MHB Diagonalizable transformation - Existence of basis

[mathmari]

Hey! :giggle:

Let $1\leq n\in \mathbb{N}$ and for $x=\begin{pmatrix}x_1\\ x_2\\ \vdots \\ x_n\end{pmatrix}, \ x=\begin{pmatrix}x_1\\ x_2\\ \vdots \\ x_n\end{pmatrix}\in \mathbb{R}^n$ and let $x\cdot y=\sum_{i=1}^nx_iy_i$ the dot product of $x$ and $y$.

Let $S=\{v\in \mathbb{R}^n\mid v\cdot v=1\}$ and for $v\in S$ let $\sigma_v$ be a map defined by $\sigma_v:\mathbb{R}^n\rightarrow \mathbb{R}^n, \ x\mapsto x-2(x\cdot v)v$.

I have shown that it holds for $v\in S$ and $x,y\in \mathbb{R}^n$ that $\sigma_v(x)\cdot \sigma_v(y)=x\cdot y$.

Let $v\in S$. I have shown that $\sigma_v^2=\text{id}_{\mathbb{R}^n}$. To show that $\sigma_v$ is diagonalizable do we have to calculate the matrix of that transformation?
Let $n=4$ and $v,w\in S$. I want to show that there is $0\leq \alpha\leq 2\pi$ and an orthogonal basis $B$ of $\mathbb{R}^4$ such that the matrix $\sigma_v\circ\sigma_w$ as for the basis $B$ is of the form $$D=\begin{pmatrix}\cos \alpha & -\sin \alpha & 0 & 0 \\ \sin \alpha & \cos \alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$ Could you give me a hint for that? Do we have to find the matrix of $\sigma_v\circ\sigma_w$ and calculate the images of the elements of the basis $B$ and write then write the result as a linear combination of the elements of $B$ and the result should be the matrix $D$? :unsure:

 

[I like Serena]

Hey mathmari!

A matrix is diagonalizable iff the eigenvectors form a basis.
This is actually the definition of diagonalizable as it applies to transformations in general.
So it suffices if we can show that $\sigma_v$ has $n$ independent eigenvectors.

Is $v$ an eigenvector?
What about a vector perpendicular to $v$?

 

[mathmari]

So it suffices if we can show that $\sigma_v$ has $n$ independent eigenvectors.

Is $v$ an eigenvector?
What about a vector perpendicular to $v$?

$v$ is an eigenvector if $\sigma_v(v)=\lambda v$ for some $\lambda\in \mathbb{R}$, right?
We have that $ \sigma_v(v)=v-2(v\cdot v)v =v-2v=-v=(-1)v $. That mens that $v$ is an eigenvector for the eigenvalue $\lambda=-1$.
Is that correct? :unsure:

Let $w$ be a vector perpendicular to $v$, then $w\cdot v=0$.
We have that $ \sigma_v(w)=w-2(w\cdot v)v =w$. That mens that $w$ is an eigenvector for the eigenvalue $\lambda=1$.
Is that correct? :unsure:

Now we have find two independent eigenvectors, but we need $n$. :unsure:

 

[I like Serena]

All correct. (Nod)

How many independent vectors can we find that are perpendicular to $v$?

 

[mathmari]

All correct. (Nod)

How many independent vectors can we find that are perpendicular to $v$?

Are there $n-1$ perpendicular vectors to $v$ since we are in $\mathbb{R}^n$ ? :unsure:

 

[I like Serena]

Are there $n-1$ perpendicular vectors to $v$ since we are in $\mathbb{R}^n$ ? :unsure:

Yes. Any orthogonal basis of $\mathbb R^n$ that includes $v$, contains $n-1$ vectors that are orthogonal to $v$.

 

[mathmari]

Yes. Any orthogonal basis of $\mathbb R^n$ that includes $v$, contains $n-1$ vectors that are orthogonal to $v$.

Ahh ok!

If we want to determine the eigenvalues of $\sigma_v$ then do we do the following?
$$\sigma_v(x)=\lambda x \Rightarrow \sigma_v\left (\sigma_v(x)\right )=\sigma_v\left (\lambda x \right ) \Rightarrow \sigma_v^2(x)=\lambda \sigma_v(x) \Rightarrow x=\lambda \sigma_v(x)\Rightarrow x=\lambda \cdot \lambda x\Rightarrow x=\lambda^2 x\Rightarrow (\lambda^2-1) x=0\Rightarrow \lambda=\pm 1$$ So the eigenvalues are $-1$ and $1$. Is that corect? :unsure:

To find the dimension of the respective eigenspace do we calculate the geometric multiplicity which has to be equal to the algebraic multiplicity since $\sigma_v$ is diagonizable? Or how do we calculate the dimension in this case? :unsure:

 

[I like Serena]

Yes. That works in both cases. (Nod)

Note that we've already found $1$ eigenvector $v$ for the eigenvalue $\lambda=-1$, and $n-1$ independent eigenvectors for the eigenvalue $\lambda=1$.
Since the dimension of the space is $n$, that implies that $\lambda=-1$ has both algebraic and geometric multiplicity of $1$.
And $\lambda=1$ has both algebraic and geometric multiplicity of $n-1$.
That is, we don't need to use the argument of diagonalizability to conclude that.

 

[mathmari]

Since the dimension of the space is $n$, that implies that $\lambda=-1$ has both algebraic and geometric multiplicity of $1$.
And $\lambda=1$ has both algebraic and geometric multiplicity of $n-1$.

How do we know that the algebraic multiplicity of $\lambda=1$ is $n-1$ ? :unsure:

 

[I like Serena]

How do we know that the algebraic multiplicity of $\lambda=1$ is $n-1$ ?

Because an eigenvalue's geometric multiplicity cannot exceed its algebraic multiplicity.

 

[mathmari]

Because an eigenvalue's geometric multiplicity cannot exceed its algebraic multiplicity.

Ah because there are $n-1$ vectors like $w$, i.e. there are $n-1$ eigenvectors for $\lambda=1$ that means that the geometric multiplicity is $n-1$ ? :unsure:

 

[I like Serena]

Ah because there are $n-1$ vectors like $w$, i.e. there are $n-1$ eigenvectors for $\lambda=1$ that means that the geometric multiplicity is $n-1$ ?

Yep. (Nod)

 

[mathmari]

Ok! :geek:

Let $n=4$ and $v,w\in S$. I want to show that there is $0\leq \alpha\leq 2\pi$ and an orthogonal basis $B$ of $\mathbb{R}^4$ such that the matrix $\sigma_v\circ\sigma_w$ as for the basis $B$ is of the form $$D=\begin{pmatrix}\cos \alpha & -\sin \alpha & 0 & 0 \\ \sin \alpha & \cos \alpha & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{pmatrix}$$ Could you give me a hint for that? Do we have to find the matrix of $\sigma_v\circ\sigma_w$ and calculate the images of the elements of the basis $B$ and write then write the result as a linear combination of the elements of $B$ and the result should be the matrix $D$? :unsure:

Could you give me a hint also for that? :unsure:

 

[I like Serena]

Could you give me a hint also for that?

Can we find a basis of $\mathbb R^4$ that has 2 vectors in it that are orthogonal to both $v$ and $w$?

 

[mathmari]

Can we find a basis of $\mathbb R^n$ that has 2 more vectors in it that are orthogonal to both $v$ and $w$?

Their dot product is one vector that it orthogonal to both, right? How can we find some more? I got stuck right now. :unsure:

 

[I like Serena]

The dot product is a scalar and not a vector. Furthermore, the cross product is not defined in 4 dimensions.

Can't we just state that such a basis must exist?
We don't have to actually find such vectors.

Either way, we can find them by starting with v and w, and by adding each of the unit vectors until we have 4 independent vectors. After that we can use the Gramm-Scmidt orthogonalization process to find 2 vectors that are orthogonal to both v and w.

 

[topsquark]

I don't mean to butt in but I would just like to say how much I am enjoying these threads. I know the material in general but I'm getting some extra details that I have missed. Both of you keep up the good work!

-Dan

 

[mathmari]

Either way, we can find them by starting with v and w, and by adding each of the unit vectors until we have 4 independent vectors. After that we can use the Gramm-Scmidt orthogonalization process to find 2 vectors that are orthogonal to both v and w.

So that means that $B$ will be then the set of $v$, $w$ and the two vectors that we get from the Gramm-Scmidt orthogonalization process? :unsure:

But just stating that such a basis exist, how can we find the form of the matrix $D$ ? I got stuck right now. :unsure:

 

[I like Serena]

What do $\sigma_v$ and $\sigma_w$ look like with respect to a basis that contains v, w, and vectors orthogonal to both v and w?

 

[mathmari]

What do $\sigma_v$ and $\sigma_w$ look like with respect to a basis that contains v, w, and vectors orthogonal to both v and w?

They are invariant, aren't they? :unsure:

 

[I like Serena]

They are invariant, aren't they?

The extra orthogonal vectors are indeed invariant with respect to both $\sigma_v$ and $\sigma_w$. (Nod)

So? (Wondering)

 

[mathmari]

The extra orthogonal vectors are indeed invariant with respect to both $\sigma_v$ and $\sigma_w$. (Nod)

So? (Wondering)

That's why we get the last two columns of the matrix $D$, right? :unsure:

 

[I like Serena]

That's why we get the last two columns of the matrix $D$, right?

Yep.
Both $\sigma_v$ and $\sigma_w$ have a matrix with respect to that basis that have the same last two columns as $D$.

 

[mathmari]

Yep.
Both $\sigma_v$ and $\sigma_w$ have a matrix with respect to that basis that have the same last two columns as $D$.

So the first two columns of $D$ correspond to the vectors $v$ and $w$ ? :unsure:

 

[I like Serena]

So the first two columns of $D$ correspond to the vectors $v$ and $w$ ?

Let's assume for now that $v$ is independent from $w$.
And let $b_3$ and $b_4$ be vectors that are orthogonal to both $v$ and $w$.
Then we have that $\sigma_v(v)=-v$, so the first column of the matrix of ${\sigma_v}$ with respect to the basis $(v,w,b_3,b_4)$ is $\begin{pmatrix}-1\\0\\0\\0\end{pmatrix}$.
We also have that $\sigma_w(w)=-w$, so the second column of he matrix of ${\sigma_w}$ with respect to the basis $(v,w,b_3,b_4)$ is $\begin{pmatrix}0\\-1\\0\\0\end{pmatrix}$.

 

[mathmari]

Let's assume for now that $v$ is independent from $w$.
And let $b_3$ and $b_4$ be vectors that are orthogonal to both $v$ and $w$.
Then we have that $\sigma_v(v)=-v$, so the first column of the matrix of ${\sigma_v}$ with respect to the basis $(v,w,b_3,b_4)$ is $\begin{pmatrix}-1\\0\\0\\0\end{pmatrix}$.
We also have that $\sigma_w(w)=-w$, so the second column of he matrix of ${\sigma_w}$ with respect to the basis $(v,w,b_3,b_4)$ is $\begin{pmatrix}0\\-1\\0\\0\end{pmatrix}$.

Ok, but shouldn't these vectors be then the first and second column of $D$? How do we get then the cosine and the sine? :unsure:

 

[I like Serena]

We can reduce the 4-dimensional problem to a 2-dimensional problem.
And we already know that the composition of two reflections in 2 dimensions is a rotation don't we? :unsure:

 

[mathmari]

We can reduce the 4-dimensional problem to a 2-dimensional problem.
And we already know that the composition of two reflections in 2 dimensions is a rotation don't we? :unsure:

Ah so we consider the rotation matrix, right? :unsure:

So $\sigma_v$ and $\sigma_w$ is the composition of two reflections, which is a rotation. Therefore the matrix $D$ must contain for these vectors the rotation matrix? :unsure:

 

[I like Serena]

Ah so we consider the rotation matrix, right?

So $\sigma_v$ and $\sigma_w$ is the composition of two reflections, which is a rotation. Therefore the matrix $D$ must contain for these vectors the rotation matrix?

Yes. (Nod)

Generally, the composition of 2 reflections is a rotation of double the angle between the normals of the planes of reflection.

 

[mathmari]

Yes. (Nod)

Generally, the composition of 2 reflections is a rotation of double the angle between the normals of the planes of reflection.

So do we have to consider that $v$ and $w$ are independent or do we have to check also the case that they are nont independent? :unsure: