(a) If $λ \neq 0$, prove the Jordan canonical form of $J^2$ is the Jordan block of size $m$ with eigenvalue $λ^2$.
(b) If $λ = 0$, prove the Jordan canonical form of $J^2$ is two Jordan blocks of sizes $\dfrac{m}{2}$, $\dfrac{m}{2}$ if $m$ is even and of sizes $\dfrac{m-1}{2}$, $\dfrac{m+1}{2}$ if $m$ is odd.
38. Determine the necessary and sufficient conditions for a matrix over $\mathbb{C}$ to have a square root.
39. Let $J$ be a Jordan block of size $m$ with eigenvalue $λ$ over a field $F$ with characteristic 2. Determine the Jordan canonical for for the matrix $J^2$. Determine the necessary conditions for a matrix over $F$ to have a square root.
Proof: We consider a more general case - determining when a matrix over $\mathbb{F}$ has an $n^{th}$ root for any natural $n$, first when $n \neq 0$ in $F$, and when $n = 0$ in $F$ and $n$ is prime in $\mathbb{Z}$.
(37)(a) Let $T$ be the linear transformation of $J$ with regard to some basis $e_1,...,e_m$. We shall show$$J^n=\begin{bmatrix}λ^n{n \choose 0} & λ^{n-1}{n \choose 1} & \cdots & {n \choose n}λ^0 & 0 & \cdots \\ 0 & λ^n{n \choose 0} & λ^{n-1}{n \choose 1} & \ddots & 0 & \cdots \\ 0 & 0 & λ^n{n \choose 0} & \ddots & 0 & \cdots \\ \ddots & & & \cdots & \cdots & \cdots \\ & \ddots & & & \cdots & \cdots \end{bmatrix}$$In the language of linear transformations, this is equivalent to showing $T^n$ is the linear transformation acting on the basis by$$T^n(e_k)=\sum_{j=0}^{k-1}λ^{n-j}{n \choose j}e_{k-j}$$This is evident for $n=1$, so we prove the proposition by induction:$$T^n(e_k)=T \circ T^{n-1}(e_k) = T(\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}e_{k-j}) =$$$$\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}T(e_{k-j})$$We note at this point that $T$ acts on the basis by $T(e_k)=λe_k+e_{k-1}$ if $k > 1$ and $T(e_1)=λe_1$, so we may continue$$\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}T(e_{k-j})=$$$$[\sum_{j=0}^{k-2}{n-1 \choose j}(λ^{n-j}e_{k-j}+λ^{n-j-1}e_{k-j-1})]+{n-1 \choose k-1}λ^{(n-1)-(k-1)+1}e_1=$$$$λ^ne_k+[\sum_{j=1}^{k-2}({n-1 \choose j} λ^{n-j}+{n-1 \choose j-1}λ^{n-1-(j-1)})e_{k-j}]+$$$$({n-1 \choose k-1}λ^{n-k+1}+{n-1 \choose k-2}λ^{n-1-(k-2)})e_1=$$$$λ^ne_k+[\sum_{j=1}^{k-2}λ^{n-j}{n \choose j}e_{k-j}]+λ^{n-(k-1)}e_1=\sum_{j=0}^{k-1}λ^{n-j}{n \choose j}e_{k-j}$$And so our claim is proven. Now, since $J^n-λ^n$ is an upper triangular matrix, it is nilpotent and thus $m_J(x) | (x-λ^n)^m$. As well, note the nullity of $T^n-λ^n$ on $V$; we have $(T^n-λ^n)V$ is generated by $λ^{n-j}{n \choose j}e_{m-j}$ for $1≤j≤m-1$, which arises as a linear combination of the linearly independent elements $e_1,...,e_{m-1}$ when $λ \neq 0$, ${n \choose 1} = n \neq 0$ and so $(T^n-λ^n)V$ is of rank $m-1$ and $T^n-λ^n$ has nullity of $1$. Thus there is one Jordan block in the Jordan canonical form of $J$ with eigenvalue $λ^n$ and by the minimal monomial this must be the only block, and hence the Jordan canonical form of $J^n$ for $J$ a Jordan block of size $m$ with eigenvalue $λ \neq 0$ is a Jordan block of size $m$ with eigenvalue $λ^n$.
(b) When $λ=0$, by the above we have $J^n$ is the matrix where the $i,j$ entry is $1$ if $j=i+n$ and all other entries are $0$. Thus $T^n$ acts on the basis by $T(e_k)=0$ if $k ≤ n$ and $T(e_k)=e_{k-n}$ when $k > n$. Write $m = an+b$ where $0 ≤ b < n$. Assume $m < n$ so $a = 0$: Then $J^n=0$ and there are $m=b$ blocks of size $1=a+1$ so the form is of the same pattern as follows when $a ≥ 1$: We may see $r_k = m-nk$ when $k ≤ a$ and $r_k = 0$ when $k > a$, where $r_k = \text{dim}(T^n)^kV$. We count the Jordan blocks by 12.3.30:$$r_{a-1}-2r_a+r_{a+1}=m-n(a-1)-2(m-na)=$$$$m-na-2(m-na)+n=n-b$$$$r_a-2r_{a+1}+r_{a+2}=m-na=b$$Thus there are $n-b$ blocks of size $a$ and $b$ blocks of size $a+1$ (all with eigenvalue $0$, remember). We count their combined dimension over $F$ by $(n-b)a+b(a+1)=na+b=m$ so this describes the Jordan form entirely.
38. We prove a matrix $A$ over $F$ wherein $n \neq 0$ and which contains all the eigenvalues of its matrices (e.g. $\mathbb{C}$) has an $n^{th}$ root if and only if:
1) Each of its nonzero eigenvalues has an $n^{th}$ root in $F$, and
2) It is possible to divide its list of exponents of its invariant factors $x^k$ into regions with generation rules:
i) Size $m$ containing only $1$s for $m < n$, and
ii) Size $n$ containing elements which all differ from each other by $1$ or $0$.
Lemma 1: Let $G$ be a ring. An element $g∈G$ has a multplicative $n^{th}$ root in $G$ if and only if each of the elements in its similarity class has an $n^{th}$ root. Proof: ($⇐$) is clear, and ($⇒$) follows by $f^n = g ⇒ (hfh^{-1})^n = hf^nh^{-1}=hgh^{-1}$.$~\square$
Lemma 2: Let $A$ be the block matrix with blocks $A_1,...,A_n$ and let $B$ be the block matrix with blocks $B_1,...,B_n$ where $A_i$ is the same size as $B_i$. Then $AB$ is the block matrix with blocks $A_1B_1,...,A_nB_n$. Proof: Expanding $AB=(A_1'+...+A_n')(B_1'+...+B_n')$ where $A_i'$ is the block matrix in position of a matrix of size $A$, and similar for $B_i'$, by viewing these individual blocks as collapses onto disjoint subspaces followed by a particular linear transformation, we see $A_i'B_j'=0$ for $i \neq j$ and $A_i'B_i'=(A_iB_i)'$. Therefore $(A_1'+...+A_n')(B_1'+...+B_n')=(A_1B_1)'+...+(A_nB_n)'$ is the block matrix described above.$~\square$
By the above lemmas and our knowledge of the Jordan canonical forms of powers of Jordan blocks, the logical equivalence now holds. $A$ has an $n^{th}$ root if and only if there is a matrix for which $B^n$ is of the same Jordan canonical form as $A$ if and only if there is such $B$ that is in Jordan canonical form itself (since $(P^{-1}BP)^n=P^{-1}B^nP$ and $B^n$ are in the same similarity class), and we see $B^n$ has blocks each of which can be separately conjugated into Jordan canonical form giving rise to a block matrix $P$ of conjugating blocks conjugating $B^n$ into its Jordan canonical form as the block sum of the Jordan canonical forms of its component blocks. By the nature of how these canonical forms arise in these blocks by (37), this completes the proof.
39. The second requirement is the same as above but applies to each class of eigenvalues as well as $0$, as when $A_λ$ is the matrix in Jordan canonical form with eigenvalue $λ$ of some fixed size, then $A_λ^n-λ^n=A_0^n$, since when $n$ is prime $n | {n \choose k}$ for all $0 < k < n$.$~\square$
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