2019-03-18T06:29:11-04:00
Let (x1,x2,x3) and (y1,y2,y3) represent the coordinates with respect to the bases B1={(1,0,0),(0,1,0),(0,0,1)}, B2={(1,0,0),(0,1,2),(0,2,1)}. If Q(x)=x1^2-2x1x2+4x2x3+x2^2+x3^2 , find the representation of Q in terms of (y1,y2,y3).
1
2019-03-22T12:12:51-0400
The following equalities are true:
( 1 , 0 , 0 ) = 1 ∗ ( 1 , 0 , 0 ) + 0 ∗ ( 0 , 1 , 0 ) + 0 ∗ ( 0 , 0 , 1 ) (1, 0, 0)=1*(1, 0, 0)+0*(0, 1, 0)+0*(0, 0, 1) ( 1 , 0 , 0 ) = 1 ∗ ( 1 , 0 , 0 ) + 0 ∗ ( 0 , 1 , 0 ) + 0 ∗ ( 0 , 0 , 1 ) ( 0 , 1 , 2 ) = 0 ∗ ( 1 , 0 , 0 ) + 1 ∗ ( 0 , 1 , 0 ) + 2 ∗ ( 0 , 0 , 1 ) (0, 1, 2)=0*(1, 0, 0)+1*(0, 1, 0)+2*(0, 0, 1) ( 0 , 1 , 2 ) = 0 ∗ ( 1 , 0 , 0 ) + 1 ∗ ( 0 , 1 , 0 ) + 2 ∗ ( 0 , 0 , 1 ) ( 0 , 2 , 1 ) = 0 ∗ ( 1 , 0 , 0 ) + 2 ∗ ( 0 , 1 , 0 ) + 1 ∗ ( 0 , 0 , 1 ) (0, 2, 1)=0*(1, 0, 0)+2*(0, 1, 0)+1*(0, 0, 1) ( 0 , 2 , 1 ) = 0 ∗ ( 1 , 0 , 0 ) + 2 ∗ ( 0 , 1 , 0 ) + 1 ∗ ( 0 , 0 , 1 ) C = ( 1 0 0 0 1 2 0 2 1 ) C=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 2 \\ 0 & 2 & 1
\end{pmatrix} C = ⎝ ⎛ 1 0 0 0 1 2 0 2 1 ⎠ ⎞ C T = ( 1 0 0 0 1 2 0 2 1 ) C^T=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 2 \\ 0 & 2 & 1
\end{pmatrix} C T = ⎝ ⎛ 1 0 0 0 1 2 0 2 1 ⎠ ⎞
Q ( X ) = x 1 2 + 2 x 1 x 2 + 4 x 2 x 3 + x 2 2 + x 3 2 Q(X)={x_1}^2+2x_1x_2+4x_2x_3+{x_2}^2+{x_3}^2 Q ( X ) = x 1 2 + 2 x 1 x 2 + 4 x 2 x 3 + x 2 2 + x 3 2 A = ( 1 1 0 1 1 2 0 2 1 ) A=\begin{pmatrix}
1 & 1 & 0 \\
1 & 1 & 2 \\ 0 & 2 & 1
\end{pmatrix} A = ⎝ ⎛ 1 1 0 1 1 2 0 2 1 ⎠ ⎞ Then the matrix of the form Q in the base B2 is equal to
B = C T A C = ( 1 0 0 0 1 2 0 2 1 ) ( 1 1 0 1 1 2 0 2 1 ) ( 1 0 0 0 1 2 0 2 1 ) B=C^TAC=\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 2 \\ 0 & 2 & 1
\end{pmatrix}\begin{pmatrix}
1 & 1 & 0 \\
1 & 1 & 2 \\ 0 & 2 & 1
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 2 \\ 0 & 2 & 1
\end{pmatrix} B = C T A C = ⎝ ⎛ 1 0 0 0 1 2 0 2 1 ⎠ ⎞ ⎝ ⎛ 1 1 0 1 1 2 0 2 1 ⎠ ⎞ ⎝ ⎛ 1 0 0 0 1 2 0 2 1 ⎠ ⎞
B = ( 1 1 2 1 5 4 0 4 5 ) ( 1 0 0 0 1 2 0 2 1 ) = ( 1 1 2 1 13 14 2 14 13 ) B=\begin{pmatrix}
1 & 1 & 2 \\
1 & 5 & 4 \\ 0 & 4 & 5
\end{pmatrix}\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 2 \\ 0 & 2 & 1
\end{pmatrix}=\begin{pmatrix}
1 & 1 & 2 \\
1 & 13 & 14 \\ 2 & 14 & 13
\end{pmatrix} B = ⎝ ⎛ 1 1 0 1 5 4 2 4 5 ⎠ ⎞ ⎝ ⎛ 1 0 0 0 1 2 0 2 1 ⎠ ⎞ = ⎝ ⎛ 1 1 2 1 13 14 2 14 13 ⎠ ⎞ So the form Q in the base B2 will
Q ( Y ) = y 1 2 + 13 y 2 2 + 13 y 3 2 + 2 y 1 y 2 + 4 y 1 y 3 + 28 y 2 y 3 Q(Y)={y_1}^2+13{y_2}^2+13{y_3}^2+2y_1y_2+4y_1y_3+28y_2y_3 Q ( Y ) = y 1 2 + 13 y 2 2 + 13 y 3 2 + 2 y 1 y 2 + 4 y 1 y 3 + 28 y 2 y 3
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