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Problem 1. Find the Jordan canonical form of the matrix$$A=\begin{pmatrix}3 & -1 & 1 \\ -1 & 3 & -1 \\ -2 & 2 & 0\end{pmatrix}$$and determine a change of basis matrix that takes the matrix $A$ to its Jordan canonical form.
Solution. The determinant of the matrix $(Iλ−A)$ reads\begin{aligned} A=\begin{vmatrix}\lambda-3 & 1 & -1 \\ 1 & \lambda-3 & 1 \\ 2 & -2 & \lambda\end{vmatrix} &=(\lambda-3)(\lambda(\lambda-3)+2)-(\lambda-2)-(-2-2(\lambda-3)) \\ &=(\lambda-3)(\lambda^{2}-3 \lambda+2)+\lambda-2 \\ &=(\lambda-3)(\lambda-2)(\lambda-1)+\lambda-2 \\ &=(\lambda-2)((\lambda-3)(\lambda-1)+1) \\ &=(\lambda-2)(\lambda^{2}-4 \lambda+4) \\ &=(\lambda-2)^{3} \end{aligned}Now,$$A-2 I=\begin{pmatrix}1 & -1 & 1 \\ -1 & 1 & -1 \\ -2 & 2 & -2\end{pmatrix} \quad \text{and} \quad \ker(A-2 I)=\operatorname{Span}\{\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix},\begin{pmatrix}0 \\ 1 \\ 1\end{pmatrix}\}$$Since $\dim(\ker(A − 2I))=2$, we conclude that the Jordan canonical form of $A$ has two Jordan blocks and thus is of the form $J=\begin{pmatrix}2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{pmatrix}$
As for Jordan basis, let us note that $(A − 2I)^2 = 0$, and choose$$e_{2}:=\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} \in \ker(A-2 I)^{2} \setminus \ker(A-2 I) \quad \text{and} \quad e_{1}:=(A-2 I) e_2=\begin{pmatrix}1 \\ -1 \\ -2\end{pmatrix}$$Finally, we complete $e_1$ to the basis of $\ker(A−2I)$ by picking $e_{3}:=\begin{pmatrix}1 \\ 1 \\ 0\end{pmatrix}$.
Then we have $P=\begin{pmatrix}1 & 1 & 1 \\ -1 & 0 & 1 \\ -2 & 0 & 0\end{pmatrix}$ and $A=P J P^{-1}$.
Problem 2. Let $J$ be a Jordan block of size 7 with eigenvalue 0. Find the Jordan canonical form of $J^2$.
Solution. Note that\begin{aligned} J^{2}=\begin{pmatrix}0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}, & J^{4}=\begin{pmatrix}0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix} \\ J^{6}=\begin{pmatrix}0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}, &J^8=\begin{pmatrix}0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}\end{aligned}Therefore, one has$$\dim \ker(J^{2})=2, \quad \dim \ker(J^{4})=4, \quad \dim \ker(J^{6})=6, \quad \dim \ker(J^{8})=7$$We conclude from the above that $J^2$ has 2 Jordan blocks, both of which are of size at least 3. Thus, the Jordan canonical form of $J^2$ is $\begin{pmatrix}0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0\end{pmatrix}$.
Problem 3. Let $\mathbb R[x]$ be the ring of polynomials, and $\mathbb R(x)$ — the field of rational functions in $x$. Show that the tensor product$$\mathbb{R}[x] /(x^{3}-x^{2}+x-1) \otimes_{\mathbb{R}[x]}\left(\mathbb{R}[x] /(x^{4}+3 x^{2}+2) \oplus \mathbb{R}[x] /(x^{4}+2 x^{2}+1) \oplus \mathbb{R}(x)\right)$$is a 4-dimensional vector space over $\mathbb R$.
Solution. Recall that for any $a(x),b(x)∈\mathbb R[x]$ one has$$\mathbb{R}[x] /(a(x)) \otimes_{\mathbb{R}[x]} \mathbb{R}(x)=0 \quad \text{and} \quad \mathbb{R}[x] /(a(x)) \otimes_{\mathbb{R}[x]} \mathbb{R}[x] /(b(x))=\mathbb{R}[x] /(d(x))$$where $d(x)$ is the greatest common divisor of $a(x)$ and $b(x)$. Now, expanding the brackets in the tensor product$$\mathbb{R}[x] /(x^{3}-x^{2}+x-1) \otimes_{\mathbb{R}[x]}\left(\mathbb{R}[x] /(x^{4}+3 x^{2}+2) \oplus \mathbb{R}[x] /(x^{4}+2 x^{2}+1) \oplus \mathbb{R}(x)\right)\tag{∗}$$and decomposing$$\begin{aligned} x^{3}-x^{2}+x-1 &=(x^{2}+1)(x-1) \\ x^{4}+3 x^{2}+2 &=(x^{2}+1)(x^2+2) \\ x^{4}+2 x^{2}+1 &=(x^{2}+1)^{2} \end{aligned}$$we see that the product (∗) is equal to$$\mathbb{R}[x] /(x^{2}+1) \oplus \mathbb{R}[x] /(x^{2}+1) \simeq \mathbb{C}^{2} \oplus \mathbb{C}^{2}$$which is a 4-dimensional space over $\mathbb R$.
Problem 4.- Show that the polynomial $p(x)=x^3 + x + 1$ is irreducible over $\mathbb F_2$.
- How many elements are there in the field $K=\mathbb F_2[x]/(p(x))$?
- Let $θ$ be the root of $p(x)$ in $K$, find $(1 + θ)^{−1}∈ K$.
Solution.- Polynomial $p(x) = x^3 + x + 1$ is of degree 3, and therefore is irreducible if and only if it has no roots. It remains to observe that $p(0) = p(1) = 1$.
- Field $K$ is an extension of degree 3 of the field $\mathbb F_2$. As such $K$ is a 3-dimensional vector space over $\mathbb F_2$ and has $2^3=8$ elements.
- Note that $(θ + 1)(θ^2 + θ) = θ^3 +\cancel{2θ^2}+ θ ≡ 1$ in $K$. Therefore, $\frac{1}{\theta+1}=\theta^{2}+\theta$.
Problem 5. Let $p(x) = x^3 + 5x^2 + 7x − 2$, and $F =\mathbb Q[x]/(p(x))$.- Show that $F$ is a finite field extension of $\mathbb Q$ and find the degree $[F :\mathbb Q]$.
- Choose a basis of $F$ over $\mathbb Q$, and write the matrix of the operator of multiplication by $x$ in this basis.
Solution.- First of all, let us prove that the polynomial $p(x) = x^3 + 5x^2 + 7x − 2$ is irreducible over $\mathbb Q$. Indeed, as a polynomial of degree 3, it is reducible if and only if it has a root in $\mathbb Q$, and the only possible roots are integers which divide the free term −2. Plugging in ±1 and ±2 we calculate that $p(1) = 11, p(−1) = −5, p(2) = 40, p(−2) = −4,$ we conclude that $p(x)$ is irreducible. Therefore, $[F :\mathbb Q] = \deg(p(x)) = 3$.
- One can choose a basis $1, x, x^2$ in which the matrix of the operator of multiplication by $x$ reads $\begin{pmatrix}0 & 0 & 2 \\ 1 & 0 & -7 \\ 0 & 1 & -5\end{pmatrix}$.
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