Find the image of a projection

  • #1
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2

Summary:

Show that the linear transformation with the matrix below with respect to a basis for ##\textbf{R}^4## is a projection on a subspace ##U'## along a subspace ##U''##. Find ##U'##.

Main Question or Discussion Point

## \dfrac{1}{2}
\left(\begin{array}{rrrr}
0 & 2 & -2 & 0 \\
3 & -2 & 2 & -3 \\
3 & -4 & 4 & -3 \\
-2 & 2 & -2 & 2
\end{array}\right)##​

The matrix satisfies ##A^2=A##, so it is a projection. To find ##U'##, one can find the ##\text{ker} \ (A-I)=\text{ker} \ (I-A)=\text{im} \ (A)=U'##. Also, ##\text{ker} \ (A-I)=\text{ker} \ (2(A-I))##.

Solving ##2(A-I)\textbf{x}=\textbf{0}## using Gauss Jordan elimination yields ##\textbf{x}=r(-2,-1,1,0)+t(-3,-3,0,1)## for ##t,r\in\textbf{R}##. But the solution given is ##\textbf{x}=r(0,3,3,-2)+t(1,-1,-2,1)##. Are these solutions equivalent? Since the columns of the matrix span the image, every vector in the image can be represented by at least the linear independent vectors in the matrix, which the latter solution does.
 

Answers and Replies

  • #2
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Summary: Show that the linear transformation with the matrix below with respect to a basis for ##\textbf{R}^4## is a projection on a subspace ##U'## along a subspace ##U''##. Find ##U'##.

## \dfrac{1}{2}
\left(\begin{array}{rrrr}
0 & 2 & -2 & 0 \\
3 & -2 & 2 & -3 \\
3 & -4 & 4 & -3 \\
-2 & 2 & -2 & 2
\end{array}\right)##​

The matrix satisfies ##A^2=A##, so it is a projection. To find ##U'##, one can find the ##\text{ker} \ (A-I)=\text{ker} \ (I-A)=\text{im} \ (A)=U'##. Also, ##\text{ker} \ (A-I)=\text{ker} \ (2(A-I))##.

Solving ##2(A-I)\textbf{x}=\textbf{0}## using Gauss Jordan elimination yields ##\textbf{x}=r(-2,-1,1,0)+t(-3,-3,0,1)## for ##t,r\in\textbf{R}##. But the solution given is ##\textbf{x}=r(0,3,3,-2)+t(1,-1,-2,1)##. Are these solutions equivalent?
I don't know whether Gauß elimination really results into the two vectors you listed. But to check whether the two solutions are equivalent, just test whether
$$
(-2,-1,1,0) \stackrel{?}{\in} r_1(0,3,3,-2)+t_1(1,-1,-2,1) \;\wedge\; (-3,-3,0,1) \stackrel{?}{\in} r_2(0,3,3,-2)+t_2(1,-1,-2,1)
$$
This is an easy system to solve. To me it doesn't look as if there were solutions ##r_i,t_i##, but I haven't checked.
Since the columns of the matrix span the image, every vector in the image can be represented by at least the linear independent vectors in the matrix, which the latter solution does.
To check whether your elimination process was correct, you could simply solve ##A(x)=x## by hand.

The given solution has a big advantage:
  1. One sees at once that ##\operatorname{rk}A =2## ...
  2. ... and that the first two or likewise last two column vectors are linearly independent,
... hence span the image of ##A##.
 
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  • #3
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Could it be that the solution given is not correct? This (scroll down) matrix calculator gives the former solution. Note that the matrix there is ##2(A-I)##, which is

##
\left(\begin{array}{rrrr}
-2 & 2 & -2 & 0 \\
3 & -4 & 2 & -3 \\
3 & -4 & 2 & -3 \\
-2 & 2 & -2 & 0
\end{array}\right)##​
 
  • #4
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Why do you want to operate with ##2(A-I)## if ##A## alone has already all answers?

Anyway, what keeps you from solving the question whether the two solutions are equivalent? It can almost be done in mind! E.g. ##t_1=-2## follows immediately, which forces ##r_1=-1## by the second coordinate, and now simply check whether these settings match coordinate three and four. Same with the second vector.

Btw., do you know why ##\operatorname{ker}(2A-2I)=\operatorname{im}(A)## holds?
 
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  • #5
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Solved it.

Btw., do you know why ##\operatorname{ker}(2A-2I)=\operatorname{im}(A)## holds?
##\text{ker} \ (A-I)=\text{ker} \ (I-A)=\text{im} \ (A)=U'##
 
  • #6
95
2
Or ##A(\textbf{u})=\textbf{u'} \iff A(\textbf{u'}+\textbf{u''})=\textbf{u'}##.

When ##\textbf{u'}=\textbf{0}##, it follows that ##A(\textbf{u''})=\textbf{0}##.
 
  • #7
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Yes, and no. The argument goes this way:
$$
A^2=A \Longrightarrow A\cdot A -A = A(A-I)=(A-I)A=0
$$
If ##v \in \operatorname{im}A## then there is a vector ##w## with ##v=A(w)##. Thus ##(A-I)(v)=(A-I)A(w)=0## and ##v\in \operatorname{ker} (A-I)##. This means we have shown that ##\operatorname{im}(A) \subseteq \operatorname{ker}(A-I)##.

Can you show why ##\operatorname{ker}(A-I) \subseteq \operatorname{im}(A)## and why multiplication by ##2## doesn't change neither kernel nor image?
 
  • #8
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Can you show why ##\operatorname{ker}(A-I) \subseteq \operatorname{im}(A)## and why multiplication by ##2## doesn't change neither kernel nor image?
How would the argument go?
 
  • #9
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By the same method. We choose an element ##v\in \operatorname{ker}(A-I)##. Now what does this mean?
 
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  • #10
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Summary: Show that the linear transformation with the matrix below with respect to a basis for ##\textbf{R}^4## is a projection on a subspace ##U'## along a subspace ##U''##. Find ##U'##.

## \dfrac{1}{2}
\left(\begin{array}{rrrr}
0 & 2 & -2 & 0 \\
3 & -2 & 2 & -3 \\
3 & -4 & 4 & -3 \\
-2 & 2 & -2 & 2
\end{array}\right)##​

The matrix satisfies ##A^2=A##, so it is a projection. To find ##U'##, one can find the ##\text{ker} \ (A-I)=\text{ker} \ (I-A)=\text{im} \ (A)=U'##. Also, ##\text{ker} \ (A-I)=\text{ker} \ (2(A-I))##.

Solving ##2(A-I)\textbf{x}=\textbf{0}## using Gauss Jordan elimination yields ##\textbf{x}=r(-2,-1,1,0)+t(-3,-3,0,1)## for ##t,r\in\textbf{R}##. But the solution given is ##\textbf{x}=r(0,3,3,-2)+t(1,-1,-2,1)##. Are these solutions equivalent? Since the columns of the matrix span the image, every vector in the image can be represented by at least the linear independent vectors in the matrix, which the latter solution does.
I am a bit confused. Since they mention it is a projection along a subspace U', I would assume U' is not the whole of ##*\mathbb R^4##. Does ##A^2=A## hold in the entire ##\mathbb R^4##,i.e., is it an identity?
 

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