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June 29th, 2014, 04:32 AM   #1
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Orthogonal projections and matrix diagonalization

In the Euclidean space $\displaystyle R^4$ with the usual inner product, let $\displaystyle U$ be the subspace given by the solutions of $\displaystyle 3x_1- x_2 -2x_3 = 0$ and $\displaystyle 2x_1 + x_3 + x_4 = 0$

a) Find a base of $\displaystyle U$ and $\displaystyle U^{\bot}$

b) Given the vector $\displaystyle v = (3,3,-2,-6)$ find a vector $\displaystyle w$ such that $\displaystyle v + w \in U$

c) Let $\displaystyle f:R^4 \to R^4$ be the linear function that associates to every vector of $\displaystyle R^4$ its orthogonal projection onto the subspace $\displaystyle U$. Find the eigenvalues, eigenvectors and find if it can be diagonalized.

A base of $\displaystyle U$ is given by the solution of the system:

$\displaystyle U =
\begin{cases}
3x_1- x_2 -2x_3 = 0 \\
2x_1 + x_3 + x_4 = 0 \\
\end{cases}=
\begin{cases}
x_1= -\frac12x_4 -\frac12x_3 \\
x_2 = -\frac72 x_3 -\frac32 x_4 \\
\end{cases}$

So a base of $\displaystyle U$ is :

$\displaystyle U = \langle (-1,-7,2,0),(-1,-3,0,2)\rangle$

Now let $\displaystyle v \in R^4$ then $\displaystyle v \in U^{\bot}$ only if:

$\displaystyle
\begin{cases}
v * (-1,-7,2,0)= 0 \\
v * (-1,-3,0,2) = 0 \\
\end{cases}=
\begin{cases}
-x_1 - 7x_2 + 2x_3=0 \\
-x_1 -3x_2 + 2x_4 = 0 \\
\end{cases}=
\begin{cases}
x_1 = - \frac32x_3 +\frac72 x_4 \\
x_2 = \frac12x_3 -\frac12 x_4 \\
\end{cases}$

So a base of $\displaystyle U^{\bot}$ is :

$\displaystyle U^{\bot}= \langle (-3,1,2,0),(7,-1,0,2)\rangle$

b) Every vector can be written as:

$\displaystyle v = P_U(v)+ P_{U^\bot}(v)$

where $\displaystyle P$ are the projections of $\displaystyle v$ onto $\displaystyle U$ and $\displaystyle U^{\bot}$

From this considerations the vector $\displaystyle w = -P_{U^\bot}(v)$

The projection can be found by solving the system:

$\displaystyle (3,3,-2,-6) = a (-1,-7,2,0)+b(-1,-3,0,2)+c(-3,1,2,0)+d(7,-1,0,2)=(-a-b-3c+7d,-7a-3b+c-d,2a+2c,2b+2d)$

$\displaystyle
\begin{cases}
-a-b-3c+7d = 3 \\
-7a-3b+c-d = 3 \\
2a+2c = -2\\
2b+2d = -6 \\
\end{cases}=
\begin{cases}
a = 1/2 \\
b = -5/2 \\
c = -3/2\\
d = -1/2 \\
\end{cases}$

$\displaystyle P_{U^\bot}(v) = c(-3,1,2,0)+d(7,-1,0,2) = -\frac32 (-3,1,2,0)-\frac12(7,-1,0,2) = (1,-1,-3,-1)$

thus

$\displaystyle w = -P_{U^\bot}(v) =(-1,1,3,1)$

c)Here I need to calculate the projections of the vectors of the standard base of $\displaystyle R^4$ onto $\displaystyle U$. The coefficients $\displaystyle a,b,c,d$ can be found quite easily by row reducing to the echelon form this matrix:

$\displaystyle
\begin{pmatrix}
-1 & -1 & -3 & 7 & 1 & 0 & 0 & 0\\
-7 & -3 & 1 & -1 & 0 & 1 & 0 & 0 \\
2 & 0 & 2 & 0 & 0 & 0 & 1 & 0\\
0 & 2 & 0 & 2 & 0 & 0 & 0 & 1\\
\end{pmatrix}
$

Which gives

$\displaystyle
\begin{pmatrix}
1 & 0 & 0 & 0 & 1/34 & -2/17 & 7/68 & -11/68\\
0 & 1 & 0 & 0 & -2/17 & -1/34 & -11/68 & 27/68\\
0 & 0 & 1 & 0 & -1/34 & 2/17 & 27/68 & 11/68\\
0 & 0 & 0 & 1 & 2/17 & 1/34 & 11/68 & 7/68\\
\end{pmatrix}
$

So the projections are:

$\displaystyle P_{U}(1,0,0,0) = 1/34 (-1,-7,2,0) -2/17 (-1,-3,0,2) = (3/34,5/34,1/17,-4/17)$

$\displaystyle P_{U}(0,1,0,0) = (5/34,31/34,-4/17,-1/17)$

$\displaystyle P_{U}(0,0,1,0) = (1/17,-4/17,7/34,-11/34)$

$\displaystyle P_{U}(0,0,0,1) = (-4/17,-1/17,-11/34,27/34)$

Thus the matrix of $\displaystyle f$ is:

$\displaystyle
\begin{pmatrix}
3/34 & 5/34 & 1/17 & -4/17\\
5/34 & 31/34 & -4/17 & -1/17\\
1/17 & -4/17 & 7/34 & -11/34\\
-4/17 & -1/17 & -11/34 & 27/34\\
\end{pmatrix}
$

Now I have to calculate the characterist polynomial

$\displaystyle
\begin{vmatrix}
3/34-t & 5/34 & 1/17 & -4/17\\
5/34 & 31/34-t & -4/17 & -1/17\\
1/17 & -4/17 & 7/34-t & -11/34\\
-4/17 & -1/17 & -11/34 & 27/34-t\\
\end{vmatrix}=0
$

Which is $\displaystyle P(t) = t^2 (t-1)^2$
So the eigenvalues are $\displaystyle t = 0, t = 1$

The enigenvectors are:

$\displaystyle V_0 = \langle (7,-1,0,2),(-3,1,2,0)\rangle$

$\displaystyle V_1 = \langle (-1,-3,0,2),(-1,-7,2,0)\rangle$

The dimensions of $\displaystyle V_0$ and $\displaystyle V_1$ equals the algebraic multiplicity of the eigenvalues thus the matrix can be diagonalized.

Now I would like to know if it is solved correctly just the logical steps (the calculations should be right because I made/check them with the computer).
This exercise was given in a final exam in my college with other 3 exercises of almost equal difficulty and the time given to solve everything is just 3 hours. I was wondering if there was a much faster way to solve it especially the part (c) where there are many calculations to find the images of the standard base and the characteristic polynomial (calculators are not allowed during the exams).
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