Exercise 1199Linear algebra › Vector spaces and subspacesL1Added 3 Nov 2026

Eigenspaces as subspaces

#eigenspaces#subspaces#dimension

Statement

Let $A=\begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ . Show that the eigenspace $E_2=\ker(A-2I)$ is a subspace of $\mathbb{R}^2$ and compute its dimension.

Hint

The kernel of a linear map is always a subspace.

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Solution

Subspace. $E_2=\ker(A-2I)$ is the kernel of the linear map $x\mapsto(A-2I)x$ , hence a subspace.

Computation.
$A-2I = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}.$
The kernel is given by $2x+y=0$ , i.e. $y=-2x$ . So $E_2=\operatorname{span}((1,-2))$ .

Dimension. $\dim E_2=1$ .

Conclusion. $E_2$ is the line spanned by $(1,-2)$ .

Eigenspaces as subspaces

Statement

Solution

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