# Find basis of fundamental subspaces with given eigenvalues and eigenvectors

Let $$\Lambda_1=0,\Lambda_2=1,\Lambda_3=2$$ and $$x_1=\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix},x_2=\begin{bmatrix} 0 \\ 1 \\ 2 \\ \end{bmatrix},x_3=\begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix}$$ are eigenvalues and eigenvectors of matrix $$A \in \Bbb{M_{3x3}}$$. Find basis of fundamental subspaces, without calculating (finding) matrix $$A$$. $$-$$ We know that different eigenvalues give different linearly independent eigenvectors,so eigenvectors $$x_1,x_2,x_3$$ are linearly independent. When $$\Lambda_1$$ is zero, it means that eigenvector $$x_1$$ is in null space of matrix $$A \ (Ax_1=\Lambda_1 x_1)$$, so we have $$\operatorname{Ker}(A)=\left\{\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}\right\}$$Because column space of $$A^{T}$$ is orthogonal on null space of matrix $$A$$, we have $$\operatorname{Im}(A^{T})=\left\{\begin{bmatrix} 0 \\ 1 \\ 0 \\ \end{bmatrix},\begin{bmatrix} 0 \\ 0 \\ 1 \\ \end{bmatrix}\right\}$$ Using rank-nullity theorem, we know that $$\dim [\operatorname{Im}(A)]=2$$, so other two eigenvectors make basis of column space of matrix $$A$$ $$\operatorname{Im}(A)=\left\{\begin{bmatrix} 0 \\ 1 \\ 2 \\ \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ 1 \\ \end{bmatrix}\right\}$$ At the end, we know that column space of $$A$$ is orthogonal on the null space of matrix $$A^{T}$$, so I think that the only vector I can use in $$\operatorname{Ker}(A^{T})$$ is vector $$\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}.$$ I'm not sure can I use the same vector for $$\operatorname{Ker}(A^{T})$$ and $$\operatorname{Ker}(A)$$. Is that actually possible?

Your reasoning is correct; by what we are given the null space of the matrix must be generated by the vector $$x_1$$, and it follows that the column space must be generated by the vectors $$x_1$$ and $$x_2$$. The conclusion that $$\operatorname{Ker}(A) = \operatorname{Ker}(A^T)$$ is no contradiction; for example, this also happens with the diagonal matrix with $$(0, 1, 1)$$ on the diagonal.