Exercise counter-image with two variables

[Kernul]

Homework Statement

I have this linear application:
$f : \mathbb {R}^4 \rightarrow \mathbb {R}^4$
$f((x, y, z, t)) = (-2x - y + t, 2x + y - t, -4x - 2y + 2t, 0)$
and then I have this vector which I have to find the counter-image of it:
$\vec v = (-k, h - 2, k, h^2 - 4)$
With $h$ and $k$ real parameters that vary.

Homework Equations

The Attempt at a Solution

I already found out that the matrix associated with the linear application had rank $1$.
So, now, I put the vector in the matrix as a column and we have:
$(A|\vec v) = \begin{pmatrix} -2 & -1 & 0 & 1 & -k\\ 2 & 1 & 0 & -1 & h - 2\\ -4 & -2 & 0 & 2 & k\\ 0 & 0 & 0 & 0 & h^2 - 4 \end{pmatrix}$
I know that at this time I have to find the determinants and see what happens. The problem here is that I don't know how it works with two parameters, since I've always done it with just one. Do I have two different counter-image because of this? What does it change from the single parameter exercise?
From what I get, the determinants are $2x2$ so the rank is max $2$, depending on the parameters. We would have:
$|A'| = h - 2 - k$
$|A''| = -2h + 4 - k$
$|A'''| = 2h^2 - 8$
If the last one is $0$ (so $h = 2$), then the rank would be the same as the rank of the linear application's matrix, and so we could proceed. But what about the $k$ in the other two cases? Should I stick with $h = 2$ and then being forced to put $k = 0$? Or I have to do it in a different way?

 

[Mark44]

Homework Statement

I have this linear application:
$f : \mathbb {R}^4 \rightarrow \mathbb {R}^4$
$f((x, y, z, t)) = (-2x - y + t, 2x + y - t, -4x - 2y + 2t, 0)$
and then I have this vector which I have to find the counter-image of it:
$\vec v = (-k, h - 2, k, h^2 - 4)$
With $h$ and $k$ real parameters that vary.

Homework Equations

The Attempt at a Solution

I already found out that the matrix associated with the linear application had rank $1$.
So, now, I put the vector in the matrix as a column and we have:
$(A|\vec v) = \begin{pmatrix} -2 & -1 & 0 & 1 & -k\\ 2 & 1 & 0 & -1 & h - 2\\ -4 & -2 & 0 & 2 & k\\ 0 & 0 & 0 & 0 & h^2 - 4 \end{pmatrix}$
I know that at this time I have to find the determinants and see what happens.

Find the determinants? I don't know what you mean by this.
Just row-reduce the matrix above, and you should see what values of h and k must have.

The problem here is that I don't know how it works with two parameters, since I've always done it with just one. Do I have two different counter-image because of this? What does it change from the single parameter exercise?
From what I get, the determinants are $2x2$ so the rank is max $2$, depending on the parameters. We would have:
$|A'| = h - 2 - k$
$|A''| = -2h + 4 - k$
$|A'''| = 2h^2 - 8$

I don't know what the equations above represent. What are A', A'', and A'''?

If the last one is $0$ (so $h = 2$), then the rank would be the same as the rank of the linear application's matrix, and so we could proceed. But what about the $k$ in the other two cases? Should I stick with $h = 2$ and then being forced to put $k = 0$? Or I have to do it in a different way?

In my work, I found that it must be true that k = 0 and h = 2. The preimage of f (what you're calling the counterimage) is a three-dimensional subspace of R⁴.

 

[Kernul]

Find the determinants? I don't know what you mean by this.

Our professor taught us to find the determinants and then the rank. If the parameter is equivalent to a number that would make the rank of the $(A|\vec v)$ equal to the rank of $A$, then we have to find the preimage by putting the number that the parameter is associated with.
For example, if $h = 2$ this implies that $rank(A) = rank (A|\vec v)$ and so we put all in a system and start the calculus to find the preimage.

I don't know what the equations above represent. What are A', A'', and A'''?

These are the three determinants that are different from $0$.
$A' = \begin{vmatrix} 1 & -k\\ -1 & h - 2 \end{vmatrix}$
$A'' = \begin{vmatrix} -1 & h - 2\\ 2 & k \end{vmatrix}$
$A''' = \begin{vmatrix} 2 & k\\ 0 & h^2 - 4 \end{vmatrix}$

Just row-reduce the matrix above, and you should see what values of h and k must have.

Should I row-reduce the matrix always in this kind of exercise?
I would have something like this if I row-reduce:
\begin{pmatrix}
-2 & -1 & 0 & 1 & -k \\
0 & 0 & 0 & 0 & h - 2 - k\\
0 & 0 & 0 & 0 & 3k\\
0 & 0 & 0 & 0 & h^2 - 4\\
\end{pmatrix}
And in a system would be:
\begin{cases}
-2x - y + t = -k \\
0 = h - 2 - k \\
0 = 3k \\
0 = h^2 - 4
\end{cases}
\begin{cases}
-2x - y + t = 0 \\
h = 2 \\
k = 0 \\
h = 2
\end{cases}
Is this right?

 

[Ray Vickson]

Find the determinants? I don't know what you mean by this.
Just row-reduce the matrix above, and you should see what values of h and k must have.
I don't know what the equations above represent. What are A', A'', and A'''?

In my work, I found that it must be true that k = 0 and h = 2. The preimage of f (what you're calling the counterimage) is a three-dimensional subspace of R⁴.

Forget determinants. In fact, why even bother with matrices? You have a simple linear system of 4 equations in 4 unknowns:
$$\begin{array}{rcl}
-2x-y+t&=&-k \\
2x+y-t&=&h-2\\
-4x-2y+2t&=& k\\
0 &=&h^2-4
\end{array}
$$
If $h = 2$ or $h = -2$ you will be left with three equations in the three unknowns $x,y,t$, which you can just solve using elementary elimination methods faster than you could even write down the matrices. If $h \neq \pm 2$ your $\vec{v}$ cannot "come from" any $(x,y,z,t)$ via $f(x,y,z,t)$. The variable $z$ does not appear anywhere in your $f(x,y,z,t)$ as you have written it.

 

[Kernul]

I don't understand how to resolve that system. I ended up with something like this:
$\begin{cases} -h + 2 = -k \\ y = -2x + t + h - 2 \\ -2h + 4 = k \\ h = \pm 2 \end{cases}$
Did I do something wrong?

If h≠±2h≠±2h \neq \pm 2 your ⃗vv→\vec{v} cannot "come from" any (x,y,z,t)(x,y,z,t)(x,y,z,t) via f(x,y,z,t)f(x,y,z,t)f(x,y,z,t)

I don't understand what you mean by "$\vec v$ cannot come from any $(x, y, z, t)$ via $f(x, y, z, t)$". You mean that it doesn't have a preimage if it's not $h \neq \pm 2$?

 

[Ray Vickson]

I don't understand how to resolve that system. I ended up with something like this:
$\begin{cases} -h + 2 = -k \\ y = -2x + t + h - 2 \\ -2h + 4 = k \\ h = \pm 2 \end{cases}$
Did I do something wrong?I don't understand what you mean by "$\vec v$ cannot come from any $(x, y, z, t)$ via $f(x, y, z, t)$". You mean that it doesn't have a preimage if it's not $h \neq \pm 2$?

Yes, that is exactly what I mean: the resulting $\vec{v}$ is not in the "range" of $f(x,y,z,t)$.

Anyway, if $h = 2$ the equations imply that $k = 0$, and so the resulting pre-image of $\vec{v}$ is $\{(x,y,z,t): t = 2x+y, x,y,z,\in R \}$. Now look at the other case, $h = -2$, to see what you get.

 

[Kernul]

Now look at the other case, h=−2h=−2h = -2, to see what you get.

I would have $k = 8$ and $k = -4$, which is impossible so it HAS to be $h = 2$, right?

 

[Ray Vickson]

I would have $k = 8$ and $k = -4$, which is impossible so it HAS to be $h = 2$, right?

Yes, I believe so.