[SOLVED]Jordan Normal Form and Root Spaces

Sudharaka

Well-known member
MHB Math Helper
Hi everyone, Here's a question that I encountered recently. I would appreciate if you could go through my solution and let me know if you see any mistakes or have any comments.

Question:

Given a linear transformation $$f:\,\mathbb{C}\rightarrow \mathbb{C}$$ with matrix $$\begin{pmatrix}3&0&8\\3&-1&6\\-2&0&5\end{pmatrix}$$. Find the Jordan normal form of $$f$$. Indicate $$X_{f}(t),\,\mu_{f}(t)$$. What are the root spaces and the eigenspaces for $$f$$?

I found the characteristic polynomial to be, $$X_{f}(t)=-(1+\lambda)(\lambda-(4+i\sqrt{15}))(\lambda-(4-i\sqrt{15}))$$. Therefore the minimal polynomial is,

$$\mu_{f}(t)=\mu_{f}(t)=(1+\lambda)(\lambda-(4+i\sqrt{15}))(\lambda-(4-i\sqrt{15}))$$

Since this matrix has three distinct eigenvalues it is diagonalizable. Therefore the Jordan normal form is,

$J=\begin{pmatrix}-1&0&0\\0&4+i\sqrt{15}&0\\0&0&4-i\sqrt{15}\end{pmatrix}$

There is only one eigenvector corresponding to $$\lambda = -1$$ and one eigenvector for both of the complex eigenvalues. The eigenspace for $$\lambda =-1$$ therefore consist of the single vector, $$\begin{pmatrix}0\\1\\0\end{pmatrix}$$ and the eigenspaces for both the complex eigenvalues consist of $$\begin{pmatrix}1\\1\\1\end{pmatrix}$$.

Now the geometric multiplicity of each of the eigenvalues is one. So the rootspace corresponding to each eigenvalue consist of all the eigenvectors for that particular eigenvalue plus the zero vector. That is the rootspace of $$\lambda =-1$$ is $$\left\{ \begin{pmatrix}0\\1\\0\end{pmatrix},\, \begin{pmatrix}0\\0\\0\end{pmatrix} \right\}$$, the rootspaces corresponding to both of the complex eigenvalues is $$\left\{ \begin{pmatrix}1\\1\\1\end{pmatrix},\, \begin{pmatrix}0\\0\\0\end{pmatrix} \right\}$$.

Klaas van Aarsen

MHB Seeker
Staff member
Hi everyone, Here's a question that I encountered recently. I would appreciate if you could go through my solution and let me know if you see any mistakes or have any comments.

Question:

Given a linear transformation $$f:\,\mathbb{C}\rightarrow \mathbb{C}$$ with matrix $$\begin{pmatrix}3&0&8\\3&-1&6\\-2&0&5\end{pmatrix}$$. Find the Jordan normal form of $$f$$. Indicate $$X_{f}(t),\,\mu_{f}(t)$$. What are the root spaces and the eigenspaces for $$f$$?

I found the characteristic polynomial to be, $$X_{f}(t)=-(1+\lambda)(\lambda-(4+i\sqrt{15}))(\lambda-(4-i\sqrt{15}))$$. Therefore the minimal polynomial is,

$$\mu_{f}(t)=\mu_{f}(t)=(1+\lambda)(\lambda-(4+i\sqrt{15}))(\lambda-(4-i\sqrt{15}))$$

Since this matrix has three distinct eigenvalues it is diagonalizable. Therefore the Jordan normal form is,

$J=\begin{pmatrix}-1&0&0\\0&4+i\sqrt{15}&0\\0&0&4-i\sqrt{15}\end{pmatrix}$

There is only one eigenvector corresponding to $$\lambda = -1$$ and no eigenvectors corresponding to the complex eigenvalues. The eigenspace for $$\lambda =-1$$ therefore consist of the single vector, $$\begin{pmatrix}0\\1\\0\end{pmatrix}$$ and the eigenspaces of the complex eigenvalues are null (no elements).

Now the geometric multiplicity of each of the eigenvalues is one. So the rootspace of corresponding to each eigenvalue consist of all the eigenvectors for that particular eigenvalue plus the zero vector. That is the rootspace of $$\lambda =-1$$ is $$\left\{ \begin{pmatrix}0\\1\\0\end{pmatrix},\, \begin{pmatrix}0\\0\\0\end{pmatrix} \right\}$$,the rootspaces corresponding to the complex eigenvalues is trivial, containing only the zero vector.
Hi Sudharaka!

Characteristic polynomial and minimal polynomial, check. Yep, diagonalizable due to distinct eigenvalues.

However, a null vector is never an eigenvector. It is excluded from the definition.
So $\lambda = -1$ has just the one eigenvector $\begin{pmatrix}0\\1\\0\end{pmatrix}$.

Now, if there is an eigenvalue, there is also always at least 1 eigenvector for that eigenvalue. This is no exception. So those complex eigenvalues must also have an eigenvector... Sudharaka

Well-known member
MHB Math Helper
Hi Sudharaka!

Characteristic polynomial and minimal polynomial, check. Yep, diagonalizable due to distinct eigenvalues.

However, a null vector is never an eigenvector. It is excluded from the definition.
So $\lambda = -1$ has just the one eigenvector $\begin{pmatrix}0\\1\\0\end{pmatrix}$.

Now, if there is an eigenvalue, there is also always at least 1 eigenvector for that eigenvalue. This is no exception. So those complex eigenvalues must also have an eigenvector... Oh, sorry. A little error in calculation. I have changed my original post. Also the root space in the original post refers to >>this<<.

Klaas van Aarsen

MHB Seeker
Staff member
Oh, sorry. A little error in calculation. I have changed my original post. Also the root space in the original post refers to >>this<<.
I'm afraid that link says that it is unavailable for viewing.

Seems to me that a root space can hardly exist of exactly 2 vectors of which one is the null vector. You can say that the root space is spanned by the eigenvector, or symbolically: [ (0,1,0) ]. This space does indeed contain the null vector.

Furthermore, the same eigenvector never appears twice. Not for the same eigenvalue and not for distinct eigenvalues.
Suppose it did, what would happen?
Is (1,1,1) really an eigenvector?

Deveno

Well-known member
MHB Math Scholar
I agree with your calculation for $\chi_f(t)$, but you have made a small mistake: $\mu_f(t)$ must be MONIC, and so:

$\mu_f(t) = -\chi_f(t) = t^3 - 7t^2 + 23t + 31 = (t + 1)(t^2 - 8t + 31)$

You have also written down $\chi_f(\lambda)$ not $\chi_f(t)$.

(The sign doesn't really matter for determining the roots, this is a minor quibble).

When I attempt to solve $(A - \lambda I)(x,y,z)^T = 0$, I get for the eigenvalue:

$\lambda = 4 + i\sqrt{15}$

The eigenvector: $\left(\dfrac{1-i\sqrt{15}}{2},\dfrac{3 - 3i\sqrt{15}}{8},1\right)$

after some laborious(!) calculations.

You should obtain for the eigenvalue:

$\lambda = 4 - i\sqrt{15}$, the eigenvector:

$\left(\dfrac{1 + \sqrt{15}}{2},\dfrac{3 + 3i\sqrt{15}}{8},1\right)$.

Each of these eigenvectors generates a root space, which all have dimension one, since the algebraic multiplicity is one.

Sudharaka

Well-known member
MHB Math Helper
I agree with your calculation for $\chi_f(t)$, but you have made a small mistake: $\mu_f(t)$ must be MONIC, and so:

$\mu_f(t) = -\chi_f(t) = t^3 - 7t^2 + 23t + 31 = (t + 1)(t^2 - 8t + 31)$

You have also written down $\chi_f(\lambda)$ not $\chi_f(t)$.

(The sign doesn't really matter for determining the roots, this is a minor quibble).

When I attempt to solve $(A - \lambda I)(x,y,z)^T = 0$, I get for the eigenvalue:

$\lambda = 4 + i\sqrt{15}$

The eigenvector: $\left(\dfrac{1-i\sqrt{15}}{2},\dfrac{3 - 3i\sqrt{15}}{8},1\right)$

after some laborious(!) calculations.

You should obtain for the eigenvalue:

$\lambda = 4 - i\sqrt{15}$, the eigenvector:

$\left(\dfrac{1 + \sqrt{15}}{2},\dfrac{3 + 3i\sqrt{15}}{8},1\right)$.

Each of these eigenvectors generates a root space, which all have dimension one, since the algebraic multiplicity is one.
Yes, you are correct, I did the calculations slowly and carefully again to obtain your answers. So the root space in this case is the same as the eigenspace for each eigenvalue, Isn't?

I'm afraid that link says that it is unavailable for viewing.
That's quite strange, the link works fine for me. Seems to me that a root space can hardly exist of exactly 2 vectors of which one is the null vector. You can say that the root space is spanned by the eigenvector, or symbolically: [ (0,1,0) ]. This space does indeed contain the null vector.

Furthermore, the same eigenvector never appears twice. Not for the same eigenvalue and not for distinct eigenvalues.
Suppose it did, what would happen?
Is (1,1,1) really an eigenvector?
You are correct. I have done some drastically silly mistakes. And yep, same eigenvector cannot appear for different eigenvalues. If $$Av=\lambda_1 v$$ and $$Av=\lambda_2 v$$ then it's clear that, $$\lambda_1=\lambda_2$$. I did the problem carefully and obtained the same answer that Deveno had obtained above.

Thank you very much for helping out. I really appreciate your help. MarkFL

Staff member
I'm afraid that link says that it is unavailable for viewing...
I get the same message. Sudharaka

Well-known member
MHB Math Helper
I get the same message. Okay, if I just copy past the url let's see what happens. Matrix And Linear Algebra 2Nd Ed. - Datta - Google Books

Do you still have problems viewing it? The weird thing is I can view it. Klaas van Aarsen

MHB Seeker
Staff member
Okay, if I just copy past the url let's see what happens. Matrix And Linear Algebra 2Nd Ed. - Datta - Google Books

Do you still have problems viewing it? The weird thing is I can view it. Now, I can view it.
However, this is Google Books. They only allow anyone to view a limited number of sections at a time, unless of course you pay for it.

Sudharaka

Well-known member
MHB Math Helper
Now, I can view it.
However, this is Google Books. They only allow anyone to view a limited number of sections at a time, unless of course you pay for it.
Yep, that's true. Anyway I wanted to quote the definition of root spaces and the page I linked has that definition. Glad that you can view it now. Klaas van Aarsen

MHB Seeker
Staff member
Yep, that's true. Anyway I wanted to quote the definition of root spaces and the page I linked has that definition. Glad that you can view it now.
For later reference, I found another link to the definition of a root space >>here<<. MarkFL

Okay, if I just copy past the url let's see what happens. Do you still have problems viewing it? The weird thing is I can view it. Yes, I can now view it too. 