Suppose $T \in \mathcal{L}(V) $ and $3$ and $8$ are eigenvalues of $T$ . Let $ \ n =\dim (V)$

Prove that $V=(\operatorname{null} T^{n-2}) \oplus (\operatorname{range} \ T^{n-2}) $


Since the eigen values of $T$ are $3,8$ and it has no zero eigenvale.

Thus $T$ is non-singular.

Clerarly $T$ is diagonalisable and the diagonal matrix is $\begin{pmatrix} 3 & 0 \\ 0 & 8 \end{pmatrix}$

Thus the eigenvalues of $T^{n-2}$ are $3^{n-2} , 8^{n-2}$ and these are distinct.

So $T^{n-2}$ is non-singular.

Thus $ \dim(\operatorname{null} T^{n-2})=0$.

But now what would be $ \dim(\operatorname{range} T^{n-2})$?

If I can show that $\dim V=\dim (\operatorname{null} T^{n-2}) \oplus \dim (\operatorname{range} T^{n-2}) $

This implies $V=\operatorname{null}(T^{n-2}) \oplus \operatorname{range} (T^{n-2})$

Help me doing this.

  • $\begingroup$ $T$ is not necessarily non-singular or diagonalizable when $n \geq 2$. $\endgroup$ Jul 25, 2018 at 10:53
  • $\begingroup$ yes It my mistake ? $\endgroup$
    – MAS
    Jul 25, 2018 at 10:56
  • $\begingroup$ Yes I forgot that dim (V)=n. . It my mistake. please help me $\endgroup$
    – MAS
    Jul 25, 2018 at 10:57

2 Answers 2


Since $3,8$ are eigenvalues of $T$, the minimal polynomial of $T$ can be written as

$$m_T(x) = x^k(x-3)(x-8)q(x)$$

for some $q \in \mathbb{C}[x]$ and $0 \le k \le n-2$. This means that the kernels and images of $T$ stabilize at $n-2$ (or sooner) so $$\operatorname{null} T^{n-2} = \operatorname{null} T^{n-1} = \operatorname{null} T = \cdots$$ $$\operatorname{range} T^{n-2} = \operatorname{range} T^{n-1} = \operatorname{range} T = \cdots$$

Assume $y \in \operatorname{null} T^{n-2} \cap \operatorname{range} T^{n-2}$. Hence $y = T^{n-2}x$ for some $x \in V$ and then $0 = T^{n-2}y = T^{2n-4}x$ which implies $x\in \operatorname{null} T^{2n-4} = \operatorname{null} T^{n-2}$ (note that $2n-4 \ge n-2$ because $n \ge 2$). Hence $y = T^{n-2}x = 0$ so we conclude $\operatorname{null} T^{n-2} \cap \operatorname{range} T^{n-2} = \{0\}$.

On the other hand for any $x \in V$ we have $T^{n-2}x \in \operatorname{range} T^{n-2} = \operatorname{range} T^{2n-4}$ so $T^{n-2}x = T^{2n-4}z$ for some $z \in V$. We have

$$x = \underbrace{(x - T^{n-2}z)}_{\in \operatorname{null} T^{n-2}} + \underbrace{T^{n-2}z}_{\in\operatorname{range}T^{n-2}}$$ because $T^{n-2}(x - T^{n-2}z) = T^{n-2}x - T^{2n-4}z = 0$.

We conclude $$\operatorname{null} T^{n-2} \oplus \operatorname{range} T^{n-2} = V$$


One approach is as follows: in the case where $T$ non-singular, the statement is trivial. If $T$ is singular, the we can put $T$ into its Jordan form. In particular, we may put $T$ into the block triangular form $$ J = \pmatrix{J_1&0&0\\0&J_2&0\\0&0&J_3} $$ where $J_1$ has $3$ as its only eigenvalue, $J_2$ has $8$ as its only eigenvalue, and $J_3$ is singular with size at most $(n-2) \times (n-2)$.

It now suffices to note that any $k \times k$ matrix $A$ satisfies $$ \Bbb C^k = \operatorname{null} A^{k} \oplus \operatorname{range} A^k $$

  • $\begingroup$ I did not get you on the last part of your answer. If I can show that $ \ \mathbb{C}^k=null \ A^k \oplus range \ A^k \ $ , then what would I get? $\endgroup$
    – MAS
    Jul 25, 2018 at 11:28

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