 \end{pmatrix}. \end{aligned}


From Standard Basis Coordinates to BB-Coordinates

Ex. If b1=(21), b2=(11), B={b1,b2},b_1=\begin{pmatrix} 2\\1 \end{pmatrix},\ b_2=\begin{pmatrix} -1\\1 \end{pmatrix},\ B=\{b_1,b_2\}, and x=(45)x=\begin{pmatrix} 4\\5 \end{pmatrix}, then find [x]B[x]_B. x=c1b1+c2b2(45)=c1(21)+c2(11)(45)=(2111)(c1c2)(214115)(115214)(115036)(115012)(103012)[x]B=(c1c2)=(32).\begin{aligned} x&=c_1b_1+c_2b_2\\ \begin{pmatrix} 4\\5 \end{pmatrix}&=c_1\begin{pmatrix} 2\\1 \end{pmatrix}+c_2\begin{pmatrix} -1\\1 \end{pmatrix}\\ \begin{pmatrix} 4\\5 \end{pmatrix}&=\begin{pmatrix} 2&-1\\1&1 \end{pmatrix}\begin{pmatrix} c_1\\c_2 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 2&-1&4\\1&1&5 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&1&5\\2&-1&4 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&1&5\\0&-3&-6 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&1&5\\0&1&2 \end{pmatrix}\\ \rightarrow &\begin{pmatrix} 1&0&3\\0&1&2 \end{pmatrix}\\ [x]_B&=\begin{pmatrix} c_1\\c_2 \end{pmatrix}\\ &=\begin{pmatrix} 3\\2 \end{pmatrix}. \end{aligned}

Theorem. Let PB=[b1...bn]P_B=\begin{bmatrix} b_1&...&b_n \end{bmatrix}. Now x=PB[x]Bx=P_B[x]_B.
The columns of PBP_B are linearly independent since they form a basis, so by the Invertible Matrix Theorem, PBP_B is invertible. This means that [x]B=PB1x[x]_B=P_B^{-1}x.


Isomorphism

Let B={b1,...,bn}B=\{b_1,...,b_n\} be a basis for VV. Assigning coordinates to VV gives a linear transformation from VV to Rn\R^n that is one-to-one and onto. This means that VV and Rn\R^n are isomorphic – they look different, but can be represented by the same mathematical object.


Ex. Let P5P_5 be the set of all polynomials of degree five or less.
{1,t,t2,t3,t4,t5}\{1,t,t^2,t^3,t^4,t^5\} is a basis for P5P_5. This means any polynomial of degree five or less PP takes the form P(x)=a5x5+a4x4+a3x3+a2x2+a1x1+a0P(x)=a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x^1+a_0.

If TT is the transformation that assigns coordinates from P5P_5 to R6\R^6, then T(P(x))=(a5a4a3a2a1a0)T(P(x))=\begin{pmatrix} a_5\\a_4\\a_3\\a_2\\a_1\\a_0 \end{pmatrix}
Thus P5P_5 is isomorphic to R6\R^6. This means we can perform computations like adding two polynomials in P5P_5 by adding their corresponding vectors in R6\R^6.


Ex. B={sint, cost}B=\{\sin{t},\ \cos{t}\} is the basis of a subspace inside of C(R)C^\infin(\R). (sint\sin{t} and cost\cos{t} are linearly independent functions.) Since BB only contains two vectors, functions within this subspace can be represented as vectors in R2\R^2.