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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?

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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.

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