Linear Algebra: linear transformation

• nateHI
In summary: S(cw_1)=cw_1## and ##S(w_2)=0##So this is a basis for the range. I don't think I need to show the other direction of this...I can't think of a reason it would be necessary.In summary, the linear transformation T(x1,x2)=(x1,0) on R2 has the matrix A = (1 0; 0 0) with respect to the standard basis and satisfies T2=T. It has been proven that if S:R2→R2 is a linear transformation satisfying S2=S, then S=0 or S=I or there exists a basis B for R2 such that the matrix of S with
nateHI

Homework Statement

We have seen that the linear transformation ##T(x_1,x_2)=(x_1,0)## on ##\mathcal{R}^2## has the matrix ##A = \left( \begin{smallmatrix} 1&0\\ 0&0 \end{smallmatrix} \right)## with respect to the standard basis. This operator satisfies ##T^2=T##. Prove that if ##S:\mathcal{R}^2 \to \mathcal{R}^2##, is a linear transformation satisfying ##S^2=S##, then ##S=0## or ##S=I## or there exists a basis ##\mathcal{B}## for ##R^2## such that the matrix of ##S## with respect to ##\mathcal{B}## is ##A##.

Homework Equations

The professor would prefer if we used basic principals to solve this since it is a undergraduate course.

This homework is past due and I will not get credit for this but it might be on the test so I would appreciate the typical, "make me think" type of replies.

The Attempt at a Solution

The fact that ##S^2=S## when ##S=0## or ##S=I## is obvious. However, looking at the dimension of the kernel and the range of ##S## in those cases gives a hint on how to solve the remaining case.
1st Case: ##S=0\implies## the ##dim(n(S))=2## and ##dim(range(S))=0##
2nd Case: ##S=I\implies## the ##dim(n(S))=0## and ##dim(range(S))=2## (I has an inverse and therefore is a bijection so the kernel is {0})

The final case must have that ##dim(n(S))=1=dim(range(S))## since that is the only option left to us.
...This is as far as I got.

nateHI said:

Homework Statement

We have seen that the linear transformation ##T(x_1,x_2)=(x_1,0)## on ##\mathcal{R}^2## has the matrix ##A = \left( \begin{smallmatrix} 1&0\\ 0&0 \end{smallmatrix} \right)## with respect to the standard basis. This operator satisfies ##T^2=T##. Prove that if ##S:\mathcal{R}^2 \to \mathcal{R}^2##, is a linear transformation satisfying ##S^2=S##, then ##S=0## or ##S=I## or there exists a basis ##\mathcal{B}## for ##R^2## such that the matrix of ##S## with respect to ##\mathcal{B}## is ##A##.

Homework Equations

The professor would prefer if we used basic principals to solve this since it is a undergraduate course.

This homework is past due and I will not get credit for this but it might be on the test so I would appreciate the typical, "make me think" type of replies.

The Attempt at a Solution

The fact that ##S^2=S## when ##S=0## or ##S=I## is obvious. However, looking at the dimension of the kernel and the range of ##S## in those cases gives a hint on how to solve the remaining case.
1st Case: ##S=0\implies## the ##dim(n(S))=2## and ##dim(range(S))=0##
2nd Case: ##S=I\implies## the ##dim(n(S))=0## and ##dim(range(S))=2## (I has an inverse and therefore is a bijection so the kernel is {0})

The final case must have that ##dim(n(S))=1=dim(range(S))## since that is the only option left to us.
...This is as far as I got.

Yes, ##S^2=S## is obviously satisfied by 0 and I. But you've still got a little work to do. Can you show if dim(range(S))=0 then S must be 0. And a little harder, can you show if dim(range(S))=2 then S must be I? You'll have to use the properties of S to show that. And in the case dim(range(S))=1 there is a single vector v which spans range(S). So? Try to use that to pick a basis.

Dick said:
Yes, ##S^2=S## is obviously satisfied by 0 and I. But you've still got a little work to do. Can you show if dim(range(S))=0 then S must be 0. And a little harder, can you show if dim(range(S))=2 then S must be I? You'll have to use the properties of S to show that. And in the case dim(range(S))=1 there is a single vector v which spans range(S). So? Try to use that to pick a basis.

##dim(range(S))=0 \implies## every element of ##R^2## is in the kernel. Using linearity, ##S(v_1+v_2)=S(v_1)+S(v_2)=0=S(R^2)## for all ##v\in R^2##(seems like there should be a better way to do this but it works)

Suppose ##M(S)=\left (\begin{matrix}a&b\\c&d\end{matrix} \right)## then ##S^2=S \implies ##
##a^2+bc=a##
##ab+bd=b##
##ac+dc=c##
##d^2+bc=d##
Since the dim(range(S))=2 the null space is empty and there is no trivial solution to the above system. This forces a=d=1 and b=c=0 ##\implies S=I##.

For the final case ##dim(n(S))=1=dim(range(S))##.

Since the kernel has dimension 1, vectors that get sent to 0 lie on a line in ##R^2##. Vectors that get sent to the range also lie on a line and are of the form ##v_1=cv_2## for any ##v_1,v_2 \in R^2##...not real sure where I'm going with this...need a push in the right direction.

Edit: I think I made some progress.
For vectors on the range
##S^2=S\implies SSv_1=Sv_1\implies Sv_1=v_1##
For vectors in the kernel
##SSv_2=Sv_2=0##
And so we want ##v_1## and ##v_2## with
##Sv_1=v_1## and ##Sv_2=0##

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nateHI said:
Edit: I think I made some progress.
For vectors on the range
##S^2=S\implies SSv_1=Sv_1\implies Sv_1=v_1##
For vectors in the kernel
##SSv_2=Sv_2=0##
And so we want ##v_1## and ##v_2## with
##Sv_1=v_1## and ##Sv_2=0##

That's getting there. But it's not generally (for any S) true that ##SSv=Sv \implies Sv=v##. What is true for this S, is that if v is in the range then ##v=Su## for some vector u. That means ##Sv=SSu=Su=v##. And you know every v can be written as the sum of a vector in the range plus a vector in the kernel. Use that to think about a nice way to pick a basis.

Dick said:
That's getting there. But it's not generally (for any S) true that ##SSv=Sv \implies Sv=v##. What is true for this S, is that if v is in the range then ##v=Su## for some vector u. That means ##Sv=SSu=Su=v##. And you know every v can be written as the sum of a vector in the range plus a vector in the kernel. Use that to think about a nice way to pick a basis.

##Sv=SSu=Su=v\implies Sv=kv## for all ##v\in R^2## that get mapped to the range. Where I used the fact that the dimension for the range are all points on a line.
##\mathcal{B}={<k,0>,<0,0>}##
##S=kA## ?
If correct, do I need to show this the other way too? Do I need to show that if the basis for the range is 1 dimensional then so is the basis for the vectors that get mapped to it?

Edit: added little more and I changed the variable for the constant to k to avoid confusion.

Let ##w_1## be in the set of vectors that get mapped to the range and ##w_2\in kernel(S)##.
Then##S(w_1+w_2)=S(w_1)+S(w_2)=cw_1+0=cw_1## What does this show?

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nateHI said:
##Sv=SSu=Su=v\implies Sv=kv## for all ##v\in R^2## that get mapped to the range. Where I used the fact that the dimension for the range are all points on a line.
##\mathcal{B}={<k,0>,<0,0>}##
##S=kA## ?
If correct, do I need to show this the other way too? Do I need to show that if the basis for the range is 1 dimensional then so is the basis for the vectors that get mapped to it?

Edit: added little more and I changed the variable for the constant to k to avoid confusion.

Let ##w_1## be in the set of vectors that get mapped to the range and ##w_2\in kernel(S)##.
Then##S(w_1+w_2)=S(w_1)+S(w_2)=cw_1+0=cw_1## What does this show?

Forget about the line thing. The dimension of a space is the number of vectors in a basis. That's all. Don't "Let ##w_1## be in the set of vectors that get mapped to the range". That's ALL vectors. Let ##w_1## be a nonzero vector in the range. So ##Sw_1=w_1##. Let ##w_2## be a nonzero vector in the kernel. Can you see why ##\{w_1, w_2\}## is a basis for the whole space?

We are given that $S: \mathbb{R}^2 \to \mathbb{R}^2$ is linear with $S^2(v) = S(v)$ for all $v \in \mathbb{R}^2$, and we are assuming that $S$ is neither the zero map nor the identity map.

Since by assumption $S$ is not the zero map, there exists $v \neq 0$ such that $u = S(v) \neq 0$. But we have $S^2 (v) = S(v)$, ie. $S(u) = u \neq 0$.

Since by assumption $S$ is not the identity map, there exists $x \neq 0$ such that $S(x) - x \neq 0$. Yet $S(S(x) - x) = S^2(x) - S(x) = 0$, so there exists $w \neq 0$ such that $S(w) = 0$.

I claim that $\{u, w\}$, where $S(u) = u$ and $S(w) = 0$, is a basis of $\mathbb{R}^2$.

Dick said:
Forget about the line thing. The dimension of a space is the number of vectors in a basis. That's all. Don't "Let ##w_1## be in the set of vectors that get mapped to the range". That's ALL vectors. Let ##w_1## be a nonzero vector in the range. So ##Sw_1=w_1##. Let ##w_2## be a nonzero vector in the kernel. Can you see why ##\{w_1, w_2\}## is a basis for the whole space?

"Can you see why ##\{w_1, w_2\}## is a basis for the whole space?"
Because ##dimension(R^2)=dimension(kernel(S))+dimension(range(S))=1+1=2##
Edit: Oh wait, they must be linearly independent. I'll check.
Edit 2: To show L.I., for any two constants ##a,b## we must have:
##aS(w_1)+bS(w_2)=0##
##aw_1+b*0=0\implies a=0##
but
##aw_1+bw_2=0*w_1+b*w_2=0\implies b=0##
and so ##\{w_1, w_2\}## are two L.I. vectors in ##R^2## forming a basis and satisfying S^2=S.

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nateHI said:
"Can you see why ##\{w_1, w_2\}## is a basis for the whole space?"
Because ##dimension(R^2)=dimension(kernel(S))+dimension(range(S))=1+1=2##
Edit: Oh wait, they must be linearly independent. I'll check.
Edit 2: To show L.I., for any two constants ##a,b## we must have:
##aS(w_1)+bS(w_2)=0##
##aw_1+b*0=0\implies a=0##
but
##aw_1+bw_2=0*w_1+b*w_2=0\implies b=0##
and so ##\{w_1, w_2\}## are two L.I. vectors in ##R^2## forming a basis and satisfying S^2=S.

Good! Now what's the matrix of ##S## is the basis ##\{w_1, w_2\}##?

nateHI said:
"Can you see why ##\{w_1, w_2\}## is a basis for the whole space?"
Because ##dimension(R^2)=dimension(kernel(S))+dimension(range(S))=1+1=2##
Edit: Oh wait, they must be linearly independent. I'll check.
Edit 2: To show L.I., for any two constants ##a,b## we must have:
##aS(w_1)+bS(w_2)=0##
##aw_1+b*0=0\implies a=0##
but
##aw_1+bw_2=0*w_1+b*w_2=0\implies b=0##
and so ##\{w_1, w_2\}## are two L.I. vectors in ##R^2## forming a basis and satisfying S^2=S.

Just to finish up. ##A## falls in this category because ##A(1,0)^t=(1,0)^t## and ##A(0,1)^t=0##
Edit: not quite finished actually.
Edit2: OK now I'm done!

##Sv_1=v_1=a_1v_1+b_1v_2 ##
##Sv_2=0=a_2v_1+b_2v_2 ##
and so S=A.

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nateHI said:
Just to finish up. ##A## falls in this category because ##A(1,0)^t=(1,0)^t## and ##A(0,1)^t=0##
Edit: not quite finished actually.
Edit2: OK now I'm done!

##Sv_1=v_1=a_1v_1+b_1v_2 ##
##Sv_2=0=a_2v_1+b_2v_2 ##
and so S=A.

Now you're done. You might want to take that same approach to show S=0 or S=I are the only other possibilities.

1. What is a linear transformation?

A linear transformation is a mathematical function that maps a vector space to another vector space while preserving the linear structure of the original space. It is represented by a matrix and can be thought of as a geometric transformation of the underlying vector space.

2. What are the properties of a linear transformation?

A linear transformation must satisfy two main properties: preservation of addition and preservation of scalar multiplication. This means that the transformation must maintain the sum of two vectors and the multiplication of a vector by a scalar. Additionally, a linear transformation must also map the zero vector to the zero vector and preserve the order of multiplication.

3. How is a linear transformation represented?

A linear transformation is represented by a matrix, where each column represents the values that the basis vectors of the original vector space are mapped to in the new vector space. The matrix can then be used to transform any vector in the original space to the new space.

4. What is the difference between a linear transformation and a linear function?

A linear transformation is a more general concept that applies to vector spaces, while a linear function is a specific type of transformation that applies to real numbers. A linear transformation can be represented by a matrix, while a linear function is typically represented by an equation in the form of y = mx + b.

5. How is a linear transformation used in real-world applications?

Linear transformations are used in a variety of fields, including physics, engineering, and computer graphics. In physics, they are used to represent physical systems and their transformations. In engineering, they are used to model and analyze systems such as electrical circuits and control systems. In computer graphics, they are used to manipulate and transform objects in 2D and 3D space.

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