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

Question:

$(i)$ 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 $-2$.

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  • $\begingroup$ 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
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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

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  • $\begingroup$ 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

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