Linear Algebra: Canonical Forms

• We’ll use rank-nullity throughout: \(\operatorname{rank}M + \dim \ker M = 7\).

• Also note that \begin{align*} Av = \lambda v \implies A^nv = A^{n-1}Av = A^{n-1}\lambda v = \cdots = \lambda^n v ,\end{align*} so if \(\lambda \in \operatorname{Spec}(A)\) then \(\lambda^n\in \operatorname{Spec}(A^n)\). Conversely, \(\lambda\in \operatorname{Spec}(A^n) \implies \lambda^{1\over n}\in \operatorname{Spec}(A)\), which we’ll use several times.

• Since \(5 = \operatorname{rank}A = \operatorname{rank}(A - 0\cdot I)\), we have \(\dim \ker(A-0\cdot I) = 2\) contributing an eigenvalue of \(\lambda = 0\) with multiplicity \(2\).

• Since \(4 = \operatorname{rank}A^2 = \operatorname{rank}(A^2 - 0\cdot \lambda) = \operatorname{rank}(A-0\cdot \lambda)^2\), we have that \(\dim \ker(A-0\cdot I)^2 = 3\). Since \(\dim \ker (A-0\cdot I)^1 = 2 < 3\), this means there is 1 generalized eigenvector associated to \(\lambda = 0\).

• Since \(6 = \operatorname{rank}(A-1\cdot I)\), \(\dim \ker (A- 1\cdot I) = 1\), contributing \(\lambda = 1\) with multiplicity 1.

• Since \(\operatorname{rank}(A^3-1\cdot I) = 4\), we have \(\dim \ker (A^3-1\cdot I) = 3\), contributing \(\lambda = 1\) now to \(\operatorname{Spec}(A^3)\) instead of \(\operatorname{Spec}(A)\). Thus some unknown cube roots of 1 are contributed to \(\operatorname{Spec}(A)\), so any of \(1=\zeta_3^0, \zeta_3, \zeta_3^2\) are possibilities at this point. Call these three contributions \(z_1, z_2, z_3\), which may not be distinct.

• Now use that \({\mathrm{tr}}(A) = \sum_{i=1}^n \lambda_i\) is the sum of the diagonal on \(\operatorname{JCF}(A)\), using that trace is a similarity invariant, to write \begin{align*} 1 = {\mathrm{tr}}(A) = (0 + 0) + (0) + (1) + (z_1 + z_2 + z_3) \implies z_1 + z_2 + z_3 = 0 ,\end{align*} which is actually enough to force \(z_1 = 1, z_2 = \zeta_3, z_3 = \zeta_3^2\), since no other combination sums to zero. That \(1 + \zeta_3 + \zeta_3^2 = 0\) is a general fact.

• Since \(\lambda=1\) occurs twice as an eigenvalue but \(\dim \ker(A-1\cdot I) = 1\), the two copies of \(\lambda = 1\) must occur in a nontrivial Jordan block.

• So we get a Jordan form \begin{align*} \operatorname{JCF}(A) = \begin{bmatrix} 0 & & & & & & \\ & 0 & 1 & & & & \\ & & 0 & & & & \\ & & & 1 & 1 & & \\ & & & & 1 & & \\ & & & & & \zeta_3 & \\ & & & & & & \zeta_3^2 \\ \end{bmatrix} .\end{align*}

Proof of (a): Let \(A\) be the matrix in the question, and \(B\) be the matrix containing 1’s in every entry.

• Noting that \(B = A+I\), we have \begin{align*} &B\mathbf{x} = \lambda \mathbf{x} \\ &\iff (A+I) \mathbf{x} = \lambda \mathbf{x} \\ &\iff A \mathbf{x} = (\lambda - 1) \mathbf{x} ,\end{align*} so we will find the eigenvalues of \(B\) and subtract one from each.

• Note that \(B\mathbf{v} = {\left[ {\sum v_i, \sum v_i, \cdots, \sum v_i} \right]}\), i.e. it has the effect of summing all of the entries of \(\mathbf{v}\) and placing that sum in each component.

• We proceed by finding \(p\) eigenvectors and eigenvalues, since the JCF and minimal polynomials will involve eigenvalues and the transformation matrix will involve (generalized) eigenvectors.

• Using that the eigenvalues of \(A\) are \(1+\lambda_i\) for \(\lambda_i\) the above eigenvalues for \(B\), \begin{align*} \operatorname{Spec}(B) \coloneqq\left\{{(\lambda_i, m_i)}\right\} &= \left\{{(p, 1), (0, p-1)}\right\} \implies \chi_{B}(x) = (x-p)x^{p-1} \\ \implies \operatorname{Spec}(A) &= \left\{{(p-1,1), (-1, p-1) }\right\} \implies \chi_{A}(x) = (x- p+1)(x+1)^{p-1} \\ \end{align*}

• The dimensions of eigenspaces are preserved, thus \begin{align*} JCF_{\mathbf{Q}}(A) = J_{p-1}^{1} \oplus (p-1)J_{-1}^1 = \left[\begin{array}{r|r|r|r|r|r} p-1 & 0 & 0 & \cdots & 0 & 0 \\ \hline 0& -1 & 0 & 0 & 0 & 0 \\ \hline 0& 0 & -1 & 0 & 0 & 0 \\ \hline 0& 0 & 0 & \ddots & \ddots & 0 \\ \hline 0& 0 & 0 & \cdots & -1 & 0 \\ \hline 0& 0 & 0 & \cdots & 0 & -1 \\ \end{array}\right] .\end{align*}

• The matrix \(P\) such that \(A = PJP^{-1}\) will have columns the bases of the generalized eigenspaces.

• In this case, the generalized eigenspaces are the usual eigenspaces, so \begin{align*} P = [\mathbf{v}_1, \mathbf{p}_1, \cdots, \mathbf{p}_{p-1}] = \left[\begin{array}{rrrrrr} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & -1 & 0 & 0 & 0 & 0 \\ 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & 0 & 0 \\ 1 & \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 0 & 0 & 0 & 0 & -1 \\ \end{array}\right] .\end{align*}

Proof of (b):

For \(F = { \mathbf{F} }_p\), all eigenvalues/vectors still lie in \({ \mathbf{F} }_p\), but now \(-1 = p-1\), making \((x-(p-1))(x+1)^{p-1} = (x+1)(x+1)^{p-1}\), so \(\chi_{A, { \mathbf{F} }_p}(x) = (x+1)^p\), and the Jordan blocks may merge.

• A computation shows that \((A+I)^2 = pA = 0 \in M_p({ \mathbf{F} }_p)\) and \((A+I) \neq 0\), so \(\min_{A, { \mathbf{F} }_p}(x) = (x+1)^2\).
  • Thus the largest Jordan block corresponding to \(\lambda = -1\) is of size 2
• Can check that \(\operatorname{det}(A) = \pm 1 \in { \mathbf{F} }_p^{\times}\), so the vectors \(\mathbf{e}_1 - \mathbf{e}_i\) are still linearly independent and thus \(\dim E_{-1} = p-1\)
  • So there are \(p-1\) Jordan blocks for \(\lambda = 0\).

Summary: \begin{align*} \min_{A, { \mathbf{F} }_p}(x) &= (x+1)^2 \\ \chi_{A, { \mathbf{F} }_p}(x) &\equiv (x+1)^p \\ \dim E_{-1} &= p-1 .\end{align*}

Thus \begin{align*} JCF_{{ \mathbf{F} }_p}(A) = J_{-1}^{2} \oplus (p-2)J_{-1}^1 = \left[\begin{array}{rr|r|r|r|r} -1 & 1 & 0 & \cdots & 0 & 0 \\ 0& -1 & 0 & 0 & 0 & 0 \\ \hline 0& 0 & -1 & 0 & 0 & 0 \\ \hline 0& 0 & 0 & \ddots & \ddots & 0 \\ \hline 0& 0 & 0 & \cdots & -1 & 0 \\ \hline 0& 0 & 0 & \cdots & 0 & -1 \\ \end{array}\right] .\end{align*}

To obtain a basis for \(E_{\lambda = 0}\), first note that the matrix \(P = [\mathbf{v}_1, \mathbf{p}_1, \cdots , \mathbf{p}_{p-1}]\) from part (a) is singular over \({ \mathbf{F} }_p\), since \begin{align*} \mathbf{v}_1 + \mathbf{p}_1 + \mathbf{p}_2 + \cdots + \mathbf{p}_{p-2} &= [p-1, 0, 0, \cdots, 0, 1] \\ &= [-1, 0,0,\cdots, 0, 1] \\ &= - \mathbf{p}_{p-1} .\end{align*}

We still have a linearly independent set given by the first \(p-1\) columns of \(P\), so we can extend this to a basis by finding one linearly independent generalized eigenvector.

Solving \((A-I\lambda)\mathbf{x} = \mathbf{v}_1\) is our only option (the others won’t yield solutions). This amounts to solving \(B\mathbf{x} = \mathbf{v}_1\), which imposes the condition \(\sum x_i = 1\), so we can choose \(\mathbf{x} = [1, 0, \cdots, 0]\).

Thus \begin{align*} P = [\mathbf{v}_1, \mathbf{x}, \mathbf{p}_1, \cdots, \mathbf{p}_{p-2}] = \left[\begin{array}{rrrrrr} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & -1 & 0 & 0 & 0 \\ 1 & 0 & 0 & -1 & 0 & 0 \\ 1 & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & 0 & 0 & -1\\ 1 & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right] .\end{align*}

Note that if \(\mathbf{x} = {\left[ {x_1,\cdots, x_n} \right]}\) then \(A\mathbf{x} = {\left[ {\sum x_i, \sum x_i, \cdots, \sum x_i} \right]}\), so \(A\) acts by summing the entries in \(\mathbf{x}\) and setting every coordinate to that sum. By inspection (or clever guessing), we can find eigenvalues and eigenvectors:

• \(\lambda = 0\) and:
  • \(\mathbf{v}_1 = {\left[ {1,0,0,\cdots,0, -1} \right]}\)
  • \(\mathbf{v}_2 = {\left[ {0,1,0,\cdots,0, -1} \right]}\)
  • \(\mathbf{v}_3 = {\left[ {0,0,1,\cdots,0, -1} \right]}\)
  • \(\cdots\)
  • \(\mathbf{v}_{n-1} = {\left[ {0,0,0,\cdots,1, -1} \right]}\)
• \(\lambda = n\) and \(\mathbf{v}_n = {\left[ {1,1,\cdots, 1} \right]}\)

Note that for \(\lambda = n\), we have \(A {\left[ {1,1,\cdots, 1} \right]} = {\left[ {n\cdot 1,\cdots, n\cdot 1} \right]}\). So for \(n\cdot 1\neq 0\), there are two eigenspaces corresponding to \(\lambda = 0, n\), and if \(n\cdot 1 = 0\) these collapse to just a single eigenspace for \(\lambda = 0\).

Assuming \(n\cdot 1\neq 0\), we get a characteristic polynomial of \((x-n)x^{n-1}\). The \(x-n\) factor corresponds to a single \(1\times 1\) Jordan block with diagonal \(n\). For the \(x^{n-1}\) factor, we’ve produced \(n-1\) distinct eigenvectors, so we get \(n-1\) Jordan blocks of size \(1\times 1\) with diagonal zero. Thus \begin{align*} \operatorname{JCF}(A) = \left[ \begin{array}{c|c|c|c|c} n & \cdot & \cdot & \cdot & \cdot \\ \hline \cdot & 0 & \cdot & \cdot & \cdot \\ \hline \cdot & \cdot & 0 & \ddots & \cdot \\ \hline \cdot & \cdot & \cdot & \ddots & \cdot \\ \hline \cdot & \cdot & \cdot & \cdot & 0 \end{array} \right] .\end{align*} One can verify this by checking directly that the minimal polynomial of \(A\) is \(p(x) = (x-n)x\), so the size of the largest Jordan block for \(\lambda = n\) is 1 and for \(\lambda = 0\) is \(n-1\), while the characteristic polynomial is \((x-n)x^{n-1}\), so the sum of the sizes of Jordan blocks for \(\lambda = n\) is 1 and for \(\lambda = 0\) is \(n-1\), forcing the \(1\times 1\) blocks everywhere.

Now consider the case when \(n\cdot 1 = 0\); then \(\mathbf{v}_n\) becomes an eigenvector for \(\lambda = 0\) instead of \(\lambda = n\). The minimal polynomial becomes \((x-0)x = x^2\) and the characteristic polynomial becomes \(x^n\), so \(\lambda = 0\) has:

• The size of the largest Jordan block is 2,
• The sum of sizes of Jordan blocks is \(n-1\),

and so this forces one block of size \(2\times 2\) and \(n-2\) blocks of size \(1\times 1\). So we now have: \begin{align*} \operatorname{JCF}(A) = \left[ \begin{array}{cc|c|c|c} 0 & 1 & \cdot & \cdot & \cdot \\ \cdot & 0 & 0 & \cdot & \cdot \\ \hline \cdot & \cdot & 0 & \ddots & \cdot \\ \hline \cdot & \cdot & \cdot & \ddots & 0 \\ \hline \cdot & \cdot & \cdot & \cdot & 0 \end{array} \right] .\end{align*}