Let $T: \mathbb{R}^7 \rightarrow \mathbb{R}^7$ be a diagonalizable linear operator with characteristic polynomial give by $p(t) = t(t-1)^2(t+2)^3(t -3)$.

Calculate $\dim(\ker(T - Id), \dim(Im(T + 2Id)), \dim(Im(T))$

I'm thinking whats the relationship between knowing that the algebraic multiplicity of each eigenvalue equals the geometric multiplicity and the dimension of the kernel and the image.

  • $\begingroup$ The geometric multiplicity of the eigenvalue $\lambda$ is the dimension of$\ker(-\lambda I)$. So, if $0$ is an eigenvalue, (which means $ker T$ is non-trivial), its geometric multiplicity is the dimension of $\ker T$. $\endgroup$
    – Bernard
    Commented Feb 19, 2017 at 16:59

1 Answer 1


Since $T$ is diagonalizable, it has an eigenbasis (in other words, you can find a linearly independent collection of eigenvectors which span $\mathbb{R}^7$). Let $p(t)$ be the characteristic polynomial of $T$ (or the matrix associated to $T$). Then,

  • $\ker(T-\lambda I)$ is a linear space of dimension equal to the multiplicity of $\lambda$ as a root of $p(t)$. The idea is that if $A$ is the matrix corresponding to $T$, then diagonalizability means that there is an invertible matrix such that $P^{-1}AP=D$ is diagonal. Then, $A-tI$ can be diagonalized as $PDP^{-1}-tPIP^{-1}=P(D-tI)P^{-1}$. If $\lambda$ is an eigenvalue then $D-\lambda I$ will have as many zeros on the diagonal as the multiplicity of the root of $p(t)$ at $\lambda$. Moreover, if the $i$-th position is zero, then $Pe_i$ will be in the kernel (and be an eigenvector). All $Pe_j$'s where the $j$-th position is nonzero are not eigenvectors.

  • The eigenbasis vectors which correspond to $\lambda$ are a basis for $\ker(T-\lambda I)$. The construction above describes how to get these.

  • The dimension of the images can be found through the rank-nullity theorem.

The first two facts fail when $T$ is not diagonalizable.


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