# Show properties of matrix A

#### mathmari

##### Well-known member
MHB Site Helper
Hey!!

We have the matrix $A=\frac{1}{3}\begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{pmatrix}$. Show that there is an unit vector $v_1$, such that $A=I-2v_1v_1^T$.

We consider an orthogonal matrix $Q=\begin{pmatrix}v_1 & v_2 & v_3\end{pmatrix}$. Show that $Q^TAQ=\begin{pmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.

Using the last relation show that $\det A=1$.

For the first part I have done the following:

Let $v_1=\begin{pmatrix}x_1 & x_2 & x_3\end{pmatrix}^T$ with $|v_1|=\sqrt{x_1^2+x_2^2+x_3^2}=1$.

Then we have that \begin{equation*}I-2v_1v_1^T=\begin{pmatrix}1-2x_1^2 & -2x_1x_2 & -2x_1x_3 \\ -2x_1x_2 & 1-2x_2^2 & -2x_2x_3 \\ -2x_1x_3 & -2x_2x_3 & 1-2x_3^2\end{pmatrix}\end{equation*} We set this equal to $A$ and we have to solve a system.

Is this correct or should we show that in an other way?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
We have the matrix $A=\frac{1}{3}\begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{pmatrix}$. Show that there is an unit vector $v_1$, such that $A=I-2v_1v_1^T$.

For the first part I have done the following:
Let $v_1=\begin{pmatrix}x_1 & x_2 & x_3\end{pmatrix}^T$ with $|v_1|=\sqrt{x_1^2+x_2^2+x_3^2}=1$.
Then we have that \begin{equation*}I-2v_1v_1^T=\begin{pmatrix}1-2x_1^2 & -2x_1x_2 & -2x_1x_3 \\ -2x_1x_2 & 1-2x_2^2 & -2x_2x_3 \\ -2x_1x_3 & -2x_2x_3 & 1-2x_3^2\end{pmatrix}\end{equation*} We set this equal to $A$ and we have to solve a system.

Is this correct or should we show that in an other way?
Hey mathmari !!

That will work yes.

Just to mention an alternative, we might try to find the eigenvalues and eigenvectors of $A$.

#### mathmari

##### Well-known member
MHB Site Helper
Just to mention an alternative, we might try to find the eigenvalues and eigenvectors of $A$.
Let $\lambda$ be an eigenvalue and $v$ be the respective eigenvector. Then we have that $(A-\lambda I)v=0$. How does this help us for the above?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Let $\lambda$ be an eigenvalue and $v$ be the respective eigenvector. Then we have that $(A-\lambda I)v=0$. How does this help us for the above?
If we have indeed that $A=I-2v_1 v_1^T$, isn't $v_1$ an eigenvector then?
If so, which eigenvalue corresponds to it?

And consider a vector $v_2$ that is perpendicular to $v_1$.
Is it an eigenvector?
If so, which eigenvalue corresponds to it?

#### mathmari

##### Well-known member
MHB Site Helper
If we have indeed that $A=I-2v_1 v_1^T$, isn't $v_1$ an eigenvector then?
If so, which eigenvalue corresponds to it?
Since the vector $v_1$ is unit it holds that $v_1^Tv_1=1$, right?

We have that \begin{align*}A=I-2v_1 v_1^T &\Rightarrow A\cdot v_1=I\cdot v_1-2v_1 v_1^T\cdot v_1 \\ & \Rightarrow A\cdot v_1= v_1-2v_1 \left (v_1^T v_1\right )\\ & \Rightarrow A\cdot v_1= v_1-2v_1 \cdot 1 \\ & \Rightarrow A\cdot v_1= v_1-2v_1 \\ & \Rightarrow A\cdot v_1= -v_1 \\ & \Rightarrow A\cdot v_1+v_1 =0 \\ & \Rightarrow (A+I)\cdot v_1 =0\end{align*}
Which means that $-1$ is the eigenvalue, right?

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#### Klaas van Aarsen

##### MHB Seeker
Staff member
Since the vector $v_1$ is unit it holds that $v_1^Tv_1=1$, right?

We have that \begin{align*}A=I-2v_1 v_1^T &\Rightarrow A\cdot v_1=I\cdot v_1-2v_1 v_1^T\cdot v_1 \\ & \Rightarrow A\cdot v_1= v_1-2v_1 \left (v_1^T v_1\right )\\ & \Rightarrow A\cdot v_1= v_1-2v_1 \cdot 1 \\ & \Rightarrow A\cdot v_1= v_1-2v_1 \\ & \Rightarrow A\cdot v_1= -v_1 \\ & \Rightarrow A\cdot v_1+v_1 =0 \\ & \Rightarrow (A+I)\cdot v_1 =0\end{align*}
Which means that $-1$ is the eigenvalue, right?
Yep. So if $A=I-2v_1v_1^T$ it must have eigenvalue $-1$.

#### mathmari

##### Well-known member
MHB Site Helper
And consider a vector $v_2$ that is perpendicular to $v_1$.
Is it an eigenvector?
If so, which eigenvalue corresponds to it?
If $v_2$ is a vector that is perpendicular to $v_1$ then it holds that $v_1^Tv_2=0$, right?

So we get \begin{align*}A=I-2v_1 v_1^T &\Rightarrow A\cdot v_2=I\cdot v_2-2v_1 v_1^T\cdot v_2 \\ & \Rightarrow A\cdot v_2= v_2-2v_1 \left (v_1^T v_2\right )\\ & \Rightarrow A\cdot v_2= v_2-2v_1 \cdot 0 \\ & \Rightarrow A\cdot v_2= v_2 \\ & \Rightarrow A\cdot v_2-v_2 =0 \\ & \Rightarrow (A-I)\cdot v_2 =0\end{align*}
Which means that $1$ is the respective eigenvalue, right?

So to show the desired result do we have to show that the eigenvalues of $A$ are $\pm 1$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
If $v_2$ is a vector that is perpendicular to $v_1$ then it holds that $v_1^Tv_2=0$, right?
So we get \begin{align*}A=I-2v_1 v_1^T &\Rightarrow A\cdot v_2=I\cdot v_2-2v_1 v_1^T\cdot v_2 \\ & \Rightarrow A\cdot v_2= v_2-2v_1 \left (v_1^T v_2\right )\\ & \Rightarrow A\cdot v_2= v_2-2v_1 \cdot 0 \\ & \Rightarrow A\cdot v_2= v_2 \\ & \Rightarrow A\cdot v_2-v_2 =0 \\ & \Rightarrow (A-I)\cdot v_2 =0\end{align*}
Which means that $1$ is the respective eigenvalue, right?

So to show the desired result do we have to show that the eigenvalues of $A$ are $\pm 1$ ?
Yep.

Or rather, more specifically, that we have the eigenvalue $+1$ with multiplicity $2$, and the eigenvalue $-1$ with multiplicity $1$.
After all, there are 2 independent vectors perpendicular to $v_1$ aren't there?

#### mathmari

##### Well-known member
MHB Site Helper
Or rather, more specifically, that we have the eigenvalue $+1$ with multiplicity $2$, and the eigenvalue $-1$ with multiplicity $1$.
After all, there are 2 independent vectors perpendicular to $v_1$ aren't there?
How do we see that?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
How do we see that?
Once we have $v_1$, which has eigenvalue $-1$, we can find 2 independent vectors $v_2$ and $v_3$ that are both perpendicular to $v_1$ can't we?

Both of them will then be eigenvectors with eigenvalue $+1$, won't they?

#### mathmari

##### Well-known member
MHB Site Helper
Once we have $v_1$, which has eigenvalue $-1$, we can find 2 independent vectors $v_2$ and $v_3$ that are both perpendicular to $v_1$ can't we?

Both of them will then be eigenvectors with eigenvalue $+1$, won't they?
How do we know that the eigenvalue $+1$ has multiplicity $2$ and the eigenvalue $-1$ has multiplicity $1$ and that it is not the other way around?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
How do we know that the eigenvalue $+1$ has multiplicity $2$ and the eigenvalue $-1$ has multiplicity $1$ and that it is not the other way around?
Because given $v_1$ we can find 2 independent vectors that are perpendicular to it.
Each of them will transform to itself won't they?