# If $A$ is an upper triangular matrix with diagonal entries are equal to $0$, why $A^n$ is equal to $0$

$${\bf A}_{n\times n} = \underbrace{ \left.\left( \begin{array}{ccccc} 0&A_{1,2}&A_{1,3}&\cdots &A_{1,n}\\ 0&0&A_{2,3}&\cdots &A_{2,n}\\ 0&0&0&\cdots &A_{3,n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots &0 \end{array} \right)\right\} }_{n\text{ columns}} \,n\text{ rows}$$

Let $A$ be an $n\times n$ upper triangular matrix and whose diagonal entries are $0$, then how could we prove that $A^n = 0$?

The eigenvalues of an upper triangular matrix are the diagonal entries. Thus, the characteristic equation of $A$ is $\lambda^n$. According to the Cayley-Hamilton theorem, this implies that $A^n = 0$.

• Potentially a useful approach. However, it is notable that this fact about strictly upper triangular matrices is often used as a means of proving the Cayley Hamilton theorem (see, e.g. Horn and Johnson). – Omnomnomnom Mar 13 '16 at 23:37

Another direction: as you may know, the $k$-th column of a matrix is the image of $(e_k)$, the $k$-th vector of the canonical basis of $\mathbb{R}^n$ (having all its coordinates zero, but the $k$-th which is 1).

Let $E_k$ be the subspace of $\mathbb{R}^n$ that contains all vectors having their $k$ last components equal to 0.

Thus, Matrix A sends $V_1,V_2,... V_n$ to vectors that are all in $E_1$.

This is the first level. Now, if we reapply $A$ to any vector or $E_1$,

$$\left( \begin{array}{ccccc} 0&A_{1,2}&A_{1,3}&\cdots &A_{1,n}\\ 0&0&A_{2,3}&\cdots &A_{2,n}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots &A_{n-1,n}\\ 0&0&0&\cdots &0 \end{array} \right) \left( \begin{array}{ccccc} x_1\\ x_2\\ \vdots\\ x_{n-1}\\ 0 \end{array} \right)$$

do you see why we obtain a vector of $E_2$, i.e., with its two last coordinates zero ? By a very simple recurrence reasoning, we can establish that there is indeed a progressive dimension shrinking, one at a time, with $E_1 \supset E_2 \supset E_3 \cdots \supset E_k \cdots$.At the end, we are left with vectors having all their components equal to zero...

The first basis vector is mapped to the zero vector; its image under the linear transformation is $0$.

The second basis vector is mapped to a scalar multiple of the first basis vector, and the first basis vector is mapped to $0$. The image of the image of the second basis vector is $0$.

The third basis vector is mapped to a linear combination of the first two basis vectors. Both of those are mapped to $0$ in two steps, so the third one is mapped to $0$ in three steps.

Continue in that way.

PS: If you prefer a slightly different rhythm of thought, you can put it like this: Call the standard basis vectors $e_1,\ldots,e_n$. Then we have $$A e_1 = 0,$$ $$A e_2 = s e_1 \text{ (where s is some scalar), so } A^2 e_2 = A(se_1) = 0.$$ $$A e_3 = s e_1 + s e_2 \text{ (where s and s are two different scalars), so } A^3 e_3 = A^2 (se_1+se_2) = 0.$$ $$A e_4 = (se_1+se_2+se_3) \text{ (where s and s and s are three different scalars, so}$$ $$A^4 e_4 = A^3(se_1+se_2+se_3) = 0$$ and so on.

What is $(A^2)_{1,2}$? What about $(A^2)_{2,3}$? Can you see what is happening here? How would this apply to $A^3$?

• I can see but I want to learn proof of that problem – onurctirtir Mar 13 '16 at 23:32
• @OnurTirtir We're trying to guide you towards making a proof on your own. – Omnomnomnom Mar 13 '16 at 23:35
• @OnurTırtır since you haven't told us about any of your thoughts on the problem, this answer is giving you a "first step" towards formulating a proof. – Omnomnomnom Mar 13 '16 at 23:38

Denote by $A^{(r)}_{i,j}$ the coefficient at place $(i,j)$ in $A^r$. Then, for $i<n$, \begin{align} A^{(2)}_{i,i+1} &=\sum_{1\le k\le n} A_{i,k}A_{k,i+1}\\[4px] &=\sum_{1\le k<i} A_{i,k}A_{k,i+1}+ A_{i,i}A_{i,i+1}+A_{i,i+1}A_{i+1,i+1}+ \sum_{i+1<k\le n} A_{i,k}A_{k,i+1} \end{align} which is $0$ because for $1\le k<i$ we have $A_{i,k}=0$, $A_{i,i}=A_{i+1,i+1}$ and for $i+1<k\le n$ we have $A_{k,i+1}=0$.

Now suppose by inductive hypothesis that all coefficients $A^{(r)}_{i,j}=0$, for $i\le j\le i+r-1\le n$. Then, for $i\le j\le i+r$, \begin{align} A^{(r+1)}_{i,j} &=\sum_{1\le k\le n}A^{(r)}_{i,k}A_{k,j}\\[4px] &=\sum_{1\le k<i}A^{(r)}_{i,k}A_{k,j}+ \sum_{i\le k\le j}A^{(r)}_{i,k}A_{k,j}+ \sum_{j<k\le n}A^{(r)}_{i,k}A_{k,j} \end{align}