Let \[ V=\{(x,y,z)\in\mathbb R^3:3x+4y+5z=0\}. \]
- Show that \(V\) is a linear subspace of \(\mathbb R^3\).
- Prove or disprove that some linear map \(S:\mathbb R^3\to\mathbb R^3\) has kernel \(V\).
- Prove or disprove that some linear map \(T:\mathbb R^3\to\mathbb R^3\) has image \(V\).
- Prove or disprove that some linear map \(U:\mathbb R^3\to\mathbb R^3\) has both kernel and image equal to \(V\).
Solution
Let \(\varphi\colon\mathbb R^3\to\mathbb R\) be \(\varphi(x,y,z)=3x+4y+5z\), so that \(V=\ker\varphi\).
1. \(V\) is a linear subspace of \(\mathbb R^3\) of dimension \(2\), with basis \(v_1=(4,-3,0)\), \(v_2=(5,0,-3)\).
\(V\) is the kernel of the linear map \(\varphi\), hence a subspace. Since \(\varphi(1,0,0)=3\neq0\), the image of \(\varphi\) is \(\mathbb R\), and rank–nullity gives \(\dim V=\dim\mathbb R^3-\dim\operatorname{im}\varphi=3-1=2\). Both \(v_1\) and \(v_2\) satisfy \(3x+4y+5z=0\), and they are linearly independent: if \(av_1+bv_2=0\), the third coordinate gives \(-3b=0\) and then the second gives \(-3a=0\). Two independent vectors in a \(2\)-dimensional space form a basis.
2. Some linear map \(S\colon\mathbb R^3\to\mathbb R^3\) has kernel \(V\).
Let \[ S=\begin{pmatrix}3&4&5\\0&0&0\\0&0&0\end{pmatrix}, \qquad S(x,y,z)^T=(3x+4y+5z,0,0)^T. \] Then \(S\mathbf x=0\) exactly when \(3x+4y+5z=0\), so \(\ker S=V\).
3. Some linear map \(T\colon\mathbb R^3\to\mathbb R^3\) has image \(V\).
Let \(T\) be the matrix with columns \(v_1\), \(v_2\), \(0\): \[ T=\begin{pmatrix}4&5&0\\-3&0&0\\0&-3&0\end{pmatrix}. \] The image of \(T\) is its column space, \(\operatorname{span}\{v_1,v_2\}=V\) by 1.
4. No linear map \(U\colon\mathbb R^3\to\mathbb R^3\) has \(\ker U=\operatorname{im}U=V\).
Rank–nullity gives \(\dim\ker U+\dim\operatorname{im}U=\dim\mathbb R^3=3\). If both equalled \(V\), the left side would be \(2\dim V=4\) by 1, a contradiction.