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*} `{=html}

  • 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/work

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.

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.

#3 #algebra/qual/completed #8 #algebra/qual/work #7 #9 #4 #6 #1 #5