$A=\begin{pmatrix} -3&-1&1\\ -1&-3&1\\ -2&-2&0 \end{pmatrix}.$


$(i)$ Determine the characteristic equation of A, hence find the eigenvalues of A.

$(ii)$ Determine the minimal polynomial of A.

$(iii)$ Write down the Jordan normal form J for A.

$(iv)$ Find a Jordan basis for A.

My attempt for $(i)$ is to find $ \text{det}(A−\lambda I)=0$. I ended up with $(\lambda + 2)^3$ so the eigenvalues are all equal $-2$.

  • $\begingroup$ Do you know how to find the determinant of a 3x3 matrix? $\endgroup$ – EpicMochi Aug 15 '14 at 4:12
  • $\begingroup$ @Petaro I always mess up with the calculation and get something really weird, I just did another calculation, Do i get correct answer this time? $\endgroup$ – user164945 Aug 15 '14 at 4:16
  • $\begingroup$ Yep, there you go. $\endgroup$ – EpicMochi Aug 15 '14 at 4:39

The characteristic polynomial of $A$ is $$\chi_A(t) = \text{det}(A-t\Bbb{1})$$

where $\Bbb{1} = \begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$.

After a few calculation you should get $\chi_A(t)=(t+2)^3$. So you have one eigenvalue with multiplicity $3$.

Note that the minimal polynomial of $A$ is $m_A(t)=(t+2)^2$ since $$(A+2\Bbb{1})^2=\begin{pmatrix}-1 & -1 &1\\-1&-1&1\\-2&-2&2\end{pmatrix}\begin{pmatrix}-1 & -1 &1\\-1&-1&1\\-2&-2&2\end{pmatrix}=\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}$$

This tells you that the Jordan Normal form of $A$ will be like this

$$J=\begin{pmatrix}-2&1&0\\0&-2&0\\0&0&-2\end{pmatrix}$$ Since the number $2$ on the minimal polynomial gives you the length of the Jordan Blocks.

Now comes the difficult part. That is the part where you need to evaluate the Jordan basis, that is a basis that express your function in the form of $J$. Normally you need to distinguish between a Nillpotent endomorphismus and a not nillpotent endomorphismus. In your case if you define $$\Phi:= A + 2\Bbb{1}$$

You have a nillpotent endomorphism, since you know that $\Phi^2 =0$ by the minimal polynomial. Wi will find a Basis $T$ with respect to which $$T\Phi T^{-1}=\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}$$

And this basis will also be the same basis that will make your $A$ look like $J$, indeed \begin{align*} T\Phi T^{-1} &= \begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix} \\ TAT^{-1}&= \begin{pmatrix}-2&0&0\\0&-2&0\\0&0&-2\end{pmatrix}+\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix} \end{align*}

I'll write you a summary of the theory that explain how to evaluate the Jordan Normal Form of a nillpotent endomorphism. You can find more detail in this book.

Let $U_l:= \text{Ker} (\Phi^l)$ Then we have

\begin{align*} U_2 &= \text{Ker}(\Phi^2) = \text{Ker}\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix} = \Bbb{R}^3 \\ U_1 &= \text{Ker}(\Phi) =\text{Ker}\begin{pmatrix}-1 & -1 &1\\-1&-1&1\\-2&-2&2\end{pmatrix} = \text{span} ( \begin{pmatrix}1 \\0\\1\end{pmatrix}, \begin{pmatrix}1\\-1\\0\end{pmatrix})\\ U_0 &= \text{Ker}(\Phi^0)= \{0\} \end{align*}

Note that we can complete each time $\Bbb{R}^3$ by completing the base. For instance

\begin{align*} \Bbb{R}^3 = U_2 &= U_1 \oplus W_2\\ \Bbb{R}^3 = U_2 &= U_0 \oplus W_1 \oplus W_2 = W_1 \oplus W_2 \end{align*} Where the subspaces $W_i$ are spanned by those vectors who complete the basis and such that $U_i \supset W_i$ for all $i$. The only thing that remains to do is to find the basis of the subspaces $W_i$ we begin with $W_2$.

You know that $\text{dim}(\Bbb{R}^3)=3$ and that $\text{dim}(U_1)=2$ since $U_1$ is spanned by two linear independent vector that we already evaluated above. This implies that $\text{dim}(W_2) = 1$ and hence that there is a vector in $W_2$ that completes the basis and that spans $\Bbb{R}^3$. Note that this vector must satisfy the property $U_2 \supset W_2$, but since $U_2 = \Bbb{R}^3$ all we need to do is find a vector $w_1^{(2)}$ that is linear independent from $( \begin{pmatrix}1 \\0\\1\end{pmatrix}, \begin{pmatrix}1\\-1\\0\end{pmatrix})$.

I've chosen the vector $w_1^{(2)} = \begin{pmatrix} 1\\0\\0\end{pmatrix}$ since this is linear independent from the other two. You can check this for instance by evaluating the determinant of this matrix

$$\text{det}\begin{pmatrix}1 & 1 & 1\\ 0 & -1 & 0 \\1 & 0 & 0\end{pmatrix} \neq 0$$

That is the matrix that contains the vectors that must be linear independent. Now you need to find a basis of $W_1$. Considering the fact that $W_2$ has dimension $1$ you know that $W_1$ must have dimension $2$ and hence that it must be spanned by two linear independent vectors. You can find one vector by evaluating $\Phi w_1^{(2)}$ and the other must be found in order to satisfy the property $U_1 \supset W_1$. Summarizing we have

Basis of $W_1 = (\Phi w_1^{(2)}, w_1^{(1)})$ where $w_1^{(1)}$ is the vector that mus satisfy the property given above. I've chosen $w_1^{(1)}= \begin{pmatrix} 1&-1&0\end{pmatrix}^T$ since this vector is linear independent from $\Phi w_1^{(2)}$ and $w_1{(2)}$, indeed we have

$$\text{det}(\Phi w_1^{(2)}, w_1^{(2)}, w_1^{(1)}) = \text{det}\underbrace{\begin{pmatrix}-1 & 1 & 1 \\ -1 & 0 & -1 \\ -2 & 0 & 0 \end{pmatrix}}_{:=T^{-1}}\neq 0$$

Finally we have found our basis of $W_1$ and $W_2$ and our basis $T^{-1}$ is exactly $(\Phi w_1^{(2)}, w_1^{(2)}, w_1^{(1)})$ Indeed you can check that

$$TAT^{-1} = \begin{pmatrix}-2&1&0\\0&-2&0\\0&0&-2\end{pmatrix}=J$$

I know it may sound a little confusing, but i always used this method and it has always worked. If you want to practice more solve the first exercises of this exercises sheet. You can find the solution here

  • $\begingroup$ Thank you :) To be honest, What really confused me is that you were using det(A−t1) instead of det(A−λI) and It took me some time to see how you did there in the beginning :P The rest are fine. Thank you again. $\endgroup$ – user164945 Aug 15 '14 at 7:34
  • $\begingroup$ @user164945 You are welcome, sorry about that, i really didn't thought about it :D $\endgroup$ – Bman72 Aug 15 '14 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.