2
$\begingroup$

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

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

$\endgroup$
2
$\begingroup$

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 $A^n = 0$.

$\endgroup$
  • 1
    $\begingroup$ 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). $\endgroup$ – Omnomnomnom Mar 13 '16 at 23:37
2
$\begingroup$

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...

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I can see but I want to learn proof of that problem $\endgroup$ – onurctirtir Mar 13 '16 at 23:32
  • $\begingroup$ @OnurTirtir We're trying to guide you towards making a proof on your own. $\endgroup$ – Omnomnomnom Mar 13 '16 at 23:35
  • $\begingroup$ @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. $\endgroup$ – Omnomnomnom Mar 13 '16 at 23:38
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.