1. **Problem Statement:** Given the linear transformation $T(x,y,z) = (2x - y, x + 4y, 3z)$, find: A) The matrix of $T$ relative to the standard basis of $\mathbb{R}^3$. B) All eigenspaces of $T$. C) Bases for each eigenspace of $T$. 2. **Matrix of $T$ relative to the standard basis:** The standard basis of $\mathbb{R}^3$ is $\{e_1 = (1,0,0), e_2 = (0,1,0), e_3 = (0,0,1)\}$. Apply $T$ to each basis vector: $$T(e_1) = T(1,0,0) = (2(1) - 0, 1 + 0, 3 \cdot 0) = (2,1,0)$$ $$T(e_2) = T(0,1,0) = (2 \cdot 0 - 1, 0 + 4(1), 3 \cdot 0) = (-1,4,0)$$ $$T(e_3) = T(0,0,1) = (2 \cdot 0 - 0, 0 + 0, 3 \cdot 1) = (0,0,3)$$ The matrix of $T$ is formed by these images as columns: $$A = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ 3. **Find eigenvalues:** The eigenvalues $\lambda$ satisfy $\det(A - \lambda I) = 0$. Compute: $$A - \lambda I = \begin{bmatrix} 2 - \lambda & -1 & 0 \\ 1 & 4 - \lambda & 0 \\ 0 & 0 & 3 - \lambda \end{bmatrix}$$ The determinant is: $$\det(A - \lambda I) = (3 - \lambda) \cdot \det \begin{bmatrix} 2 - \lambda & -1 \\ 1 & 4 - \lambda \end{bmatrix}$$ Calculate the $2 \times 2$ determinant: $$(2 - \lambda)(4 - \lambda) - (-1)(1) = (2 - \lambda)(4 - \lambda) + 1$$ Expand: $$= 8 - 2\lambda - 4\lambda + \lambda^2 + 1 = \lambda^2 - 6\lambda + 9$$ So, $$\det(A - \lambda I) = (3 - \lambda)(\lambda^2 - 6\lambda + 9)$$ Note that $\lambda^2 - 6\lambda + 9 = (\lambda - 3)^2$. Therefore, $$\det(A - \lambda I) = (3 - \lambda)(\lambda - 3)^2 = -(\lambda - 3)^3$$ The only eigenvalue is $\lambda = 3$ with algebraic multiplicity 3. 4. **Find eigenspace for $\lambda = 3$:** Solve $(A - 3I)\mathbf{v} = \mathbf{0}$. $$A - 3I = \begin{bmatrix} -1 & -1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ Let $\mathbf{v} = (x,y,z)$. The system is: $$-x - y = 0$$ $$x + y = 0$$ $$0 = 0$$ The first two equations are the same, so $-x - y = 0 \Rightarrow y = -x$. The third equation imposes no restriction on $z$. Therefore, the eigenspace is: $$\{(x, -x, z) : x,z \in \mathbb{R}\}$$ 5. **Find a basis for the eigenspace:** Express vectors as linear combinations: $$x(1, -1, 0) + z(0, 0, 1)$$ Hence, a basis is: $$\{(1, -1, 0), (0, 0, 1)\}$$ **Final answers:** - Matrix of $T$ relative to standard basis: $$\begin{bmatrix} 2 & -1 & 0 \\ 1 & 4 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ - Eigenvalue: $\lambda = 3$ (with multiplicity 3). - Eigenspace: $$\{(x, -x, z) : x,z \in \mathbb{R}\}$$ - Basis for eigenspace: $$\{(1, -1, 0), (0, 0, 1)\}$$