Let $A$ be an $n \times n$ matrix. We know that $A^{23} = 0$. What are the eigenvalues of $A$?   I think it's just $0$, but I'm not sure. How should I do this problem?
180
I need to prove that:      if a linear operator $\phi : V \rightarrow V$ on a vector space is nilpotent, then its only eigenvalue is $0$.   I know how to prove that this for a nilpotent matrix, but I'm not sure in the case of an operator. How would I be able to relate $\phi$ to a matrix?  Note: A nilpotent operator $\phi$ has been defined as an operator that satisfies $\phi^{n} = 0$ for some $n \geq 1$.
26315
Give an example of $2\times2$ matrix $A$ such that $A$ has one independent eigenvector while $A^{2}$ has two independent eigenvectors.  I would like to know a systematic answer of how to get this. My guess was that any matrix that has 1 independent eigenvector will have the same number of independent eigenvector for the squared matrix. I have tried random selections but obviously couldn't come up with one. What's more useful will be to get the intuition behind this problem.
95482
Suppose A is a 3*3  symmetric matrix s.t.  $$\begin{pmatrix} x &amp; y &amp; 1 \\   \end{pmatrix} A  \begin{pmatrix} x  \\  y\\ 1\end{pmatrix}  = xy -1 $$  let p be the no. of positive eigen value of A and q = rank(A)-p, then p=? q=?  what matrix A should be I can't find??
28059
It is known that all $B$, $C$ and $D$ are $3 \times 3$ matrices. And the eigenvalues of $B$ are $1, 2, 3$; $C$ are $4, 5, 6$; and $D$ are $7, 8, 9$. What are the eigenvalues of the $6 \times 6$ matrix  $$\begin{pmatrix} B &amp; C\\0 &amp; D \end{pmatrix}$$ where $0$ is the $3 \times 3$ matrix whose entries are all $0$.  From my intuition, I think the eigenvalues of the new $6 \times 6$ matrix are the eigenvalues of $B$ and $D$. But how can I show that?
94030
I f $$A= \begin{pmatrix}1&amp; 0 &amp; 0 \\  1 &amp; 0 &amp; 1\\ 0 &amp; 1 &amp; 0 \end{pmatrix}$$  then $ A^{50} $  is   $$ \begin{pmatrix}1&amp; 0 &amp; 0 \\  50 &amp; 1 &amp; 0\\ 50 &amp; 0 &amp; 1 \end{pmatrix}$$ $$\begin{pmatrix}1&amp; 0 &amp; 0 \\  48 &amp; 1 &amp; 0\\ 48 &amp; 0 &amp; 1 \end{pmatrix}$$ $$\begin{pmatrix}1&amp; 0 &amp; 0 \\  25 &amp; 1 &amp; 0\\ 25 &amp; 0 &amp; 1 \end{pmatrix}$$ $$\begin{pmatrix}1&amp; 0 &amp; 0 \\  24 &amp; 1 &amp; 0\\ 24 &amp; 0 &amp; 1\end{pmatrix}$$   I am stuck on this problem. Can anyone help me please...............
53084
Let $A$ be $n \times n$ matrix and $A^2=A^*A$.  Why is $A$ a Hermitian matrix?
71598
We have to find $\det(A^{2}+A^{T})$. It is given that eigenvalues of $A$ are $1,2,3$.  My attempt: Since the question implicitly states that the answer would be same for all $A$ with eigenvalues $1,2,3$ we find that the answer is $(1+1)*(2+4)*(3+9)$ by taking $A$ as a diagonal matrix :P  I am unable to prove it for general $A$. Any help will be appreciated.
27051
Problem: We have matrix $A\in M_{n-1\times n}(\mathbb Z)$ so that the sum of elements in each row is zero. Prove that $\det(AA^T)=nk^2$, where $k\in \mathbb Z$.  What have I considered so far:   First I thought, since sum of all elements in each row is zero, zero is eigenvalue of $A$ but $A\in M_{(n-1)\times n}(\mathbb Z)$ confused me. I see that $AA^T$ will be $(n-1)$ by $(n-1)$, so I tried calculating $AA^T$ but I failed to see any connection.   I did some research and found  question but we had no mention of it, which makes me believe I am on the wrong track.  Thank you all for your help.
2966
How to find the eigenvalues of $$\begin{bmatrix} 0 &amp; 1   &amp;   &amp;  &amp;\\ k &amp; 0   &amp; 2 &amp;  &amp;\\   &amp; k-1 &amp; 0 &amp; 3 &amp;\\   &amp;     &amp;\ddots&amp;\ddots&amp;\ddots\\   &amp;     &amp;      &amp;2 &amp; 0 &amp;k \\   &amp;     &amp;      &amp; &amp; 1 &amp;0 \end{bmatrix}$$ I tried recurrence equation, but it doesn't work to find the characteristic polynomial. Any hint or solution are welcomed, thanks for your help!  PS:The missing parts are all zeroes.
78808
Let matrix $A_{n\times n}$, be such that there exists a matrix $B$ for which $$AB-BA=A$$  Prove or disprove that there exists $m\in \mathbb N^{+}$such  $$A^m=0,$$  I know $$tr(A)=tr(AB)-tr(BA)=0$$ then I can't.Thank you
64706
Let $A$ be a $n\times n$ matrix with entries $a_{ij}=i+j $ . Calculate rank of $A$.  My work : I noticed that A is symmetric . Hence all of its eigen vectors are real .. That is all i have got .   Your help will be highly appreciated .Thank you .
77384
The matrix is   $$\begin{pmatrix} 0&amp;2&amp;2\\ 2&amp;0&amp;2\\ 2&amp;2&amp;0 \end{pmatrix} $$  I have calculated the equation for eigenvalues, which is $\lambda^3-4\lambda-16=0$, but how do I solve this equation? Can this matrix be diagonalized?
8575
$A$ is an $n\times n$ matrix of complex numbers. Prove that if $\lambda$ is an eigenvalue of $A^2,$ then $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A.$    If $\lambda$ is an eigenvalue of $A^2,$ we have $\lambda\alpha=A^2\alpha$ for some $\alpha.$ Then how can we find a $\beta$ s.t. $\sqrt{\lambda}\beta=A\beta?$
123000
What would be the eigenvalues for a Householder matrix defined as: $H = I - 2 u u^T$? Can someone explain it  to me intuitively or with a simple proof?
45885
Suppose $A=uv^T$ where $u$ and $v$ are non-zero column vectors in ${\mathbb R}^n$, $n\geq 3$. $\lambda=0$ is an eigenvalue of $A$ since $A$ is not of full rank. $\lambda=v^Tu$ is also an eigenvalue of $A$ since $$Au = (uv^T)u=u(v^Tu)=(v^Tu)u.$$  Here is my question:     Are there any other eigenvalues of $A$?   Added:  Thanks to Didier's comment and anon's answer, $A$ can not have other eigenvalues than $0$ and $v^Tu$. I would like to update the question:     Can $A$ be diagonalizable?
115239
The matrix $A$ below is a block diagonal matrix where each block $A_i$ is a  $4 \times 4$ matrix with known eigenvalues.   $$A= \begin{pmatrix}A_1 &amp; 0 &amp; \cdots &amp; 0 \\ 0 &amp; A_2 &amp; \cdots &amp; 0 \\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ 0 &amp; 0 &amp; \cdots &amp; A_n  \end{pmatrix}$$  How do I find the eigenvalues of the block diagonal matrix $A$? Does this mean that I will have $4 n$ eigenvalues?   Am I correct in thinking that the eigenvalues of the block diagonal matrix $A$ above are just a list of the individual eigenvalues of each $A_i$ and not the product of everything?
21751
If $A$ and $B$ are $3\times 3$ matrices and $A$ is invertible, then can we say that there exists an integer $n$ such that $A+nB$ invertible?  I was trying to show this by choosing $n$ such that eignevalues of $A+nB$ are non-zero. In the case where $B = I$ we can find the eigenvalues of $A+nB$ that would be $\lambda + nB$ (though I am not certain about its proof). This choosing of $n$ such that $\lambda$ is not equal to $-n$ times an eigenvalue of $B$ will serve the purpose. But I am not sure about general $B$. What if I take arbitrary matrices $A$ and $B$.