I found the titular question     Let $A$ be a $10\times 10$ matrix and $A^{10,000} = 0$. Show $A^{10} = 0$   On a practice qualifier. I am very rusty on linear algebra and could not remember any specific theories other than maybe finding the characteristic polynomial of $A^{10}$ and reasoning that if it is not equal to the $\lambda^{10}$ then the characteristic polynomial of $A^{10,000}$ is certainly not equal to $\lambda^{10,000}$ but drew a blank when trying to rationalize why this would be so and why it would matter. So maybe that is a dead end.  Then I saw this post  which included many unique ways of solving that (possibly similar) question. So what are some possible proofs for this question? Ideally I would start completely reviewing linear-algebra but I am catching up on a lot of subjects and won't have the time to dive deep into it until the summer. In the mean time I'd like to do a little targeting refreshing on practice qualifiers like this one. Thank you for any help!  p.s. I included an abstract algebra tag because that is where I have the strongest background so feel free to use generalized theorems and techniques from abstract algebra along with any linear-algebra proofs you like.
122678
I want to use the fact that for a $(n \times n)$ nilpotent matrix $A$, we have that $A^n=0$, but we haven't yet introduced the minimal polynomials -if we had, I know how to prove this.  The definition for a nilpotent matrix is that there exists some $k\in \mathbb{N}$ such that $A^k=0$.  Any ideas?
121588
I have matrices $A,B$ of dimension $n$ with real coefficients which satisfy the following: $A^2-B^2=c(AB-BA)$ where $c$ is a real number. If $c\neq0$ , prove that $(AB-BA)^n = 0$.   So far, I've been able to show that $AB-BA$ is singular. Can someone help?
36949
Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification.
36774
Well, the original task was to figure out what the following expression evaluates to for any $n$.  $$\begin{pmatrix} 1 &amp; 1 \\ 0 &amp; 1\end{pmatrix}^{\large n}$$  By trying out different values of $n$, I found the pattern: $$\begin{pmatrix} 1 &amp; 1 \\ 0 &amp; 1\end{pmatrix}^{\large n} = \begin{pmatrix} 1 &amp; n \\ 0 &amp; 1\end{pmatrix}$$  But I have yet to figure out how to prove it algebraically.   Suggestions?
84843
Prove that if $A$ is a square matrix with integer entries and $\det(A) = \pm 1$, then $A^{-1}$ contains all integer entries.  I'm really thrown off by this one, its unlike all the examples I've seen.. I just dont know where to start... how can I begin to prove this?
99575
Why is $a^0=1$  $\forall a \in Z, a\neq0$. I understand $2^4=2\cdot2\cdot2\cdot2$ How can I express $a^0$. I am serious about the practical proof of this
39380
$A$ is a $n\times n$ matrix such that $ A^m = 0 $ for some positive integer $m$.   Show that $A^n = 0$.   My attempt: For $n &gt; m$, it's obvious since matrix multiplication is associative.  For $n &lt; m$, $A^n\times A^{m-n} = 0$; not sure what to do next. Also I know that $\det A = 0$.
108649
Let $A$ and $B$ be $m \times n$ and $n \times m$ real matrices.  I proved that if $\lambda$ is a nonzero eigenvalue of the $m \times m$ matrix $AB$ then it is also an eigenvalue of the $n \times n$ matrix $BA$. But I am having trouble finding an example showing that this need not be true if $\lambda=0$  Thank you in advance
41104
I have to show that all $n \times n$ nilpotent matrices of order n are similar. My initial approach was to show that for all nilpotent matrices their minimun characteristic polynomial is of the form: $$\lambda^n$$  Is this sufficient?  Can someone show me a formal approach to this problem?  Thanks!
77261
I'm working on the following problem:     Let $A \in M_3 (\mathbb {R})$ be such that $A^8=I$. Then           the minimal polynomial of $A$ can only be of degree $2$.     the minimal polynomial of $A$ can only be of degree $3$.     either $A = I$ or $ A = -I$.     there are uncountably many such $A$.        By taking $A=I $ we can eliminate options (1) &amp; (2). For the minimal polynomial of $A$ in that case is of degree $1$. Now take   $$A = \begin{bmatrix}      1 &amp; 0 &amp; 0 \\      0 &amp; -1 &amp; 0 \\      0 &amp; 0 &amp; -1 \end{bmatrix}$$  Then $A^8=I $ but $A$ is neither $I$ nor $-I$. So option (3) is eliminated. Now I don't know how to proceed about with option  (4). Can someone help me please? Thanks in advance.
11445
Prove that if $a&gt;0$ then $1+a^9 \leq \frac{1}{a}+a^{10} $.   Using the fact that $a \gt 0$, multiply by $a$ on both sides and get everything to one side we have; $a^{11}-a^{10}-a+1 \geq 0$. By factoring $(a^{10}-1)(a-1) \geq 0 $.  I am not sure how to proceed any further.
90729
Say A is an n by n matrix over the complex numbers so that A raised to the kth power is the identity I. How do we show A can be diagonalized?  Also, if alpha is an element of a field of characteristic p, how do we show that the matrix A=[1, alpha; 0, 1] satisfies A raised to the pth power equals the identity I and cannot be diagonalized if alpha is nonzero.  Please be detailed. I really have no idea how to start on this one.
89209
Determine the number of possible values for $\det(A)$, given that $A$ is an $n \times n$ matrix with real entries such that $A^3 - A^2 -3A +2I=0$.   is the source of the problem. In the last comment, I don't understand why we can say for sure that $A$ has only three distinct eigenvalues. Can anyone explain to me?  Remark: I understand that the roots of the equation $x^3-x^2-3x+2=0$ are eigenvalues of $A$. My problem is why the matrix $A$ cannot have another eigenvalue?
97671
Let $A$ be a real square matrix en let $I$ be the identity matrix with the same size. Let $n&gt;0$ be the smallest natural number for which there exist  $a_0,a_1,...a_{n-1} \in {R}$ such that $A^n+a_{n-1}A^{n-1}+....+a_1A+a_0I=O$.  Show that A is invertible if and only if $a_0\neq0 $.  So i need to proof    $A$ invertible $\rightarrow$$a_0 \neq 0$ $a_0 \neq0 \rightarrow$ $A$ invertible    I don't really know where to start with 1, and with 2 I tried to prove it with a contradiction but I don't think it's right either.
4604
Why $0^0=I$?   I'd tried prove that considering $N^0$ where N is a Nilpotent matrix and then using the Cayley -Hamilton theorem  Thanks in advance.
99442
Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $\operatorname{rank} A=2$ and $A^3=A^2\neq 0$  Suppose $A$ is not diagonalizable.  Then which one is true   $A^2=A\neq 0$ $\exists$ a vector $v$ such that $Av\neq 0$ but $A^2v= 0$ the characteristic polynomial of $A$ is $x^4-x^3$   My try:  If $A^2=A$ then minimal polynomial of $A$ has two distinct roots and hence $A$ is diagonalisable hence false. Since $A^3=A^2\neq 0$ so the characteristic roots of $A$ are $0,0,1$ Since rank of $A$ is $2$ $1$ must appear twice so the characteristic polynomial of $A=x^4-(0+0+1+1)x^3+(0.0.1+0.1.1+1.1.0+1.0.1)x^2+(0.0+0.1+1.1+...)x+(0.0.1.1)$ hence 3 is false  2 I am not sure Please find the error in my solution and don't refer it to some previous solution
76155
If $A\in M_{n}$ has $n$ distinct eigenvalues and if $A$ commutes with a given matrix $B\in M_{n}$, how can I show that $B$ is a polynomial in $A$ of degree at most $n-1$? I think first I need to show that $A$ and $B$ are simultaneously diagonalizable, but I don't know how to begin. Any advice is much appreciated.