1. a) State the Second Fundamental Theorem of Subspace: If $W$ is a subspace of $V$, then $W$ must be closed under vector addition and scalar multiplication, and contain the zero vector. (i) Check if $W = \{(a,b,c) : a \geq 0\}$ is a subspace of $V=\mathbb{R}^3$: - Zero vector $(0,0,0)$ has $a=0 \geq 0$, so $0 \in W$. - Check closure under addition: Take $u = (1,0,0)$ and $v = (-1,0,0)$ both in $W$? $u$ has $a=1\geq0$ so $u \in W$, but $v$ has $a=-1<0$, so $v \notin W$. - Thus addition of two vectors in $W$ is not always defined because $v$ is not in $W$, so consider addition only for $u, u$ in $W$. - However, scalar multiplication fails: Multiply $u=(1,0,0) $ by scalar $-1$, we get $(-1,0,0)$ which has $a=-1<0$, not in $W$. Therefore, $W$ is not closed under scalar multiplication, so $W$ is not a subspace. (ii) Check if $W=\{(a,b,c): a^2 + b^2 + c^2 \leq 1\}$ is a subspace: - Check zero: $(0,0,0)$ satisfies $0\leq1$, so $0\in W$. - Closure under addition: Take $u=(1,0,0), v=(1,0,0)$ in $W$ because each has norm $1$. - Their sum is $(2,0,0)$ with norm $2 > 1$, not in $W$. - Closure under scalar multiplication: Multiply $u$ by scalar $2$, get $(2,0,0)$ not in $W$. Therefore, $W$ is not a subspace. b) Express $v=(5,9,5)$ as a linear combination of $v_1=(2,1,4)$, $v_2=(1,-1,3)$ and $v_3=(3,2,5)$: Let $v = x v_1 + y v_2 + z v_3$. So, $5 = 2x + y + 3z$ $9 = x - y + 2z$ $5 = 4x + 3y + 5z$ Solve system: From first: $5 = 2x + y + 3z$ From second: $9 = x - y + 2z$ Add first and second: $14 = 3x + 5z$ From third: $5 = 4x + 3y + 5z$ Express $y$ from first: $y = 5 - 2x -3z$ Substitute $y$ into third: $5=4x +3(5-2x-3z) + 5z = 4x + 15 -6x -9z +5z = -2x +15 - 4z$ Rearranged: $-2x - 4z = 5 -15 = -10$ Divide by -2: $x + 2z =5$ From earlier: $14=3x +5z$ Substitute $x=5 -2z$ into this: $14 = 3(5 -2z) + 5z = 15 -6z +5z = 15 - z$ $z = 15 - 14 =1$ Then $x = 5 - 2(1) = 3$ $y = 5 - 2(3) - 3(1) = 5 - 6 -3 = -4$ Therefore, $v = 3 v_1 - 4 v_2 + 1 v_3$. 2. a) Show $S=\{(1,0,0),(0,1,0),(0,0,1),(1,1,1)\}$ spans $\mathbb{R}^3$ but is not independent: - Standard basis vectors $(1,0,0), (0,1,0), (0,0,1)$ span $\mathbb{R}^3$. - Adding $(1,1,1)$ does not reduce span. - Check dependence: Suppose $a(1,0,0) + b(0,1,0) + c(0,0,1) + d(1,1,1) = (0,0,0)$. This gives: $a + d =0$ $b + d =0$ $c + d =0$ From these, $a=b=c = -d$. If $d\neq 0$, then not all coefficients are zero but sum zero vector, so vectors are dependent. b) Check if $u_1 = (1,2,3)$ and $u_2 = (2,4,6)$ form a basis of $\mathbb{R}^3$. - Number of vectors is 2, dimension of $\mathbb{R}^3$ is 3, so cannot be a basis. - Also, $u_2 = 2 u_1$, so vectors are linearly dependent. Hence, $\{u_1, u_2\}$ is not a basis of $\mathbb{R}^3$. 3. a) Show $T: \mathbb{R}^3 \to \mathbb{R}^3, T(x,y,z) = (x+2y -z, y - x, x - z)$ is linear: Check additivity and homogeneity: Let $u=(x_1,y_1,z_1)$ and $v=(x_2,y_2,z_2)$. $T(u+v) = T(x_1+x_2, y_1+y_2, z_1+z_2) = ((x_1+x_2)+2(y_1+y_2)-(z_1+z_2), (y_1+y_2)-(x_1+x_2), (x_1+x_2) - (z_1+z_2)) = (x_1+2y_1 -z_1, y_1 - x_1, x_1 - z_1) + (x_2 + 2y_2 - z_2, y_2 - x_2, x_2 - z_2) = T(u) + T(v)$. Similarly, for scalar $c$, $T(cu) = cT(u)$. Hence, $T$ is linear. Find basis and dimension of $Im(T)$: Matrix of $T$ w.r.t standard basis is: $$ \begin{bmatrix} 1 & 2 & -1 \\ -1 & 1 & 0 \\ 1 & 0 & -1 \end{bmatrix} $$ The image is span of its columns. Check rank: By Gaussian elimination, rank is 3, so $\dim(Im(T))=3$, so image is $\mathbb{R}^3$. Find $Ker(T)$: Solve $T(x,y,z) = (0,0,0)$: Equations: $x + 2y - z = 0$ $y - x = 0$ $x - z = 0$ From second: $y = x$ From third: $z = x$ Substitute into first: $x + 2x - x = 2x = 0 \implies x=0$ Thus $x=y=z=0$ Therefore, $Ker(T) = \{0\}$ and its dimension is 0. b) Let $T(x,y,z) = (2x, 4x - y, 2x + 3y - z)$. Show $T$ invertible: Write matrix: $$ A = \begin{bmatrix} 2 & 0 & 0 \\ 4 & -1 & 0 \\ 2 & 3 & -1 \end{bmatrix} $$ Check $det(A)$: $det(A) = 2(-1 \cdot -1 - 0 \cdot 3) - 0 + 0 = 2(1) = 2 \neq 0$ Since determinant non-zero, $T$ is invertible. Find $T^{-1}$: Let $T(x,y,z) = (a,b,c)$, so $a = 2x$ $b = 4x - y$ $c = 2x + 3y - z$ From first: $x = \frac{a}{2}$. From second: $b = 4x - y \implies y = 4x - b = 2a - b$. From third: $c = 2x + 3y - z \implies z = 2x + 3y - c = a + 3(2a - b) - c = a + 6a - 3b - c = 7a - 3b - c$. Hence, $$ T^{-1}(a,b,c) = \left(\frac{a}{2}, 2a - b, 7a - 3b - c \right) $$ 4. a) Find eigenvalues and eigenvectors for $$ A = \begin{bmatrix}1 & 4 \\ 9 & 1 \end{bmatrix} $$ Eigenvalues satisfy: $$ \det(A - \lambda I) = 0 \implies (1 - \lambda)^2 - 36 = 0 $$ Expand: $$ (1 - \lambda)^2 = 36 \implies (1-\lambda) = \pm 6 $$ So, $1 - \lambda = 6 \implies \lambda = -5$, and $1 - \lambda = -6 \implies \lambda = 7$. Eigenvalues: $\lambda_1 = -5$, $\lambda_2 = 7$. Find eigenvectors: For $\lambda = -5$: $(A + 5I)v = 0 \implies \begin{bmatrix}6 & 4 \\ 9 & 6\end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$ Equations: $6x + 4y = 0$ $9x + 6y = 0$ From the first: $4y = -6x \implies y = -\frac{3}{2}x$. Eigenvector: $$ v_1 = \begin{bmatrix}1 \\ -\frac{3}{2} \end{bmatrix} $$ For $\lambda = 7$: $A - 7I = \begin{bmatrix} -6 & 4 \\ 9 & -6 \end{bmatrix}$ Equations: $-6x + 4y = 0 \implies 4y = 6x \implies y = \frac{3}{2}x$ Similarly from second equation confirm. Eigenvector: $$ v_2 = \begin{bmatrix}1 \\ \frac{3}{2} \end{bmatrix} $$ b) For $$ A = \begin{bmatrix}2 & -1 \\ 1 & 0 \end{bmatrix} $$ verify Cayley-Hamilton theorem. Characteristic polynomial: $$ \det(A - \lambda I) = (2 - \lambda)(0 - \lambda) - (-1)(1) = -\lambda(2 - \lambda) + 1 = \lambda^2 - 2 \lambda + 1 $$ The polynomial is: $$ \chi_A(\lambda) = \lambda^2 - 2\lambda + 1 $$ According to Cayley-Hamilton, matrix satisfies: $$ A^2 - 2A + I = 0 $$ Calculate $A^2$: $$ A^2 = \begin{bmatrix}2 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix}2 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} (2)(2) + (-1)(1) & (2)(-1) + (-1)(0) \\ (1)(2) + (0)(1) & (1)(-1) + (0)(0) \end{bmatrix} = \begin{bmatrix}4 -1 & -2 + 0 \\ 2 + 0 & -1 + 0 \end{bmatrix} = \begin{bmatrix}3 & -2 \\ 2 & -1 \end{bmatrix} $$ Now compute $A^2 - 2A + I$: $$ \begin{bmatrix}3 & -2 \\ 2 & -1 \end{bmatrix} - 2 \begin{bmatrix}2 & -1 \\ 1 & 0 \end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix}3 & -2 \\ 2 & -1 \end{bmatrix} - \begin{bmatrix}4 & -2 \\ 2 & 0 \end{bmatrix} + \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} $$ $$ = \begin{bmatrix}3 - 4 + 1 & -2 + 2 + 0 \\ 2 - 2 + 0 & -1 - 0 + 1 \end{bmatrix} = \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix} $$ Hence verified. 5. Find LU decomposition and solve system: $$ \begin{cases} 2x + 6y + 2z = 2 \\ -3x - 8y = 2 \\ 4x + 9y + 2z = 3 \end{cases} $$ Coefficient matrix: $$ A = \begin{bmatrix}2 & 6 & 2 \\ -3 & -8 & 0 \\ 4 & 9 & 2 \end{bmatrix}, \quad b = \begin{bmatrix}2 \\ 2 \\ 3 \end{bmatrix} $$ Step 1: LU decomposition (Doolittle method): Initialize $L = \begin{bmatrix}1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix}, U = \begin{bmatrix}u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix}$ Compute $u_{11} = a_{11} = 2$ $l_{21} = a_{21} / u_{11} = -3 / 2 = -1.5$ $l_{31} = a_{31} / u_{11} = 4 / 2 = 2$ $u_{12} = a_{12} = 6$ $u_{13} = a_{13} = 2$ Calculate $u_{22} = a_{22} - l_{21} u_{12} = -8 - (-1.5)(6) = -8 + 9 = 1$ $u_{23} = a_{23} - l_{21} u_{13} = 0 - (-1.5)(2) = 3$ $l_{32} = (a_{32} - l_{31} u_{12}) / u_{22} = (9 - 2*6)/1 = (9 - 12)/1 = -3$ $u_{33} = a_{33} - l_{31} u_{13} - l_{32} u_{23} = 2 - 2 * 2 - (-3) * 3 = 2 - 4 + 9 = 7$ Final matrices: $$ L = \begin{bmatrix}1 & 0 & 0 \\ -1.5 & 1 & 0 \\ 2 & -3 & 1 \end{bmatrix}, U = \begin{bmatrix}2 & 6 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 7 \end{bmatrix} $$ Solve $Ly = b$: $y_1 = 2$ $-1.5 y_1 + y_2 = 2 \implies -1.5(2) + y_2 = 2 \implies y_2 = 2 + 3 = 5$ $2 y_1 -3 y_2 + y_3 = 3 \implies 2(2) - 3(5) + y_3 = 3 \implies 4 -15 + y_3 = 3 \implies y_3 = 3 + 11 = 14$ Solve $Ux = y$: $7 z = 14 \implies z = 2$ $1 y + 3 z = 5 \implies y + 3(2) = 5 \implies y = -1$ $2 x + 6 y + 2 z = 2 \implies 2 x + 6(-1) + 2(2) = 2 \implies 2x -6 +4 = 2 \implies 2x -2 = 2 \implies 2x=4 \implies x=2$ Solution: $x=2, y=-1, z=2$