MHB Zero matrix with non-zero above diagonal

[rputra]

I am working on this problem which has been baffling me since the beginning: Prove that $A^k = 0$ and $A^{k-1} \neq 0$ if $A_{k \times k}$ is a zero matrix but with entries of 1's right above its diagonal. For example, if $k = 3$ the it will look like this

$$\begin{pmatrix}
0 &1 &0\\
0 &0 &1\\
0 &0 &0\\
\end{pmatrix},$$

with small diagonal of 1's slanted parallel to the main diagonal. The hint I have been getting suggests me to use standard basis vectors. For example, in the case of $k = 3$, we have $e_1 = (1,0,0), \ e_2 = (0,1,0), \ e_3 = (0,0,1)$. Then I need to find out what are $Ae_1, \ Ae_2, \ Ae_3$, etc. After that I have to continue with $A^2 e_1, \ A^2 e_2, \ A^2 e_3,$ etc. But even with these hints, I am still confused. Any help or guidance will be very much appreciated. Thank you in advance for your help and effort.

 

[Opalg]

I am working on this problem which has been baffling me since the beginning: Prove that $A^k = 0$ and $A^{k-1} \neq 0$ if $A_{k \times k}$ is a zero matrix but with entries of 1's right above its diagonal. For example, if $k = 3$ the it will look like this

$$\begin{pmatrix}
0 &1 &0\\
0 &0 &1\\
0 &0 &0\\
\end{pmatrix},$$

with small diagonal of 1's slanted parallel to the main diagonal. The hint I have been getting suggests me to use standard basis vectors. For example, in the case of $k = 3$, we have $e_1 = (1,0,0), \ e_2 = (0,1,0), \ e_3 = (0,0,1)$. Then I need to find out what are $Ae_1, \ Ae_2, \ Ae_3$, etc. After that I have to continue with $A^2 e_1, \ A^2 e_2, \ A^2 e_3,$ etc. But even with these hints, I am still confused. Any help or guidance will be very much appreciated. Thank you in advance for your help and effort.

The standard basis has the property that for any $k\times k$ matrix $A$, the vectors $Ae_1,\,Ae_2,\ldots,Ae_k$ are given by the columns of $A$. So for example in the case of the matrix $\begin{bmatrix} 0 &1 &0\\ 0 &0 &1\\ 0 &0 &0\\ \end{bmatrix},$ you can see that $Ae_1$ is the left column of $A$ which is the zero vector. The second column of $A$ tells you that $Ae_2 = e_1$, and the third column tells you that $Ae_3 = e_2.$

Using that information, you can then see what happens to higher powers of $A$. For example, $A^2e_2 = A(Ae_2) = Ae_1 = 0.$

 

[rputra]

The standard basis has the property that for any $k\times k$ matrix $A$, the vectors $Ae_1,\,Ae_2,\ldots,Ae_k$ are given by the columns of $A$. So for example in the case of the matrix $\begin{bmatrix} 0 &1 &0\\ 0 &0 &1\\ 0 &0 &0\\ \end{bmatrix},$ you can see that $Ae_1$ is the left column of $A$ which is the zero vector. The second column of $A$ tells you that $Ae_2 = e_1$, and the third column tells you that $Ae_3 = e_2.$

Using that information, you can then see what happens to higher powers of $A$. For example, $A^2e_2 = A(Ae_2) = Ae_1 = 0.$

Thank you! I did note the properties you mentioned above, however, the big question is: How do you use the properties to arrive at the conclusion that $A^k = 0$ while $A^{k-1} \neq 0$? Thanks again and again, I am sincerely looking forward to hear from you again.

 

[Opalg]

The big question is: How do you use the properties to arrive at the conclusion that $A^k = 0$ while $A^{k-1} \neq 0$? Thanks again and again, I am sincerely looking forward to hear from you again.

The effect of $A$ on the basis vector $e_j$ is that $Ae_j= e_{j-1}$ except when $j=1$, in which case $Ae_1=0.$ When you iterate $A$ $n$ times, you find that $A^ne_j = e_{j-n}$ unless $j-n\leqslant0$, in which case $A^ne_j = 0$. (To prove that carefully, you'll probably have to use induction on $n$.)

Now putting $n=k-1$ you see that $A^{k-1}e_k = e_1 \ne0$, so that $A^{k-1} \neq 0$. However, when you put $n=k$ you find that $j-k\leqslant0$, and hence $A^ke_j = 0$, for all $j$. Thus $A^k$ takes all the basis vectors to zero, which means that $A^k=0.$

 

[rputra]

The effect of $A$ on the basis vector $e_j$ is that $Ae_j= e_{j-1}$ except when $j=1$, in which case $Ae_1=0.$ When you iterate $A$ $n$ times, you find that $A^ne_j = e_{j-n}$ unless $j-n\leqslant0$, in which case $A^ne_j = 0$. (To prove that carefully, you'll probably have to use induction on $n$.)

Now putting $n=k-1$ you see that $A^{k-1}e_k = e_1 \ne0$, so that $A^{k-1} \neq 0$. However, when you put $n=k$ you find that $j-k\leqslant0$, and hence $A^ke_j = 0$, for all $j$. Thus $A^k$ takes all the basis vectors to zero, which means that $A^k=0.$

Thanks again! I think I should have posted this problem earlier, I have been struggling with this problem for days. This form has been useful for me, I am all for it! Thank you again.