1. **Problem statement:** Find the generalized eigenspaces $G(3,T)$ and $G(5,T)$, the characteristic and minimal polynomials of $T$, and Jordan bases corresponding to two given Jordan normal forms for the operator $T \in L(\mathbb{C}^5)$ defined by $$T(z_1,z_2,z_3,z_4,z_5) = (2z_1 - z_2 + z_4, z_1 + 4z_2, 5z_3, 3z_4, 5z_5).$$ 2. **Generalized eigenspaces $G(\lambda,T)$:** Recall $G(\lambda,T) = \ker((T - \lambda I)^m)$ for some $m$ large enough. - For $\lambda=3$: Compute $T - 3I$: $$T - 3I = \begin{pmatrix} 2-3 & -1 & 0 & 1 & 0 \\ 1 & 4-3 & 0 & 0 & 0 \\ 0 & 0 & 5-3 & 0 & 0 \\ 0 & 0 & 0 & 3-3 & 0 \\ 0 & 0 & 0 & 0 & 5-3 \end{pmatrix} = \begin{pmatrix} -1 & -1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 2 \end{pmatrix}.$$ We want to find $\ker((T-3I)^m)$ for $m$ large enough. Since the matrix is block upper-triangular, focus on the $2\times 2$ block in the upper-left and the $3$rd, $4$th, and $5$th diagonal entries. - The $3$rd and $5$th diagonal entries are $2$, so $(T-3I)$ acts invertibly on those coordinates, so no generalized eigenvectors there. - The $4$th diagonal entry is $0$, so the $4$th coordinate is in the kernel of $T-3I$. For the $2\times 2$ block: $$A = \begin{pmatrix} -1 & -1 \\ 1 & 1 \end{pmatrix}.$$ Compute $A^2$: $$A^2 = \begin{pmatrix} -1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -1 & -1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}.$$ So $A$ is nilpotent of index 2. Therefore, $(T-3I)^2$ kills the first two coordinates. Hence, $$G(3,T) = \ker((T-3I)^2) = \{(z_1,z_2,z_3,z_4,z_5) : (T-3I)^2(z) = 0\} = \{(z_1,z_2,0,z_4,0)\}.$$ Dimension is 3 (since $z_1,z_2,z_4$ free). - For $\lambda=5$: $$T - 5I = \begin{pmatrix} 2-5 & -1 & 0 & 1 & 0 \\ 1 & 4-5 & 0 & 0 & 0 \\ 0 & 0 & 5-5 & 0 & 0 \\ 0 & 0 & 0 & 3-5 & 0 \\ 0 & 0 & 0 & 0 & 5-5 \end{pmatrix} = \begin{pmatrix} -3 & -1 & 0 & 1 & 0 \\ 1 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{pmatrix}.$$ We want $\ker((T-5I)^m)$ for some $m$. - The $3$rd and $5$th coordinates correspond to zero diagonal entries, so they are in the kernel. - The $4$th coordinate has $-2$ on diagonal, so invertible there. Check nilpotency on the $2\times 2$ block: $$B = \begin{pmatrix} -3 & -1 \\ 1 & -1 \end{pmatrix}.$$ Compute $B^2$: $$B^2 = \begin{pmatrix} -3 & -1 \\ 1 & -1 \end{pmatrix}^2 = \begin{pmatrix} 8 & 4 \\ -4 & 0 \end{pmatrix} \neq 0,$$ so not nilpotent. Since $B$ is invertible (determinant $(-3)(-1) - (-1)(1) = 3 + 1 = 4 \neq 0$), no generalized eigenvectors here. Thus, $$G(5,T) = \ker(T-5I) = \{(0,0,z_3,0,z_5)\}.$$ Dimension is 2. 3. **Characteristic polynomial $p_T(\lambda)$:** The matrix of $T$ is upper-triangular with diagonal entries $2,4,5,3,5$ except the first two rows have off-diagonal entries. The characteristic polynomial is $$p_T(\lambda) = \det(T - \lambda I) = \det \begin{pmatrix} 2-\lambda & -1 & 0 & 1 & 0 \\ 1 & 4-\lambda & 0 & 0 & 0 \\ 0 & 0 & 5-\lambda & 0 & 0 \\ 0 & 0 & 0 & 3-\lambda & 0 \\ 0 & 0 & 0 & 0 & 5-\lambda \end{pmatrix}.$$ Since the matrix is block upper-triangular, determinant is product of determinants of blocks: - The $2\times 2$ block: $$M = \begin{pmatrix} 2-\lambda & -1 \\ 1 & 4-\lambda \end{pmatrix},$$ determinant: $$(2-\lambda)(4-\lambda) - (-1)(1) = (2-\lambda)(4-\lambda) + 1 = (8 - 2\lambda - 4\lambda + \lambda^2) + 1 = \lambda^2 - 6\lambda + 9.$$ - The other diagonal entries contribute $(5-\lambda)(3-\lambda)(5-\lambda) = (5-\lambda)^2 (3-\lambda)$. So, $$p_T(\lambda) = (\lambda^2 - 6\lambda + 9)(5-\lambda)^2 (3-\lambda).$$ Note $\lambda^2 - 6\lambda + 9 = (\lambda - 3)^2$. Hence, $$p_T(\lambda) = (\lambda - 3)^2 (5-\lambda)^2 (3-\lambda) = (\lambda - 3)^3 (5-\lambda)^2.$$ 4. **Minimal polynomial $m_T(\lambda)$:** The minimal polynomial divides the characteristic polynomial and annihilates $T$. From the Jordan form given, the largest Jordan block for eigenvalue 3 is size 3, so minimal polynomial factor for 3 is $(\lambda - 3)^3$. For eigenvalue 5, largest Jordan block size is 1, so factor is $(\lambda - 5)$. Therefore, $$m_T(\lambda) = (\lambda - 3)^3 (\lambda - 5).$$ 5. **Jordan basis for the first Jordan form:** The matrix is $$J_1 = \begin{pmatrix} 3 & 1 & 0 & 0 & 0 \\ 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 5 & 0 \\ 0 & 0 & 0 & 0 & 5 \end{pmatrix}.$$ This corresponds to one Jordan block of size 3 for eigenvalue 3 and two Jordan blocks of size 1 for eigenvalue 5. To find a Jordan basis: - Find a chain of generalized eigenvectors $v_1,v_2,v_3$ for eigenvalue 3 such that $$(T - 3I)v_1 = 0,$$ $$(T - 3I)v_2 = v_1,$$ $$(T - 3I)v_3 = v_2.$$ - Find eigenvectors $w_1,w_2$ for eigenvalue 5. From step 2, $G(3,T)$ has dimension 3, so such a chain exists. Explicit construction: - Solve $(T-3I)v_1=0$ for eigenvector $v_1$: From $(T-3I)$ matrix, $$\begin{cases} -z_1 - z_2 + z_4 = 0 \\ z_1 + z_2 = 0 \\ 2 z_3 = 0 \\ 0 = 0 \\ 2 z_5 = 0 \end{cases}$$ From second equation, $z_2 = -z_1$. Substitute into first: $-z_1 - (-z_1) + z_4 = 0 \Rightarrow 0 + z_4 = 0 \Rightarrow z_4=0$. Also $z_3=0$, $z_5=0$. So eigenvectors for 3 are of form $$v_1 = (z_1, -z_1, 0, 0, 0) = z_1(1,-1,0,0,0).$$ - Find $v_2$ such that $(T-3I)v_2 = v_1$: Let $v_2 = (a,b,c,d,e)$. Then $$(T-3I)v_2 = ( -a - b + d, a + b, 2 c, 0, 2 e ) = (1,-1,0,0,0)$$ (taking $z_1=1$ for $v_1$). Equations: $$-a - b + d = 1,$$ $$a + b = -1,$$ $$2 c = 0 \Rightarrow c=0,$$ $$0=0,$$ $$2 e = 0 \Rightarrow e=0.$$ From second: $b = -1 - a$. Substitute into first: $$-a - (-1 - a) + d = 1 \Rightarrow -a + 1 + a + d = 1 \Rightarrow 1 + d = 1 \Rightarrow d=0.$$ So $v_2 = (a, -1 - a, 0, 0, 0)$ with $a$ free. - Find $v_3$ such that $(T-3I)v_3 = v_2$: Let $v_3 = (p,q,r,s,t)$. Then $$(T-3I)v_3 = (-p - q + s, p + q, 2 r, 0, 2 t) = (a, -1 - a, 0, 0, 0).$$ Equations: $$-p - q + s = a,$$ $$p + q = -1 - a,$$ $$2 r = 0 \Rightarrow r=0,$$ $$0=0,$$ $$2 t = 0 \Rightarrow t=0.$$ From second: $q = -1 - a - p$. Substitute into first: $$-p - (-1 - a - p) + s = a \Rightarrow -p + 1 + a + p + s = a \Rightarrow 1 + s = a \Rightarrow s = a - 1.$$ So $v_3 = (p, -1 - a - p, 0, a - 1, 0)$ with $p$ free. Choose convenient values to get explicit vectors: - Take $a=0$, $p=0$: $$v_2 = (0, -1, 0, 0, 0),$$ $$v_3 = (0, -1, 0, -1, 0).$$ So Jordan chain for eigenvalue 3 is: $$v_1 = (1,-1,0,0,0),$$ $$v_2 = (0,-1,0,0,0),$$ $$v_3 = (0,-1,0,-1,0).$$ - For eigenvalue 5, eigenvectors satisfy $(T-5I)w=0$: From step 2, eigenvectors are of form $$w = (0,0,z_3,0,z_5).$$ Choose basis vectors: $$w_1 = (0,0,1,0,0), \quad w_2 = (0,0,0,0,1).$$ The Jordan basis for $J_1$ is $$\{v_3, v_2, v_1, w_1, w_2\}$$ (order reversed for Jordan chain). 6. **Jordan basis for the second Jordan form:** The matrix is $$J_2 = \begin{pmatrix} 5 & 0 & 0 & 0 & 0 \\ 0 & 3 & 1 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 \\ 0 & 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 0 & 5 \end{pmatrix}.$$ This corresponds to two Jordan blocks for eigenvalue 5 (each size 1) and one Jordan block of size 3 for eigenvalue 3. The Jordan chain for eigenvalue 3 is the same as above. For eigenvalue 5, eigenvectors are as before: $$w_1 = (0,0,1,0,0), \quad w_2 = (0,0,0,0,1).$$ The Jordan basis for $J_2$ is $$\{w_1, v_3, v_2, v_1, w_2\}.$$ **Final answers:** - $G(3,T) = \{(z_1,z_2,0,z_4,0)\}$, dimension 3. - $G(5,T) = \{(0,0,z_3,0,z_5)\}$, dimension 2. - Characteristic polynomial: $$p_T(\lambda) = (\lambda - 3)^3 (\lambda - 5)^2.$$ - Minimal polynomial: $$m_T(\lambda) = (\lambda - 3)^3 (\lambda - 5).$$ - Jordan bases as constructed above for the two Jordan normal forms.