Exercise with Linear Transformation

[Kernul]

Homework Statement

Being $f : \mathbb R^4\rightarrow\mathbb R^4$ the endomorphism defined by:
$$f((x,y,z,t)) = (13x + y - 2z + 3t, 10y, 9z + 6t, 6z + 4t)$$

1) Determine the basis and dimension of $Ker(f)$ and $Im(f)$. Complete the base chosen in $Ker(f)$ into a base of $\mathbb R^4$;
2) Determine the basis and dimension of the subspaces of $\mathbb R^4 Ker(f) \cap Im(f)$ and $Ker(f) + Im(f)$;
3) Determine the preimage of the vector $\vec v = (0, 2h, 1, h - 4)$, with the variation of the real parameter $h$;
4) Find, if they exist, the values of the $h$ parameter with $[2\vec e_1 + \vec e_2, 3\vec e_3, \vec v, -\vec e_2 + \vec e_4]$ being a base of $\mathbb R^4$. $ε = (\vec e_1, \vec e_2, \vec e_3, \vec e_4)$ is the canonic base of $\mathbb R^4$.
5) Having with $A$ the matrix $M_{εε}(f)$ associated to $f$ in respect to the canonic base, find out if $A$ is diagonalizable and, if so, determine a diagonalizing (?) matrix $P$ and the corresponding diagonal matrix $D$ to which $A$ is similar.
6) Find out if the matrix $B = \begin{pmatrix} 3 & 0 & 0\\ 1 & 3 & 0\\ 1 & 0 & 1\end{pmatrix}$ is diagonalizable and show a maximum independent system of eigenvectors of $B$

Homework Equations

Rank-Nullity Theorem
Operation on the vector subspaces (intersection and sum)
Canonic Base
Diagonalization
Eigenvectors and eigenvalues

The Attempt at a Solution

So, I know how to do the first point of the exercise but I found myself stuck.
In order to find the dimension and basis of both $Ker(f)$ and $Im(f)$ I first have to find the rank of the matrix $$A = \begin{pmatrix} 13 & 1 & -2 & 3\\ 0 & 10 & 0 & 0\\ 0 & 0 & 9 & 6\\ 0 & 0 & 6 & 4 \end{pmatrix}$$
From what I see, the matrix' rank is 4. Knowing that $ρ(A) = dim(Im(f))$, the base of $Im(f)$ is made of all the columns of $A$. So: $$B_{Im(f)} = [A^1, A^2, A^3, A^4]$$
Now, knowing that $dim(Ker(f)) = dim(A) - dim(Im(f))$, I have $dim(Ker(f)) = 4 - 4 = 0$. This means that $Ker(f)$ has no basis, right? Then how do I complete the basis of $Ker(f)$(as the exercise asks) if there is none?
For now I would like to concentrate on this first point and then ask other questions for the following points, if needed.

 

[Mark44]

Homework Statement

Being $f : \mathbb R^4\rightarrow\mathbb R^4$ the endomorphism defined by:
$$f((x,y,z,t)) = (13x + y - 2z + 3t, 10y, 9z + 6t, 6z + 4t)$$

1) Determine the basis and dimension of $Ker(f)$ and $Im(f)$. Complete the base chosen in $Ker(f)$ into a base of $\mathbb R^4$;
2) Determine the basis and dimension of the subspaces of $\mathbb R^4 Ker(f) \cap Im(f)$ and $Ker(f) + Im(f)$;
3) Determine the preimage of the vector $\vec v = (0, 2h, 1, h - 4)$, with the variation of the real parameter $h$;
4) Find, if they exist, the values of the $h$ parameter with $[2\vec e_1 + \vec e_2, 3\vec e_3, \vec v, -\vec e_2 + \vec e_4]$ being a base of $\mathbb R^4$. $ε = (\vec e_1, \vec e_2, \vec e_3, \vec e_4)$ is the canonic base of $\mathbb R^4$.
5) Having with $A$ the matrix $M_{εε}(f)$ associated to $f$ in respect to the canonic base, find out if $A$ is diagonalizable and, if so, determine a diagonalizing (?) matrix $P$ and the corresponding diagonal matrix $D$ to which $A$ is similar.

Item 6 below seems unrelated to this problem. Please start a new thread with that problem.

6) Find out if the matrix $B = \begin{pmatrix} 3 & 0 & 0\\ 1 & 3 & 0\\ 1 & 0 & 1\end{pmatrix}$ is diagonalizable and show a maximum independent system of eigenvectors of $B$

Homework Equations

Rank-Nullity Theorem
Operation on the vector subspaces (intersection and sum)
Canonic Base
Diagonalization
Eigenvectors and eigenvalues

The Attempt at a Solution

So, I know how to do the first point of the exercise but I found myself stuck.
In order to find the dimension and basis of both $Ker(f)$ and $Im(f)$ I first have to find the rank of the matrix $$A = \begin{pmatrix} 13 & 1 & -2 & 3\\ 0 & 10 & 0 & 0\\ 0 & 0 & 9 & 6\\ 0 & 0 & 6 & 4 \end{pmatrix}$$
From what I see, the matrix' rank is 4.

No.
It's fairly obvious (at least to me) that the rank of A is 3. To see this for yourself, row-reduce the matrix.

Knowing that $ρ(A) = dim(Im(f))$, the base of $Im(f)$ is made of all the columns of $A$. So: $$B_{Im(f)} = [A^1, A^2, A^3, A^4]$$
Now, knowing that $dim(Ker(f)) = dim(A) - dim(Im(f))$, I have $dim(Ker(f)) = 4 - 4 = 0$. This means that $Ker(f)$ has no basis, right? Then how do I complete the basis of $Ker(f)$(as the exercise asks) if there is none?
For now I would like to concentrate on this first point and then ask other questions for the following points, if needed.

 

[Kernul]

Yeah, I know it wasn't part of the exercise but it was put there. I don't know why.

Item 6 below seems unrelated to this problem. Please start a new thread with that problem.
No.
It's fairly obvious (at least to me) that the rank of A is 3. To see this for yourself, row-reduce the matrix.

Oh yeah, I'm sorry. I tried that but it seems I didn't quite try enough...
Anyway, doing the calculations, I find myself with: $B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}]$
In order to complete the base of $Ker(f)$ I'll put it together with the base of $Im(f)$, so:$$B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}, \begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}]$$
I hope I'm right until now.

Now the second point.
I have to find the base and dimension of $Ker(f) \cap Im(f)$ and $Ker(f)+Im(f)$, knowing that $dim(Im(f)) = 3$ and $dim(Ker(f)) = 1$. At this point, do I have to find a vector that is in both of them? If yes, should I do something like this:$$\vec v = αA^1 + βA^2 +γA^3 + ωKer(f)$$

 

[Mark44]

Yeah, I know it wasn't part of the exercise but it was put there. I don't know why.Oh yeah, I'm sorry. I tried that but it seems I didn't quite try enough...
Anyway, doing the calculations, I find myself with: $B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}]$

No, this isn't right. If you multiply A times the vector above, you don't get the zero vector. Check your work in row reducing A.

In order to complete the base of $Ker(f)$ I'll put it together with the base of $Im(f)$, so:$$B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}, \begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}]$$
I hope I'm right until now.

?
I don't know what you're doing here. As it turns out, dim(Ker(A)) = 1, so a basis for the kernel would consist of one vector.

Now the second point.
I have to find the base and dimension of $Ker(f) \cap Im(f)$ and $Ker(f)+Im(f)$, knowing that $dim(Im(f)) = 3$ and $dim(Ker(f)) = 1$. At this point, do I have to find a vector that is in both of them?

For the first part of item 2, $Ker(f) \cap Im(f)$, yes you need to find the vectors that are in both subspaces.
For the second part, $Ker(f)+Im(f)$, how is '+' defined for two sets?

If yes, should I do something like this:$$\vec v = αA^1 + βA^2 +γA^3 + ωKer(f)$$

I'm not sure you have the right vectors for your basis of Im(f).

 

[Kernul]

No, this isn't right. If you multiply A times the vector above, you don't get the zero vector. Check your work in row reducing A.

I made some miscalculations. It's $-1/3$ instead of $-1/13$ and $-2/3$ instead of $-1$.

I don't know what you're doing here. As it turns out, dim(Ker(A)) = 1, so a basis for the kernel would consist of one vector.

The exercise asks me

Complete the base chosen in $Ker(f)$ into a base of $\mathbb R^4$;

so I know that I have to put together both $Ker(f)$ and $Im(f)$ in order to have a basis of $\mathbb R^4$. Or am I doing something wrong? Because it's something I learned recently.

For the first part of item 2, Ker(f)∩Im(f)Ker(f)∩Im(f)Ker(f) \cap Im(f), yes you need to find the vectors that are in both subspaces.
For the second part, Ker(f)+Im(f)Ker(f)+Im(f)Ker(f)+Im(f), how is '+' defined for two sets?

I know that the dimension of $Ker(f) + Im(f) = dim(Ker(f)) + dim(Im(f)) - dim(Ker(f) \cap Im(f))$ and that the sum is the subspace $$Ker(f) + Im(f) = {\vec v + \vec w : \vec v \in Ker(f), \vec w \in Im(f)}$$. I have to first find the dimension of the intersection in order to find the dimension of the sum.

I'm not sure you have the right vectors for your basis of Im(f).

Why? Now that I know that the dimension of $Im(f)$ is 3, the basis would be $$B_{Im(f)} = [A^1, A^2, A^3]$$ So these are the vectors I will have to use in order to find the intersection.

 

[Mark44]

I made some miscalculations. It's $-1/3$ instead of $-1/13$ and $-2/3$ instead of $-1$.

OK, that's better. You could just as well have chosen $\begin{bmatrix} 1 \\ 0 \\ 2 \\ -3 \end{bmatrix}$ for your basis vector.

The exercise asks me so I know that I have to put together both $Ker(f)$ and $Im(f)$ in order to have a basis of $\mathbb R^4$. Or am I doing something wrong? Because it's something I learned recently.

For the four vectors, I would use the kernel basis vector you found, plus three linearly independent vectors that form a basis for the image of f.

I know that the dimension of $Ker(f) + Im(f) = dim(Ker(f)) + dim(Im(f)) - dim(Ker(f) \cap Im(f))$ and that the sum is the subspace $$Ker(f) + Im(f) = {\vec v + \vec w : \vec v \in Ker(f), \vec w \in Im(f)}$$. I have to first find the dimension of the intersection in order to find the dimension of the sum.

Can any vector be in both the kernel and the image?

Why? Now that I know that the dimension of $Im(f)$ is 3, the basis would be $$B_{Im(f)} = [A^1, A^2, A^3]$$ So these are the vectors I will have to use in order to find the intersection.

Writing them as $A^1, A^2, A^3$ is confusing, as it suggests you are raising them to powers. It's better to write them as $A_1, A_2, A_3$, with subscripts, not exponents. Again, these three vectors should be a basis for Im(f).

 

[Kernul]

For the four vectors, I would use the kernel basis vector you found, plus three linearly independent vectors that form a basis for the image of f.

Yes, that's exactly what I wrote.

$$B_{Ker(f)} = [\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}, \begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}]$$

with $\begin{pmatrix} -1/13\\ 0\\ -1\\ 1 \end{pmatrix}$ being the kernel basis vector and $\begin{pmatrix} 13\\ 0\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 1\\ 10\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} -2\\ 0\\ 9\\ 6 \end{pmatrix}$ being the image basis vectors which are $A^1, A^2, A^3$ respectively.

Can any vector be in both the kernel and the image?

I guess no? I don't know how to check that. Do I have to calculate something?

Writing them as A1,A2,A3A1,A2,A3A^1, A^2, A^3 is confusing, as it suggests you are raising them to powers. It's better to write them as A1,A2,A3A1,A2,A3A_1, A_2, A_3, with subscripts, not exponents. Again, these three vectors should be a basis for Im(f).

I'm sorry but my lecturer taught us that when it is a column of a matrix we write $A^x$, if it is a row of a matrix we write $A_x$.
And yes, that's what I wrote:

$$B_{Im(f)} = [A^1, A^2, A^3]$$

 

[Mark44]

Can any vector be in both the kernel and the image?

I guess no? I don't know how to check that. Do I have to calculate something?

How are kernel and image defined?

 

[HallsofIvy]

I would NOT write the first problem in terms of matrices. To find the kernel simply means finding x, y, z, and t such that
13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0. From the second equation, y= 0 and then the other equations are easy.

 

[Kernul]

How are kernel and image defined?

$Ker(f) = \{\vec v \in V : f(\vec v) = \vec 0\}$
$Im(f) = \{\vec w \in W : \exists \vec v \in V, f(\vec v) = \vec w\} = \{f(\vec v) : \vec v \in V\}$
So it is a vector $\vec v = \vec 0$? The null vector?

And yeah, I know I wasn't bound to use the matrix. I actually put the first three rows in a system and assigned $t = a$.

 

[HallsofIvy]

For any linear transformation, T, T(0)= 0 so the 0 vectors is always in the kernel- in fact the kernel is a subspace. But it is not necessarily the only vector in the kernel. As I said before, solve the equations 13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0 for x, y, z, and t. Since you have only three equations in three unknowns you will have at least some of the variables in terms of others.

 

[Kernul]

For any linear transformation, T, T(0)= 0 so the 0 vectors is always in the kernel- in fact the kernel is a subspace. But it is not necessarily the only vector in the kernel. As I said before, solve the equations 13x+y−2z+3t= 0,10y= 0, 9z+6t= 0, and 6z+4t= 0 for x, y, z, and t. Since you have only three equations in three unknowns you will have at least some of the variables in terms of others.

I have something like this:
\begin{array}{l}
13x + y - 2z + 3t = 0\\
10y = 0\\
9z + 6t = 0\\
6z + 4t = 0
\end{array}
\begin{array}{l}
x = + \frac{2z}{13} - \frac{3t}{13}\\
y = 0\\
z = - \frac{6t}{9} = - \frac{2t}{3}\\
t = - \frac{6z}{4} = - \frac{3z}{2}
\end{array}

 

[HallsofIvy]

So finish it! If x= 2z/13- 3t/13 and z= -2t/3, what is x in terms or 6 only?

 

[Kernul]

I get this, which doesn't give me a number:
$\begin{array}{l} x = - \frac{1z}{3}\\ y = 0\\ z = - \frac{2t}{3}\\ t = - \frac{3z}{2} \end{array}$
And if I replace $z$ in $t$ as $- \frac{2t}{3}$ I get $t = t$.