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Find Matrix P for Quadratic Transformation

You can do it by the ratios of the eigen vector components.\$\$H = left( beginarrayrr 3 & -2 sqrt 2 -2 sqrt 2 & 5 endarray

ight) tag1\$\$\$\$(H-lambda I)v = 0tag2\$\$\$\$ det left( beginarrayrr 3 - lambda& -2 sqrt 2 -2 sqrt 2 & 5 - lambda endarray

ight) = 0 tag3\$\$\$\$ (3 - lambda)(5-lambda) - 8 = 0 tag4\$\$\$\$ lambda^2 - 8 lambda 7 = 0 tag5\$\$\$\$ (lambda - 1)(lambda - 7) = 0 tag6\$\$The eigen values are: \$\$lambda_1 = 1 :,: lambda_2=7\$\$Now find the eigen vectors for \$lambda_1\$ see \$(2)\$\$\$left( beginarrayrr 3-1 & -2 sqrt 2 -2 sqrt 2 & 5-1 endarray

ight) left( beginarrayrr v_11 v_12 endarray

ight) = left( beginarrayrr 2 & -2 sqrt 2 -2 sqrt 2 & 4 endarray

ight) left( beginarrayrr v_11 v_12 endarray

ight) = 0tag7\$\$This gives:\$\$ 2 v_11 -2 sqrt2 v_12 = 0 tag8\$\$\$\$ fracv_11v_12 = sqrt2 tag9\$\$Note that the relationship is a ratio so it can have many scalar solutions.Now find the eigen vectors for \$lambda_2\$ see \$(2)\$\$\$left( beginarrayrr 3-7 & -2 sqrt 2 -2 sqrt 2 & 5-7 endarray

ight) left( beginarrayrr v_21 v_22 endarray

ight) = left( beginarrayrr -4 & -2 sqrt 2 -2 sqrt 2 & -2 endarray

ight) left( beginarrayrr v_21 v_22 endarray

ight) = 0tag8\$\$This gives:\$\$ -4 v_21 -2 sqrt2 v_22 = 0 tag9\$\$\$\$ fracv_21v_22 = -frac1sqrt2 tag10\$\$The order of the eigen values in the diagonal matrix is arbitrary but the order of eigen values and their vectors must be the same.So you are looking for \$P\$ with top and bottom eigen vector ratios in each column of:\$\$ left( beginarrayrr sqrt2 & 1 1 & -sqrt2 endarray

ight) or left( beginarrayrr 1 & sqrt2 -sqrt2 & 1 endarray

ight) tag11\$\$\$\$b.P = beginbmatrix sqrt 2 & sqrt2 -2 &1 endbmatrix\$\$Has this ratio.Sanity check. Octave:This gives the correct eigen values in the diagonal matrix.\$\$D = P^-1HP tag12\$\$\$\$ PDP^-1 = H tag13\$\$\$\$ PDP^-1x = Hx tag14\$\$\$\$ P^-1x = y tag15\$\$\$\$ x = Py tag16\$\$Your answer is a correct solution. Factoring out \$frac1sqrt3\$ and swapping the order of the eigen vectors matches the ratios within the columns of \$b\$.\$\$ beginbmatrix sqrt2/sqrt3 & -sqrt3/3 1/sqrt3 & sqrt2/sqrt3 endbmatrix = frac1sqrt3 beginbmatrix sqrt2 & -1 1 & sqrt2 endbmatrix xrightarrow[textcolumns]textswap beginbmatrix -1 & sqrt2 sqrt2 & 1 endbmatrix xrightarrow[texttimes: -sqrt2]1_st :textcolumn beginbmatrix sqrt 2 & sqrt2 -2 &1 endbmatrix \$\$ 1. Determine whether the transformation is a linear transformation

To show something is a linear transformation you must show that \$T(ax y) = aT(x) T(y)\$. If you try this you will find your answer

2. Jacobian of transformation

We have \$\$ fracpartial(x,y)partial(alpha, beta) = frac 12 beginpmatrix defIdmathrmid_3Id & -Id Id & Id endpmatrix \$\$ Hence \$\$ det fracpartial(x,y)partial(alpha, beta) = frac 12^6 detbeginpmatrix defIdmathrmid_3Id & -Id Id & Id endpmatrix = frac 12^6 detbeginpmatrix Id & -Id 0 & 2Id endpmatrix = frac 12^6 cdot 8 = frac 18 \$\$ 3. Is my reasoning on Transformation Matrices right?

I think I got it and i can answer myself:In other means: \$\$[v]_B = underbraceB^-1cdot A_textM? cdot [v]_A\$\$\$T(overrightarrow v)\$ is the Linear Transformation that maps \$overrightarrow v\$ to coordinate tuple on basis \$B\$As described before this is: \$\$T(overrightarrow v) = [v]_B\$\$Since \$overrightarrow v\$ is expressed as linear combination on basis \$A\$, the linear transformation acts on each vector of basis \$A\$, yielding the associated Transformation Matrix \$M\$ of the linear transformation.The column vectors of \$M\$ are the vectors of basis \$A\$ transformed to coordinate tuple on basis \$B\$. Since \$T(overrightarrowv_i) = B^-1cdot overrightarrowv_i\$Then \$[v]_B = Mcdot [v]_A\$ as we already know that \$[v]_B = B^-1cdot A cdot [v]_A\$ the proof is concluded.

4. Interpretation of a homogeneous transformation matrix of the plane

Work backwards. The translation of \$y = x - 1\$ to \$y = x\$:\$\$beginbmatrix x' y' endbmatrix = beginbmatrix 1 & 0 0 & 1 endbmatrix beginbmatrix x y endbmatrix beginbmatrix 0 -1 endbmatrix\$\$Affine to linear:\$\$beginbmatrix x' y' 1 endbmatrix = beginbmatrix 1 & 0 & 0 0 & 1 & -1 0 & 0 & 1 endbmatrix beginbmatrix x y 1 endbmatrix\$\$So it seems the book is representing the vectors as row vectors, \$beginbmatrix x & y endbmatrix\$, so your matrix is transposed

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