Upper Triangular Matrix with zeros up I have a linear algebra homework, but honestly I'vee been trying to solve this problem for some days without success:
Suppose T $\in$ $M_{5 \times 5}$ (K) is a triangulable matrix. Be $\beta = \{ \vec{v_1}, \vec{v_2}, \vec{v_3}, \vec{v_4}, \vec{v_5} \}$ a base of $K^5$ such that:
$$[T]_\beta = \begin{pmatrix}
4 & * & a & b & c \\
0 & 4 & d & e & f \\
0 & 0 & 5 & * & * \\
0 & 0 & 0 & 5 & * \\
0 & 0 & 0 & 0 & 5
\end{pmatrix}$$
And I must prove that exists a base $\beta'$ such that:
$$[T]_{\beta'} = \begin{pmatrix}
4 & * & 0 & 0 & 0 \\
0 & 4 & 0 & 0 & 0 \\
0 & 0 & 5 & * & * \\
0 & 0 & 0 & 5 & * \\
0 & 0 & 0 & 0 & 5
\end{pmatrix}$$
How can I prove it?
Thoughts: If a=b=c=d=e=f=0 means that there is nothing in the coordinates corresponding to the vectors related to 4, but I don't know how to prove that there is a basis with vectors that satisfy this.
 A: Hint:
Both matrices are Upper Triangular.
UT matrices are closed under addition and multiplication.
The matrix $R$ establishing the similarity $T'= R T R^{-1}$ will be UT as well.
You can partition $T, T', R$ in blocks anf for $R^{-1}$ you can apply Block Inversion
A: The generalized eigenspaces $\ker(T-4)^2$ and $\ker(T-5)^3$ intersect at only the zero vector.
In general, if $(T-\lambda)^nv=0=(T-\mu)^mv$ with $\lambda\ne\mu$ and $n,m\ge1$, then by writing the gcd $I$ of the coprime factors $(T-\lambda)^n$, $(T-\mu)^m$, as a Bezout sum of them, $$v=p_1(T)(T-\lambda)^nv+p_2(T)(T-\mu)^mv=0$$
Hence it is possible to find a spanning set $\{w_1,w_2\}$ for $\ker(T-4)^2$ and another $\{w_3,w_4,w_5\}$ for $\ker(T-5)^3$, such that together form a basis $\beta'$ for $K^5$.
The coefficients $a,\ldots,f=0$ since generalized eigenspaces are $T$-invariant. For example, suppose $Tw_3=\sum_{i=1}^5\alpha_iw_i$, then $$0=(T-5)^3Tw_3=\alpha_1(T-5)^3w_1+\alpha_2(T-5)^3w_2$$ $$\therefore\ (T-5)^3(\alpha_1w_1+\alpha_2w_2)=0$$ which forces $\alpha_1=0=\alpha_2$ by the above.
