Let \(A = \begin{bmatrix} 1 & -1 & 2 \\ -1 & 2 & -5 \\ 2 & -1 & 1 \end{bmatrix} \in M_3(\mathbb{R})\), and let \(T_A: \mathbb{R}^3 \to \mathbb{R}^3\) be the linear transformation \(T_A(\vec{x}) = A\vec{x}\). Find \(\dim_{\mathbb{R}}(\ker(A))\), and use this to determine \(\dim_{\mathbb{R}}(\operatorname{im}(A))\).
Solution
1. Row reduction gives\[ \begin{pmatrix} 1&-1&2\\ -1&2&-5\\ 2&-1&1 \end{pmatrix} \sim \begin{pmatrix} 1&-1&2\\ 0&1&-3\\ 0&1&-3 \end{pmatrix} \sim \begin{pmatrix} 1&-1&2\\ 0&1&-3\\ 0&0&0 \end{pmatrix}. \]
Use \(R_2\leftarrow R_2+R_1\), \(R_3\leftarrow R_3-2R_1\), and then \(R_3\leftarrow R_3-R_2\).
2. The kernel is one-dimensional:\[ \ker A=\operatorname{span}\left\{\begin{pmatrix}1\\3\\1\end{pmatrix}\right\}. \]
The reduced equations are \[ y-3z=0, \qquad x-y+2z=0. \] Thus \(y=3z\) and \(x=z\), so every kernel vector is \[ z(1,3,1)^t. \] Hence \(\dim\ker A=1\).
3. Therefore\[ \boxed{\dim\ker A=1, \qquad \dim\operatorname{im}A=2.} \]
The rank–nullity theorem for \(T_A:\mathbb R^3\to\mathbb R^3\) gives \[ \dim\ker A+\dim\operatorname{im}A=3. \] Using 2 yields \(1+\dim\operatorname{im}A=3\).