# Linear Algebra: Canonical Forms

## Fall 2021 #3#algebra/qual/completed

What is the Jordan normal form over $$\mathbb{C}$$ of a $$7 \times 7$$ matrix $$A$$ which satisfies all of the following conditions:

• $$A$$ has real coefficients,

• $$\mathrm{rk} A=5$$,

• $$\mathrm{rk} A^{2}=4$$,

• $$\mathrm{rk} A-I=6$$,

• $$\mathrm{rk} A^{3}-I=4$$,

• $$\operatorname{tr} A=1 ?$$

• 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*}

## $$\star$$ Spring 2012 #8#algebra/qual/work

Let $$V$$ be a finite-dimensional vector space over a field $$k$$ and $$T:V\to V$$ a linear transformation.

• Provide a definition for the minimal polynomial in $$k[x]$$ for $$T$$.

• Define the characteristic polynomial for $$T$$.

• Prove the Cayley-Hamilton theorem: the linear transformation $$T$$ satisfies its characteristic polynomial.

## $$\star$$ Spring 2020 #8#algebra/qual/work

Let $$T:V\to V$$ be a linear transformation where $$V$$ is a finite-dimensional vector space over $${\mathbf{C}}$$. Prove the Cayley-Hamilton theorem: if $$p(x)$$ is the characteristic polynomial of $$T$$, then $$p(T) = 0$$. You may use canonical forms.

## $$\star$$ Spring 2012 #7#algebra/qual/work

Consider the following matrix as a linear transformation from $$V\coloneqq{\mathbf{C}}^5$$ to itself: \begin{align*} A=\left(\begin{array}{ccccc} -1 & 1 & 0 & 0 & 0 \\ -4 & 3 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 2 \end{array}\right) .\end{align*}

• Find the invariant factors of $$A$$.

• Express $$V$$ in terms of a direct sum of indecomposable $${\mathbf{C}}[x]{\hbox{-}}$$modules.

• Find the Jordan canonical form of $$A$$.

## Fall 2019 Final #8#algebra/qual/work

Exhibit the rational canonical form for

• $$A\in M_6({\mathbf{Q}})$$ with minimal polynomial $$(x-1)(x^2 + 1)^2$$.
• $$A\in M_{10}({\mathbf{Q}})$$ with minimal polynomial $$(x^2+1)^2(x^3 + 1)$$.

## Fall 2019 Final #9#algebra/qual/work

Exhibit the rational and Jordan canonical forms for the following matrix $$A\in M_4({\mathbf{C}})$$: \begin{align*} A=\left(\begin{array}{cccc} 2 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ -2 & -2 & 0 & 1 \\ -2 & 0 & -1 & -2 \end{array}\right) .\end{align*}

## Spring 2016 #7#algebra/qual/work

Let $$D = {\mathbf{Q}}[x]$$ and let $$M$$ be a $${\mathbf{Q}}[x]{\hbox{-}}$$module such that \begin{align*} M \cong \frac{\mathbb{Q}[x]}{(x-1)^{3}} \oplus \frac{\mathbb{Q}[x]}{\left(x^{2}+1\right)^{3}} \oplus \frac{\mathbb{Q}[x]}{(x-1)\left(x^{2}+1\right)^{5}} \oplus \frac{\mathbb{Q}[x]}{(x+2)\left(x^{2}+1\right)^{2}} .\end{align*}

Determine the elementary divisors and invariant factors of $$M$$.

## Spring 2020 #7#algebra/qual/work

Let \begin{align*} A=\left[\begin{array}{ccc} 2 & 0 & 0 \\ 4 & 6 & 1 \\ -16 & -16 & -2 \end{array}\right] \in M_{3}(\mathrm{C}) .\end{align*}

• Find the Jordan canonical form $$J$$ of $$A$$.

• Find an invertible matrix $$P$$ such that $$P^{-1}A P = J$$.

• Write down the minimal polynomial of $$A$$.

You should not need to compute $$P^{-1}$$.

## Spring 2019 #7#algebra/qual/completed

Let $$p$$ be a prime number. Let $$A$$ be a $$p \times p$$ matrix over a field $$F$$ with 1 in all entries except 0 on the main diagonal.

Determine the Jordan canonical form (JCF) of $$A$$

• When $$F = {\mathbf{Q}}$$,

• When $$F = { \mathbf{F} }_p$$.

Hint: In both cases, all eigenvalues lie in the ground field. In each case find a matrix $$P$$ such that $$P^{-1}AP$$ is in JCF.

• Work with matrix of all ones instead.
• Eyeball eigenvectors.
• Coefficients in minimal polynomial: size of largest Jordan block
• Dimension of eigenspace: number of Jordan blocks
• We can always read off the characteristic polynomial from the spectrum.

• Todo

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.

Each vector of the form $$\mathbf{p}_i \coloneqq\mathbf{e}_1 - \mathbf{e}_{i+1} = {\left[ {1, 0, 0,\cdots, 0 -1, 0, \cdots, 0 } \right]}$$ where $$i\neq j$$ is also an eigenvector with eigenvalues $$\lambda_0 = 0$$, and this gives $$p-1$$ linearly independent vectors spanning the eigenspace $$E_{\lambda_0}$$

$$\mathbf{v}_1 = {\left[ {1, 1, \cdots, 1} \right]}$$ is an eigenvector with eigenvalue $$\lambda_1 = p$$.

• 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*}

• Compute \begin{align*}B \mathbf{p}_i = {\left[ { 1 + 0 + \cdots + 0 + (-1) + 0 + \cdots + 0} \right]} = {\left[ {0, 0, \cdots, 0} \right]}\end{align*}
• So every $$\mathbf{p}_i \in \ker(B)$$, so they are eigenvectors with eigenvalue 0.
• Since the first component is fixed and we have $$p-1$$ choices for where to place a $$-1$$, this yields $$p-1$$ possibilities for $$\mathbf{p}_i$$
• These are linearly independent since the $$(p-1)\times (p-1)$$ matrix $${\left[ { \mathbf{p}_1^t, \cdots, \mathbf{p}_{p-1}^t} \right]}$$ satisfies \begin{align*} \operatorname{det} \begin{bmatrix} 1 & 1 & 1 & \cdots & 1\\ -1 & 0 & 0 & \cdots & 0\\ 0 & -1 & 0 & \cdots & 0\\ 0 & 0 & -1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1\\ \end{bmatrix} &= (1) \cdot \operatorname{det} \begin{bmatrix} -1 & 0 & 0 & \cdots & 0\\ 0 & -1 & 0 & \cdots & 0\\ 0 & 0 & -1 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & -1\\ \end{bmatrix} \\ &= (-1)^{p-2} \neq 0 .\end{align*}

where the first equality follows from expanding along the first row and noting this is the first minor, and every other minor contains a row of zeros.

• Compute \begin{align*}B\mathbf{v} = {\left[ {\sum_{i=1}^p 1, \sum_{i=1}^p 1, \cdots, \sum_{i=1}^p 1} \right]} = {\left[ {p, p, \cdots, p} \right]} = p {\left[ {1, 1, \cdots, 1} \right]} = p\mathbf{v}_1,\end{align*} thus $$\lambda_1 = p$$
• $$\dim E_{\lambda_1} = 1$$ since the eigenspaces are orthogonal and $$E_{\lambda_0} \oplus E_{\lambda_1} \leq F^p$$ is a subspace, so $$p > \dim(E_{\lambda_0}) + \dim E_{\lambda_1} = p-1 + \dim E_{\lambda_1}$$ and it isn’t zero dimensional.

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*}

## Spring 2018 #4#algebra/qual/work

Let \begin{align*} A=\left[\begin{array}{lll}{0} & {1} & {-2} \\ {1} & {1} & {-3} \\ {1} & {2} & {-4}\end{array}\right] \in M_{3}(\mathbb{C}) \end{align*}

• Find the Jordan canonical form $$J$$ of $$A$$.

• Find an invertible matrix $$P$$ such that $$P^{-1}AP = J$$.

You should not need to compute $$P^{-1}$$.

## Spring 2017 #6#algebra/qual/work

Let $$A$$ be an $$n\times n$$ matrix with all entries equal to $$0$$ except for the $$n-1$$ entries just above the diagonal being equal to 2.

• What is the Jordan canonical form of $$A$$, viewed as a matrix in $$M_n({\mathbf{C}})$$?

• Find a nonzero matrix $$P\in M_n({\mathbf{C}})$$ such that $$P^{-1}A P$$ is in Jordan canonical form.

## Spring 2016 #1#algebra/qual/work

Let \begin{align*} A=\left(\begin{array}{ccc} -3 & 3 & -2 \\ -7 & 6 & -3 \\ 1 & -1 & 2 \end{array}\right) \in M_{3}(\mathrm{C}) .\end{align*}

• Find the Jordan canonical form $$J$$ of $$A$$.

• Find an invertible matrix $$P$$ such that $$P^{-1}A P = J$$. You do not need to compute $$P^{-1}$$.

## Spring 2015 #6#algebra/qual/completed

Let $$F$$ be a field and $$n$$ a positive integer, and consider \begin{align*} A=\left[\begin{array}{ccc} 1 & \dots & 1 \\ & \ddots & \\ 1 & \dots & 1 \end{array}\right] \in M_{n}(F) .\end{align*}

Show that $$A$$ has a Jordan normal form over $$F$$ and find it.

Hint: treat the cases $$n\cdot 1 \neq 0$$ in $$F$$ and $$n\cdot 1 = 0$$ in $$F$$ separately.

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*}

## Fall 2014 #5#algebra/qual/work

Let $$T$$ be a $$5\times 5$$ complex matrix with characteristic polynomial $$\chi(x) = (x-3)^5$$ and minimal polynomial $$m(x) = (x-3)^2$$. Determine all possible Jordan forms of $$T$$.

## Spring 2013 #5#algebra/qual/work

Let $$T: V\to V$$ be a linear map from a 5-dimensional $${\mathbf{C}}{\hbox{-}}$$vector space to itself and suppose $$f(T) = 0$$ where $$f(x) = x^2 + 2x + 1$$.

• Show that there does not exist any vector $$v\in V$$ such that $$Tv = v$$, but there does exist a vector $$w\in V$$ such that $$T^2 w= w$$.

• Give all of the possible Jordan canonical forms of $$T$$.

## Spring 2021 #1#algebra/qual/completed

Let m \begin{align*} A \coloneqq \begin{bmatrix} 4 & 1 & -1 \\ -6 & -1 & 2 \\ 2 & 1 & 1 \end{bmatrix} \in \operatorname{Mat}(3\times 3, {\mathbf{C}}) .\end{align*}

• Find the Jordan canonical form $$J$$ of $$A$$.

• Find an invertible matrix $$P$$ such that $$J = P ^{-1}A P$$.

• Write down the minimal polynomial of $$A$$.

You should not need to compute $$P^{-1}$$

• $$\chi_A(t) = t^n - {\mathrm{tr}}\qty{\bigwedge\nolimits^1 A}t^{n-1} + {\mathrm{tr}}\qty{\bigwedge\nolimits^2 A}t^{n-2} - \cdots \pm \operatorname{det}(A)$$
• Finding generalized eigenvectors: let $$B = A-\lambda I$$, get eigenvector $$v$$, solve $$Bw_1 = v, Bw_2 = w_1, \cdots$$ to get a Jordan block. Repeat with any other usual eigenvectors.
• Convention: construct Jordan blocks in decreasing order of magnitude of eigenvalues.
• Polynomial exponent data:
• Minimal polynomial exponents: sizes of largest Jordan blocks.
• Characteristic polynomial exponents: sum of sizes of Jordan blocks, i.e. how many times $$\lambda$$ is on the diagonal of $$\operatorname{JCF}(A)$$.

• Write $$\chi_A(t) = t^3 - T_1 t^2 + T_2 t - T_3$$ where $$T_i \coloneqq{\mathrm{tr}}\qty{\bigwedge\nolimits^i A}$$:
• $$T_1 = {\mathrm{tr}}(A) = 4-1+1=4$$.
• $$T_2 = (-1-2) + (4+2) + (-4-6) = 5$$.
• $$T_3 = \operatorname{det}(A) = 4(-1-2) -1(-10) + (-1)(-6+2) = 2$$.
• So $$\chi_A(t) = t^3 - 4t^2 + 5t-2$$.
• Try rational roots test: $$r \in \left\{{\pm 2/1}\right\}$$, and check that 2 is root.
• By polynomial long division, $$\chi_A(t) / (t-2) = t^2-2t+1 = (t-1)^2$$.
• So the eigenvalues are $$\lambda = 2, 1$$.
• $$\lambda = 2$$:
• Set $$U\coloneqq A-\lambda I$$, then find $$\operatorname{RREF}(U)$$ to compute its kernel: \begin{align*} U \coloneqq \begin{bmatrix} 2 & 1 & -1 \\ -6 & -3 & 2 \\ 2 & 1 & -1 \end{bmatrix} \leadsto \begin{bmatrix} 2 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix} ,\end{align*} which yields $$v_1 = [1,-2,0]$$.
• $$\lambda = 2$$:
• Similarly, \begin{align*} U \coloneqq \begin{bmatrix} 3 & 1 & -1 \\ -6 & -2 & 2 \\ 2 & 1 & 0 \end{bmatrix} \leadsto \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} ,\end{align*} which yields $$v_2 = [1,-2,1]$$.

• Solve $$Uw = v_3$$: \begin{align*} \begin{bmatrix} 3 & 1 & -1 & 1 \\ -6 & -2 & 2 & -2 \\ 2 & 1 & 0 & 1 \end{bmatrix} \leadsto \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} ,\end{align*} so take $$v_3 = [0,1,0]$$.

• Putting things together: \begin{align*} A &= P^{-1}J P \text{ where } \\ J = J_1(\lambda = 2) \oplus J_2(\lambda = 1) &= \begin{bmatrix} 2 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} \\ P = [v_1, v_2, v_3] &= \begin{bmatrix} 1 & 1 & 0 \\ -2 & -2 & 1 \\ 0 & 1 & 0 \end{bmatrix} .\end{align*}

• Write $$\min_A(t) = (t-2)(t-1)^{\ell_1}$$, then since $$\min_A(t)$$ divides $$\chi_A(t)$$ either $$\ell_1 = 1, 2$$.
• $$\ell_1$$ is the size of the largest block corresponding to $$\lambda = 1$$, which is size 2, so $$\lambda_1=2$$.
• Thus \begin{align*} \min_A(t) = (t-2)(t-1)^2 .\end{align*}

## Fall 2020 #5#algebra/qual/work

Consider the following matrix: \begin{align*} B \coloneqq \begin{bmatrix} 1 & 3 & 3 \\ 2 & 2 & 3 \\ -1 & -2 & -2 \end{bmatrix} .\end{align*}

• Find the minimal polynomial of $$B$$.

• Find a $$3\times 3$$ matrix $$J$$ in Jordan canonical form such that $$B = JPJ^{-1}$$ where $$P$$ is an invertible matrix.

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