MHB Is the Matrix A Diagonalizable with Real Eigenvalues?

[mathmari]

Hey! :giggle:

We consider the $4\times 4$ matrix $$A=\begin{pmatrix}0 & 1 & 1 & 0\\ a & 0 & 0 & 1\\ 0 & 0 & b & 0 \\ 0 & 0 & 0 & c\end{pmatrix}$$
(a) For $a=1, \ b=2, \ c=3$ check if $A$ is diagonalizable and find a basis of $\mathbb{R}^4$ where the elements are eigenvectors of $A$.
(b) Show that if $a>0$ and $b^2\neq a\neq c^2$ then $A$ is diagonalizable.
(c) Show that if $a\leq 0$ then $A$ is not diagonalizable. For (a) I have done the following :

The eigenvalues of $A$ are $\lambda_1=2, \ \lambda_2=3, \ \lambda_3=-1, \ \lambda_4=1$, each has th algebraic multiplicity $1$.
The characteristic polynomial $(2-\lambda)\cdot (3-\lambda)\cdot (\lambda-1)\cdot (\lambda+1)$ can be written as a product of linear factors.

Now we calculate the eigenvectors.

- $\lambda_1=2$ :
\begin{equation*}\begin{pmatrix}-2& 1 & 1 & 0 \\ 1 & -2 & 0 & 1\\ 0 & 0 & 2-2& 0 \\ 0 & 0 & 0 & 3-2\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}-2& 1 & 1 & 0 \\ 1 & -2 & 0 & 1\\ 0 & 0 & 0& 0 \\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix}\end{equation*}
Using Gauss-algorithm we get:
\begin{equation*}\begin{pmatrix}-2& 1 & 1 & 0 \\ 1 & -2 & 0 & 1\\ 0 & 0 & 0& 0 \\ 0 & 0 & 0 & 1\end{pmatrix} \longrightarrow \begin{pmatrix} 1 & -2 & 0 & 1 \\ -2& 1 & 1 & 0\\ 0 & 0 & 0& 0 \\ 0 & 0 & 0 & 1\end{pmatrix}\longrightarrow \begin{pmatrix} 1 & -2 & 0 & 1 \\ 0& -3 & 1 & 2\\ 0 & 0 & 0& 0 \\ 0 & 0 & 0 & 1\end{pmatrix}\longrightarrow \begin{pmatrix} 1 & -2 & 0 & 1 \\ 0& -3 & 1 & 2 \\ 0 & 0 & 0& 1 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{equation*}
So we get \begin{align*}x-2y+w=0& \\ -3y+z+2w=0& \\ w=0& \end{align*}
Therefore, $w=0$, $z=3y$, $x=2y$.
The eigenspace is then \begin{equation*}\left \{\begin{pmatrix}2y\\ y\\ 3y\\ 0\end{pmatrix}: y\in \mathbb{R}\right \}=\left \{y\cdot \begin{pmatrix}2\\ 1\\ 3\\ 0\end{pmatrix}: y\in \mathbb{R}\right \}\end{equation*}
The geometric multiplicity, i.e. the dimension of the eigenspace, is $1$. So Algebraic multiplicity = Geometric Multiplicity. - $\lambda_2=3$ :
\begin{equation*}\begin{pmatrix}-3& 1 & 1 & 0 \\ 1 & -3 & 0 & 1\\ 0 & 0 & 2-3& 0 \\ 0 & 0 & 0 & 3-3\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}-3& 1 & 1 & 0 \\ 1 & -3 & 0 & 1\\ 0 & 0 & -1& 0 \\ 0 & 0 & 0 & 0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix}\end{equation*}
Using Gauss-algorithm we get:
\begin{equation*}\begin{pmatrix}-3& 1 & 1 & 0 \\ 1 & -3 & 0 & 1\\ 0 & 0 & -1& 0 \\ 0 & 0 & 0 & 0\end{pmatrix} \longrightarrow \begin{pmatrix} 1 & -3 & 0 & 1\\-3& 1 & 1 & 0 \\ 0 & 0 & -1& 0 \\ 0 & 0 & 0 & 0\end{pmatrix}\longrightarrow \begin{pmatrix} 1 & -3 & 0 & 1\\0& -8 & 1 & 3 \\ 0 & 0 & -1& 0 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{equation*}
So we get \begin{align*}x-3y+w=0& \\ -8y+z+3w=0& \\ -z=0& \end{align*}
Therefore, $z=0$, $y=\frac{3}{8}w$, $x=\frac{1}{8}w$.
The eigenspace is then \begin{equation*}\left \{\begin{pmatrix}\frac{1}{8}w\\ \frac{3}{8}w\\ 0\\ w\end{pmatrix}: w\in \mathbb{R}\right \}=\left \{w\cdot \begin{pmatrix}\frac{1}{8}\\ \frac{3}{8}\\ 0\\ 1\end{pmatrix}: w\in \mathbb{R}\right \}\end{equation*}
The geometric multiplicity, i.e. the dimension of the eigenspace, is $1$. So Algebraic multiplicity = Geometric Multiplicity.- $\lambda_3=1$ :
\begin{equation*}\begin{pmatrix}-1& 1 & 1 & 0 \\ 1 & -1 & 0 & 1\\ 0 & 0 & 2-1& 0 \\ 0 & 0 & 0 & 3-1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}-1& 1 & 1 & 0 \\ 1 & -1 & 0 & 1\\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 2\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix}\end{equation*}
Using Gauss-algorithm we get:
\begin{equation*}\begin{pmatrix}-1& 1 & 1 & 0 \\ 1 & -1 & 0 & 1\\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 2\end{pmatrix} \longrightarrow \begin{pmatrix} -1& 1 & 1 & 0 \\ 0 & 0 & 1 & 1\\ 0 & 0 & 1& 0 \\ 0 & 0 & 0 & 2\end{pmatrix}\longrightarrow \begin{pmatrix} -1& 1 & 1 & 0 \\ 0 & 0 & 1 & 1\\ 0 & 0 & 0& -1 \\ 0 & 0 & 0 & 2\end{pmatrix}\longrightarrow \begin{pmatrix} -1& 1 & 1 & 0 \\ 0 & 0 & 1 & 1\\ 0 & 0 & 0& -1 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{equation*}
So we get \begin{align*}-x+y+z=0& \\ z+w=0& \\ -w=0& \end{align*}
Therefore, $w=0$, $z=0$, $x=y$.
The eigenspace is then \begin{equation*}\left \{\begin{pmatrix}y\\ y\\ 0\\ 0\end{pmatrix}: y\in \mathbb{R}\right \}=\left \{y\cdot \begin{pmatrix}1\\ 1\\ 0\\ 0\end{pmatrix}: y\in \mathbb{R}\right \}\end{equation*}
The geometric multiplicity, i.e. the dimension of the eigenspace, is $1$. So Algebraic multiplicity = Geometric Multiplicity.- $\lambda_4=-1$ :
\begin{equation*}\begin{pmatrix}1& 1 & 1 & 0 \\ 1 & 1 & 0 & 1\\ 0 & 0 & 2+1& 0 \\ 0 & 0 & 0 & 3+1\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix} \Rightarrow \begin{pmatrix}1& 1 & 1 & 0 \\ 1 & 1 & 0 & 1\\ 0 & 0 & 3& 0 \\ 0 & 0 & 0 & 4\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix}\end{equation*}
Using Gauss-algorithm we get:
\begin{equation*}\begin{pmatrix}1& 1 & 1 & 0 \\ 1 & 1 & 0 & 1\\ 0 & 0 & 3& 0 \\ 0 & 0 & 0 & 4\end{pmatrix} \longrightarrow \begin{pmatrix} 1& 1 & 1 & 0 \\ 0 & 0 & -1 & 1\\ 0 & 0 & 3& 0 \\ 0 & 0 & 0 & 4\end{pmatrix}\longrightarrow \begin{pmatrix} 1& 1 & 1 & 0 \\ 0 & 0 & -1 & 1\\ 0 & 0 & 0& 3 \\ 0 & 0 & 0 & 4\end{pmatrix}\longrightarrow \begin{pmatrix} 1& 1 & 1 & 0 \\ 0 & 0 & -1 & 1\\ 0 & 0 & 0& 3 \\ 0 & 0 & 0 & 0\end{pmatrix}\end{equation*}
So we get \begin{align*}x+y+z=0& \\ -z+w=0& \\ 3w=0& \end{align*}
Therefore, $w=0$, $z=0$, $x=-y$.
The eigenspace is then \begin{equation*}\left \{\begin{pmatrix}-y\\ y\\ 0\\ 0\end{pmatrix}: y\in \mathbb{R}\right \}=\left \{y\cdot \begin{pmatrix}-1\\ 1\\ 0\\ 0\end{pmatrix}: y\in \mathbb{R}\right \}\end{equation*}
The geometric multiplicity, i.e. the dimension of the eigenspace, is $1$. So Algebraic multiplicity = Geometric Multiplicity.That measn that $A$ is diagonalizable. Is everything correct so far? Could you give me a hint for questions (b) and (c) ? :unsure: Eine Basis von $\mathbb{R}^3$ ist \begin{equation*}\left \{\begin{pmatrix}2\\ 1\\ 3\\ 0\end{pmatrix},\ \begin{pmatrix}\frac{1}{8}\\ \frac{3}{8}\\ 0\\ 1\end{pmatrix},\ \begin{pmatrix}1\\ 1\\ 0\\ 0\end{pmatrix}, \ \begin{pmatrix}-1\\ 1\\ 0\\ 0\end{pmatrix}\right \}\end{equation*}

 

[I like Serena]

Hey mathmari!

Your eigenvalues and eigenvectors in (a) are correct. (Sun)

A matrix with distinct eigenvalues is diagonalizable. So we could immediateĺy see from the eigenvalues that the matrix in (a) was diagonalizable.
Did you already check which eigenvalues we get in (b)? (Wondering)

If the matrix is not diagonalizable, then there must be either a duplicate eigenvalue, or an imaginary eigenvalue.
And if there is a duplicate eigenvalue, then the corresponding independent eigenvectors must be fewer than the algebraic multiplicity.

 

[mathmari]

Your eigenvalues and eigenvectors in (a) are correct. (Sun)

A matrix with distinct eigenvalues is diagonalizable. So we could immediateĺy see from the eigenvalues that the matrix in (a) was diagonalizable.

So in (a) we have to show just that there are $4$ distinct eigenvalues and this implies that $A$ is diagonalizable. To find a basis we need also the remaining part where I calculate the eigenvalues and the set of all eigenvalues forms a basis, right? :unsure:

Did you already check which eigenvalues we get in (b)? (Wondering)

If the matrix is not diagonalizable, then there must be either a duplicate eigenvalue, or an imaginary eigenvalue.
And if there is a duplicate eigenvalue, then the corresponding independent eigenvectors must be fewer than the algebraic multiplicity.

The eigenvalues of $A$ are \begin{align*}\det (A-\lambda I)=0 &\Rightarrow \begin{vmatrix}-\lambda & 1 & 1 & 0 \\ a & -\lambda & 0 & 1\\ 0 & 0 & b-\lambda& 0 \\ 0 & 0 & 0 & c-\lambda\end{vmatrix}=0 \\ & \Rightarrow (-\lambda)\begin{vmatrix} -\lambda & 0 & 1\\ 0 & b-\lambda& 0 \\ 0 & 0 & c-\lambda\end{vmatrix}-a\begin{vmatrix} 1 & 1 & 0 \\ 0 & b-\lambda& 0 \\ 0 & 0 & c-\lambda\end{vmatrix}=0\\ & \Rightarrow (-\lambda)\cdot (-\lambda)\begin{vmatrix} b-\lambda& 0 \\ 0 & c-\lambda\end{vmatrix}-a\begin{vmatrix} b-\lambda& 0 \\ 0 & c-\lambda\end{vmatrix}=0 \\ & \Rightarrow (-\lambda)\cdot (-\lambda)\cdot (b-\lambda)\cdot (c-\lambda)-a\cdot (b-\lambda)\cdot (c-\lambda)=0 \\ & \Rightarrow (b-\lambda)\cdot (c-\lambda)\cdot (\lambda^2-a)=0 \\ & \Rightarrow \lambda_1=b, \ \lambda_2=c, \ \lambda_3=-\sqrt{a}, \ \lambda_4=\sqrt{a}\end{align*}
Because of $b^2\neq a\neq c^2$ and $a>0$ we have $4$ dinstinct real eigenvalues and so $A$ is diagonalizable. As for (c), we have imaginary eigenvalue since $a\leq 0$ which means that $A$ is not diagonalizable. Is everything correct? :unsure:

 

[I like Serena]

All correct except (c) for the case that $a=0$.
Because then the eigenvalues are real.

 

[mathmari]

All correct except (c) for the case that $a=0$.
Because then the eigenvalues are real.

Ah in that case we have a duplicate eigenvalueand so weget again that $A$ isnot diagonalizable, right? :unsre:

In general, does it hold that a $n\times n$ matrix has to have $n$ distrinct eigenvalues to be diagobalizable? :unsure:

 

[I like Serena]

Not exactly. (Shake)

Each eigenvalue has at least 1 eigenvector by definition.
That's why distinct eigenvalues must have independent eigenvectors.
However, if an eigenvalue is duplicated, then only 1 corresponding independent eigenvector is guaranteed. (Nerd)

So for (c) we still have to check how many independent eigenvectors we can find for eigenvalue $0$. (Sweating)

 

[mathmari]

Each eigenvalue has at least 1 eigenvector by definition.
That's why distinct eigenvalues must have independent eigenvectors.
However, if an eigenvalue is duplicated, then only 1 corresponding independent eigenvector is guaranteed. (Nerd)

Could you explain that further to me? I haven't really understood that. :unsure:

 

[I like Serena]

Let's start with the definition.
A scalar $\lambda$ is called an eigenvalue of a matrix $A$ if there is a nonzero vector $v$ such that $Av=\lambda v$.
The vector $v$ is then called an eigenvector that corresponds to the eigenvalue $\lambda$.

Suppose $\lambda_1$ and $\lambda_2$ are 2 eigenvalues.
Then there must be corresponding eigenvectors $v_1$ respectively $v_2$ to satisfy the definition.
If $\lambda_1\ne\lambda_2$, then it follows that $v_1$ must be independent from $v_2$, doesn't it? Can you say why?
However, if $\lambda_1=\lambda_2$, then we don't necessarily have $v_1\ne v_2$.As an example, consider $\begin{pmatrix}2&0\\0&2\end{pmatrix}$.
This matrix has the duplicate eigenvalue $2$. And the eigenvectors are $\begin{pmatrix}1\\0\end{pmatrix}$ and $\begin{pmatrix}0\\1\end{pmatrix}$.
It is obviously diagonalizable since it is already in diagonal form. Alternatively, consider $\begin{pmatrix}2&1\\0&2\end{pmatrix}$.
It has again the duplicate eigenvalue $2$, but this time there is only 1 independent eigenvector, which is $\begin{pmatrix}1\\0\end{pmatrix}$.
That is, any vector $v$ that is independent from $\begin{pmatrix}1\\0\end{pmatrix}$ will not have $Av=2v$.
Consequently it is not diagonalizable.

 

[I like Serena]

So for (c) we still have to check how many independent eigenvectors we can find for eigenvalue $0$.

Let's see...

For $a=0$ we have the eigenvalues $0,\,0,\,b,\,c$.

Solving for the eigenvalue $\lambda=0$ we get:
\begin{equation*}\begin{pmatrix}0-0& 1 & 1 & 0 \\ 0 & 0-0 & 0 & 1\\ 0 & 0 &b-0& 0 \\ 0 & 0 & 0 & c-0\end{pmatrix}\begin{pmatrix}x\\ y\\ z\\ w\end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0\\ 0\end{pmatrix} \Rightarrow
\begin{cases}z=-y\\w=0\\bz=0\\cw=0\end{cases}\Rightarrow
\begin{cases}z=-y\\w=0\\b\ne 0\land z=0\end{cases}\lor \begin{cases}z=-y\\w=0\\b=0\end{cases}
\end{equation*}
If $b\ne0$, then we the set of independent eigenvectors for $\lambda=0$ is $\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\right\}$, while the algebraic multiplicity of $\lambda=0$ is at least $2$.
And if $b=0$, then the algebraic multiplicity of $\lambda=0$ is at least $3$, while the set of corresponding independent eigenvectors is $\left\{\begin{pmatrix}1\\0\\0\\0\end{pmatrix},\,\begin{pmatrix}0\\1\\-1\\0\end{pmatrix}\right\}$.

In both cases, we have less independent eigenvectors than the algebraic multiplicity of $\lambda=0$.
Therefore if $a=0$, then the matrix is not diagonalizable. :geek:

 

[mathmari]

If the matrix is not diagonalizable, then there must be either a duplicate eigenvalue, or an imaginary eigenvalue.
And if there is a duplicate eigenvalue, then the corresponding independent eigenvectors must be fewer than the algebraic multiplicity.

If we know that the eigenvalues are real, does it hold that the matrix diagonalizable iff all eigenvalues are distinct? :unsure:

 

[I like Serena]

If we know that the eigenvalues are real, does it hold that the matrix diagonalizable iff all eigenvalues are distinct?

It's not 'iff'.
If the eigenvalues are distinct, then the matrix is diagonalizable.
The converse does not hold in general.