Exercise with intersection and sum

[Kernul]

My professor did this exercise that I didn't quite get how she went through all of it.
We have a $U = {(x, y, z, t) : x+y+z+t = 0}$ and $B_{Im(f)} = \left[ \begin{pmatrix} 7 \\ -3 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 3 \\ -3 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 5 \\ 0 \\ 1 \\ -5 \end{pmatrix}\right]$ with $dim(Im(f)) = 3$.
The exercise asks the base and dimension of $Im(f) \cap U$ and of $Im(f) + U$.

Now she starts the exercise doing this:
$x = - y - z - t$
and
$\left\{ \begin{array}{l} y = a\\ z = b\\ t = c\\ x = - a - b - c \end{array} \right.$
After this she writes
$U = \left\{a \begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix} + b \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + c \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \end{pmatrix} : a, b, c \in \mathbb{R}\right\}$
then she writes $dim(U) = 3$ and $B_U = \left[ \begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} -1 \\ 0 \\ 0 \\ 1 \end{pmatrix}\right]$
So here they come my first questions:
1) Why did she put $x = - y - z - t$?
2) Why did she put everything in a system and put $y = a, z = b, t = c$?

Then she starts with the intersection saying
$\vec v \in Im(f) \Rightarrow \vec 0$
and then she writes
$\vec v = \alpha \begin{pmatrix} 7 \\ -3 \\ 0 \\ 0 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ -3 \\ 0 \\ 0 \end{pmatrix} + \gamma \begin{pmatrix} 5 \\ 0 \\ 1 \\ -5 \end{pmatrix} = \begin{pmatrix} 7\alpha + 3\beta + 5\gamma \\ -3\alpha - 3\beta \\ \gamma \\ -5\gamma \end{pmatrix}$
then she put all these in a single equation
$7\alpha + 3\beta + 5\gamma -3\alpha - 3\beta + \gamma -5\gamma = 0$
that becomes $\gamma = -4\alpha$.
After finding that, she substitutes to all the $\gamma$ in the last matrix and does the operations and she writes
$Im(f) \cap U = \left\{\begin{pmatrix} - 13\alpha + 3\beta \\ -3\alpha - 3\beta \\ -4\alpha \\ 20\alpha \end{pmatrix} = \alpha \begin{pmatrix} -13 \\ -3 \\ -4 \\ 20 \end{pmatrix} + \beta \begin{pmatrix} 3 \\ -3 \\ 0 \\ 0 \end{pmatrix} : \alpha, \beta \in \mathbb{R}\right\}$
then she writes $dim(Im(f) \cap U) = 2$ and using a formula(that I know it's the Graussmann formula) writes $dim(Im(f) + U) = dim(Im(f)) + dim(U) - dim(Im(f) \cap U) = 3 + 3 - 2 = 4$ and then she writes $Im(f) + U = \mathbb{R}^4$
My questions are:
3) Why did she put the vector $\vec v$ only belonging to $Im(f)$? Shouldn't it belong to $U$ too in order to know the intersection between the two subspaces?
4) Why didn't she write, in the $\vec v$ equation, the independent vectors of $U$ too?
5) Why at one point she sum all the frist, second, third and fourth rows of the vectors in one matrix? And why after that she put everything in a normal equation?
6) And in the end, what does $Im(f) + U = \mathbb{R}^4$ mean?

Sorry for all these questions but really I can't understand how she did all this. I know how to solve it differently but I want to know why she did like this.

 

[haruspex]

1) that's the condition for (x,y,z,t) to be a point in U.
2) the idea is to express this point in terms of three parameters
3) how do you think she got that 'single equation'? She had a column vector representing v, then somehow produced a scalar equation from it by adding up the elements of the vector and saying that sum is zero. That is the condition for v being an element of U, is it not?
I'll leave it at that, since the rest may become clear to you now.

 

[Kernul]

1) that's the condition for (x,y,z,t) to be a point in U.

Why? Is this the only way? What would have happen if it had been something like $U = {(x, y, z, t) : x +2y +z = 0}$? I would have simply done $x = -2y - z$ and then put $y = a, z = b, t = c$? Does it work like this?

2) the idea is to express this point in terms of three parameters

This in order to have three linearly independent vectors of $U$, right? And does it have to be three parameters only?

3) how do you think she got that 'single equation'? She had a column vector representing v, then somehow produced a scalar equation from it by adding up the elements of the vector and saying that sum is zero. That is the condition for v being an element of U, is it not?

I know that the subspace of a general intersection $U \cap V$ is made of vectors $\vec v$ that belong to both subspaces and that you can write them as $\vec v = \alpha_1 \vec u_1 + \alpha_2 \vec u_2 = \beta_1 \vec v_1 + \beta_2 \vec v_2$. This means that
$\alpha_1 \vec u_1 + \alpha_2 \vec u_2 - \beta_1 \vec v_1 - \beta_2 \vec v_2 = \vec 0$
but in the exercise the professor didn't put the vectors of the other subspace, she only put those of $Im(f)$.
I know that we have to check if they are linearly independent. Like, for example, $\vec u_1, \vec u_2$ are independent and we have to check if $\vec v_1$ is dependent to $\vec u_1, \vec u_2$, checking this with $\vec v_1 = \alpha_1 \vec u_1 + \alpha_2 \vec u_2$. Some thing for $\vec v_2$. This is what I know about the intersection. Am I wrong?

 

[Mark44]

Why?

Because U was defined like this: $U = {(x, y, z, t) : x+y+z+t = 0}$

Is this the only way? What would have happen if it had been something like $U = {(x, y, z, t) : x +2y +z = 0}$? I would have simply done $x = -2y - z$ and then put $y = a, z = b, t = c$? Does it work like this?

Yes, pretty much. Alternatively you could do it like this:
x = -2y - z
y = y
z = ... z
t = ... t
If you stare at this long enough, you start to see that a vector v has this form:
$\vec{v} = \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$
Here, the parameters are y, z, and t.
The equation x + 2y + z = 0 has three variables, so two of the variables are going to be free (or parameters), plus your equation doesn't involve t at all, so it is free as well. I chose y and z (and t) to be the free variables, but you could choose any other two. With three free variables, geometrically the solution is going to be a 3-D space in $\mathbb{R}^43$. This space takes three vectors to describe.

This in order to have three linearly independent vectors of $U$, right? And does it have to be three parameters only?

I know that the subspace of a general intersection $U \cap V$ is made of vectors $\vec v$ that belong to both subspaces and that you can write them as $\vec v = \alpha_1 \vec u_1 + \alpha_2 \vec u_2 = \beta_1 \vec v_1 + \beta_2 \vec v_2$. This means that
$\alpha_1 \vec u_1 + \alpha_2 \vec u_2 - \beta_1 \vec v_1 - \beta_2 \vec v_2 = \vec 0$
but in the exercise the professor didn't put the vectors of the other subspace, she only put those of $Im(f)$.
I know that we have to check if they are linearly independent. Like, for example, $\vec u_1, \vec u_2$ are independent and we have to check if $\vec v_1$ is dependent to $\vec u_1, \vec u_2$, checking this with $\vec v_1 = \alpha_1 \vec u_1 + \alpha_2 \vec u_2$. Some thing for $\vec v_2$. This is what I know about the intersection. Am I wrong?

 

[haruspex]

$\vec{v} = \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

$\vec{v} = \begin{bmatrix} x \\ y \\ z \\ 0 \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ 1 \\0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\1\end{bmatrix}$

 

[Mark44]

$\vec{v} = \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$

$\vec{v} = \begin{bmatrix} x \\ y \\ z \\ 0 \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ 1 \\0 \end{bmatrix} + t\begin{bmatrix} 0 \\ 0 \\ 0 \\1\end{bmatrix}$

It took me a while to get the glitch out of my LaTeX, but I figured it out.
I don't see any reason why the t coordinate of $\vec{v}$ should be 0. $\vec{v} \in \mathbb{R}^4$, so it has four coordinates, but no restriction on t.

 

[haruspex]

in the exercise the professor didn't put the vectors of the other subspace, she only put those of Im(f)

She wrote v as a linear sum of the basis vectors of Im(f). She then applied the condition for membership of U to get a restriction on the coefficients. Therefore v has been defined as an arbitrary member of the intersection.

 

[Mark44]

She wrote v as a linear sum of the basis vectors of Im(f).

Gotcha. I got a little confused between the original problem and the other example the OP posted and asked about.

 

[Kernul]

So, in order to get $\vec v$ we can simply use one of the two subspaces? I mean, if I wanted to, could I have done something like this
$$\vec v = \alpha \begin{pmatrix}
-1 \\
1 \\
0 \\
0
\end{pmatrix} + \beta \begin{pmatrix}
-1 \\
0 \\
1 \\
0
\end{pmatrix} + \gamma \begin{pmatrix}
-1 \\
0 \\
0 \\
1
\end{pmatrix}$$
and solve it like the professor did with the vectors of $Im(f)$?

 

[haruspex]

So, in order to get $\vec v$ we can simply use one of the two subspaces? I mean, if I wanted to, could I have done something like this
$$\vec v = \alpha \begin{pmatrix}
-1 \\
1 \\
0 \\
0
\end{pmatrix} + \beta \begin{pmatrix}
-1 \\
0 \\
1 \\
0
\end{pmatrix} + \gamma \begin{pmatrix}
-1 \\
0 \\
0 \\
1
\end{pmatrix}$$
and solve it like the professor did with the vectors of $Im(f)$?

Not sure what you meant by the last phrase, but if you mean solve it by applying the conditions for being in Im(f), then yes.

 

[Kernul]

What do you mean "solve it by applying the conditions for being in $Im(f)$"? Are you saying that $U \subseteq Im(f)$?

 

[haruspex]

What do you mean "solve it by applying the conditions for being in $Im(f)$"? Are you saying that $U \subseteq Im(f)$?

No. You want v to be in both subspaces. Your prof started by taking it to be in Im(f), and expressed that by writing it as a sum of basis vectors. She then needed to make it a member of U as well. She did that by applying the condition for membership of U to the coefficients in the sum.
You can equally do it the other way around: write it as a sum of basis vectors of U then apply the condition for membership of Im(f).

 

[Kernul]

I don't understand what is the condition of membership. Where did she use it? Is the equation? Because all I see is calculations with vectors of $Im(f)$.

 

[haruspex]

I don't understand what is the condition of membership. Where did she use it? Is the equation? Because all I see is calculations with vectors of $Im(f)$.

After "and then she writes", there is an equation showing v as a linear sum of the basis vectors of Im(f). That makes v a member of Im(f).
Next, look at "and then she put all these in a single equation". To get from the equation for v to that equation she has simply added up the elements of v and set the result to zero. That is the condition for membership of U: x+y+z+t=0. So now v is in the intersection.

 

[Kernul]

Oooh! Now I got it!
And what if I had, instead of $U$ written like the exercise I showed, something like this
$$B_U = \left[\begin{pmatrix}
2 \\
-5 \\
3 \\
0 \\
\end{pmatrix}, \begin{pmatrix}
-7 \\
0 \\
9 \\
3 \\
\end{pmatrix}\right]$$
This means that I don't know that, for example, $x + y + z + t = 0$ or anything else like that, so I can't do the final equation.

 

[haruspex]

Oooh! Now I got it!
And what if I had, instead of $U$ written like the exercise I showed, something like this
$$B_U = \left[\begin{pmatrix}
2 \\
-5 \\
3 \\
0 \\
\end{pmatrix}, \begin{pmatrix}
-7 \\
0 \\
9 \\
3 \\
\end{pmatrix}\right]$$
This means that I don't know that, for example, $x + y + z + t = 0$ or anything else like that, so I can't do the final equation.

I don't understand... that second vector is not in U, and to form a basis for U you need three vectors.

 

[Kernul]

No, no! I mean like a new exercise!
Let's say that I have $U$ and $V$. $dim(U) = 2$ and $B_U$ with two vectors, then $dim(V) = 3$ and $B_V$ with three vectors. Let's say that I want to know $U \cap V$. In this case none of the two has a description like $x + y + z + t = 0$, right? So, how should you proceed in this case? Like I said here?

I know that the subspace of a general intersection U∩VU∩VU \cap V is made of vectors ⃗vv→\vec v that belong to both subspaces and that you can write them as ⃗v=α1⃗u1+α2⃗u2=β1⃗v1+β2⃗v2v→=α1u→1+α2u→2=β1v→1+β2v→2\vec v = \alpha_1 \vec u_1 + \alpha_2 \vec u_2 = \beta_1 \vec v_1 + \beta_2 \vec v_2. This means that
α1⃗u1+α2⃗u2−β1⃗v1−β2⃗v2=⃗0α1u→1+α2u→2−β1v→1−β2v→2=0→\alpha_1 \vec u_1 + \alpha_2 \vec u_2 - \beta_1 \vec v_1 - \beta_2 \vec v_2 = \vec 0
but in the exercise the professor didn't put the vectors of the other subspace, she only put those of Im(f)Im(f)Im(f).
I know that we have to check if they are linearly independent. Like, for example, ⃗u1,⃗u2u→1,u→2\vec u_1, \vec u_2 are independent and we have to check if ⃗v1v→1\vec v_1 is dependent to ⃗u1,⃗u2u→1,u→2\vec u_1, \vec u_2, checking this with ⃗v1=α1⃗u1+α2⃗u2v→1=α1u→1+α2u→2\vec v_1 = \alpha_1 \vec u_1 + \alpha_2 \vec u_2. Some thing for ⃗v2v→2\vec v_2. This is what I know about the intersection. Am I wrong?