Show properties of matrix A

• MHB
• mathmari
In summary, the conversation revolved around proving that the matrix $A$ can be written as $I-2v_1v_1^T$ where $v_1$ is a unit vector. Two methods were discussed - one using a system of equations and the other using eigenvalues and eigenvectors. It was concluded that in order to show the desired result, it is necessary to prove that the eigenvalues of $A$ are $\pm 1$, with the eigenvalue $+1$ having multiplicity $2$ and the eigenvalue $-1$ having multiplicity $1$.

mathmari

Gold Member
MHB
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? (Wondering)

mathmari said:
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. (Nod)

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

Klaas van Aarsen said:
Just to mention an alternative, we might try to find the eigenvalues and eigenvectors of $A$. (Thinking)

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? (Wondering)

mathmari said:
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? (Wondering)
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? (Wondering)

Klaas van Aarsen said:
If we have indeed that $A=I-2v_1 v_1^T$, isn't $v_1$ an eigenvector then? (Wondering)
If so, which eigenvalue corresponds to it?

Since the vector $v_1$ is unit it holds that $v_1^Tv_1=1$, right? (Wondering)

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? (Wondering)

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mathmari said:
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 $-1$. (Nod)

Klaas van Aarsen said:
And consider a vector $v_2$ that is perpendicular to $v_1$.
Is it an eigenvector?
If so, which eigenvalue corresponds to it? (Wondering)

If $v_2$ is a vector that is perpendicular to $v_1$ then it holds that $v_1^Tv_2=0$, right? (Wondering)

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? (Wondering)

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

mathmari said:
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. (Nod)

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

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

How do we see that? (Wondering)

mathmari said:
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? (Thinking)

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

Klaas van Aarsen said:
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? (Thinking)

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

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? (Wondering)

mathmari said:
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? (Thinking)