Linear Algebra: linear transformation

[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.

 

[Dick]

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.

 

[nateHI]

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 $Sv_2=0$

 

[Dick]

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.

 

[nateHI]

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?

 

[Dick]

$Sv=SSu=Su=v\implies Sv=kv$ 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?

 

[pasmith]

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 $\mathbb{R}^2$.

 

[nateHI]

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.

 

[Dick]

"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]

"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.

 

[Dick]

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.