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Homework 1

1-Homework-AlgebraB.pdf

  • Problem 1

Consider \(A = \begin{pmatrix} 2 & -k & 0 \\ -1 & 2 & k \\ 0 & 1 & 2 \end{pmatrix}\) with \(k \in \mathbb{R}\).

  • For which values of \(k\) is \(A\) diagonalizable?

    Consider \(\chi_{A}(x)=\det(xId-A)=\det \begin{pmatrix} x-2 & k & 0 \\ 1 & x-2 & -k \\ 0 & -1 & x-2 \end{pmatrix}=\left(x-2\right)^{3}-k\left(x-2\right)-\left(x-2\right)k=0\)​

    Then \((x-2)((x-2)^{2}-2k)=0\implies x=2,\left(x-2\right)^{2}=2k\implies x=2,\sqrt{2k}+ 2,-\sqrt{2k}+2\)​

    Thus if \(k<0\), Since the algebraic multiplicity of \(2\) is \(1\), then geometric multiplicity less than 1, but there at least one eigenvectors, thus \(\dim V_{\lambda =2}=1\). And we only have one eigenvalue, thus \(1<\dim V=3\), it is impossible

    Thus if \(k=0\), then the algebraic multiplicity of \(2\) is \(3\) but \(\dim V_{\lambda =2}=1\), thus impossible

    Thus \(k>0\). We will have three different eigenvalue and three linearly independent eigenvectors. Possible!

    Thus if \(k>0\), \(A\) is diagonalizable - Find the minimal polynomial of \(A\) for any value \(k \in \mathbb{R}\).

    If \(k>0\), then \(A\) is diagonalizable, then \(m_{A}(x)=(x-2)\left(x-\left(\sqrt{2k}+2\right)\right)\left(x-\left(-\sqrt{2k}+2\right )\right)\)​

    If \(k<0\), then \(m_{A}(x)=\chi_{a}(x)=(x-2)\left((x-2)^{2}-2k\right)\)​

    If \(k=0\), then \(m_{A}(x)=\chi_{a}(x)=(x-2)^{3}\)​ - Problem 2

Define \(A = \begin{pmatrix} 3 & -1 & -1 & -1 \\ 0 & 2 & 0 & 0 \\ -1 & -1 & 3 & -1 \\ 0 & 0 & 0 & 2 \end{pmatrix}\).

  • Find the minimal polynomial of \(A\).

    Consider \(\chi_{A}(x)=\det(xId-A)=\det \begin{pmatrix} x-3 & 1 & 1 & 1 \\ 0 & x-2 & 0 & 0 \\ 1 & 1 & x-3 & 1 \\ 0 & 0 & 0 & x-2 \end{pmatrix}=\left(x-2\right)\det \begin{pmatrix} x-3 & 1 & 1 \\ 0 & x-2 & 0 \\ 1 & 1 & x-3 \end{pmatrix}=\left(x-2\right)^{2}\det \begin{pmatrix} x-3 & 1 \\ 1 & x-3 \end{pmatrix}\)

    \(=\left(x-2\right)^{2}\left(\left(x-3\right)^{2}-1\right)=\left(x-2\right)^{2}\left (x^{2}-6x+8\right)=\left(x-2\right)^{3}\left(x-4\right)=0\)​

    Thus \(x=2,4\) is eigenvalue.

    \(2-A= \begin{pmatrix} -1 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & -1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\rightarrow \begin{pmatrix} 1 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}\), thus rank is 2 and \(\dim V_{\lambda = 2}=2\)

    And the algebraic multiplicity of \(4\) is \(1\)​

    Thus we have \(3\) linearly independent eigenvector, thus it is not diagonalizable

    Then after calculating \((A-2)^{2}(A-4)=0\), the minimal polynomial is \(\left(x-2\right)^{2}\left(x-4\right)\) - Write \(A^{-1}\) as a linear combination of \(\{I, A, A^2, A^3\}\).

    Since \(A\) is invertible, then \(\chi_{A}(0)=\left(-1\right)^{4}\det A=\det A=a_{0}\) and \(\det A\neq 0\)​\, then \(\chi_{A}(A)=0=A^{4}+a_{3}A^{3}+\cdots+a_{0}I\) where \(a_0\neq 0\)

    Thus \(\frac{1}{a_{0}}\left(A^{4}+a_{3}A^{3}+\ldots+a_{1}A\right)=-I\), then \(A\left(-\frac{1}{a_{0}}\left(A^{3}+a_{3}A^{2}+\ldots+a_{1}I\right)\right)=I\)​

    Thus \(A^{-1}=-\frac{1}{a_{0}}\left(A^{3}+a_{3}A^{2}+\ldots+a_{1}I\right)\)​

    Since \(\chi_{A}\left(x\right)=\left(x-2\right)^{3}\left(x-4\right)\), then \(\chi_{A}(x)=x^4-10x^3+36x^2-56x+32\)​

    Thus \(A^{-1}=-\frac{1}{32}A^{3}+\frac{5}{16}A^{2}-\frac{9}{8}A+\frac{7}{4}I\) - Problem 3

Indicate whether the following statements are true or false. Justify your response.

  • A matrix \(A\) is diagonalizable if and only if its transpose \(A^T\) is diagonalizable.

    True.

    \(\det(xId-A)=\det(xId-A)^{T}=\det(xId-A^{T})\), thus they have same eigenvalues

    And \(\text{rank}(xId-A)=\text{rank}(xId-A)^{T}\), thus they have same \(\dim(\ker(xId-A))\)​

  • If \(0\) is an eigenvalue of \(AB\), then \(0\) is not an eigenvalue of \(BA\).

    False

    We have \(\det(-AB)=\left(-1\right)^{n}\det AB=0\), then \(\det(AB)=0\Rightarrow\det BA=0\Rightarrow\det\left(-BA\right)=0\)

    Thus \(0\) is an eigenvalue of \(BA\)​ - If \(A^3 - A^2 - 4A + 4I = 0\), then \(A\) is invertible and non-diagonalizable on \(\mathbb{R}\).

    Flase

    Since \(A^{3}-A^{2}-4A+4I=0=\left(A-1\right)\left(A^{2}-4I\right)=0\), then \(\det\left(A-1\right)\left(A^{2}-4I\right)=0\Rightarrow\det\left(A-1\right)=0\text{ or }\det\left(A-2\right)=0\text{ or }\det\left(A+2\right)=0\)​

    Thus the eigenvalue is \(1,2,-2\). Since there is no eigenvalue 0, thus it is invertible

    But \(A\) can be diagonalizable on \(\R\) since we have three linearly independent eigenvectors: \(\mathcal{B}=\{v_{1},v_{2},v_{-2}\}\), then it is diagonalizable: \(\left\lbrack A\right\rbrack_{\mathcal{B}}= \begin{pmatrix} 1 & \placeholder{} & \placeholder{} \\ \placeholder{} & 2 & \placeholder{} \\ \placeholder{} & \placeholder{} & -2 \end{pmatrix}\)​

    Thus this is false - Let \(A\) and \(B\) be two matrices in \(\mathbb{R}^{4 \times 4}\) such that the minimal polynomials of \(A\) and \(B\) divide \(p(x) = x^3 - 2x^2 + 4x\). Then \(A\) and \(B\) are similar and diagonalizable.

    False

    Since \(p(x)=x(x^2-2x+4)\), thus \(m_A(x),m_B(x)=x\) or \(x^{2}-2x+4\) or \(x^{3}-2x^{2}+4x\)​

    If \(m_{A}(x),m_{B}(x)=x^{2}-2x+4\), then it is not diagonalizable since \(m_A(x)=0,m_B(x)=0\) has no solutions, then by theorem there is no eigenvalue

    Thus they are not diagonalizable

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