This is a multipart linear algebra textbook question. Let $A, P_1, P_2, N_1, N_2$ be $n$ × $n$ matrices with real entries such that $I_n$ = $(t I_n − A) \big(\frac{N_1}{(t-1)^3} + \frac{N_2}{(t-1)^2} + \frac{P_1}{(t-1)} + \frac{P_2}{(t-2)}\big)$ Then, I'm asked to show the following:  If $\lambda$ is an eigenvalues of $A$, then λ ∈ {1, 2}.  I'm not really sure what my first steps here should be. Given the initial equation, I don't really know what matrix algebra I can apply. I'm wondering first of all how I can even take the determinant of both sides to connect it to eigenvalues Any assistance on this question would be much appreciated!
123550
Thanks everyone for viewing my question! I'm getting stucked on this problem: Suppose $A,B,C,D,E$ are real $n\times n$ matrices such that for any real number $t\neq 1,2$: $$ I_n=(tI_n-A)(\frac{B}{(t-1)^3}+\frac{C}{(t-1)^2}+\frac{D}{t-1}+\frac{E}{t-2}) $$ The problem asks me to prove $C=(A-I)D=D(A-I)$. My progress: Multiply $t-1$ to both sides we obtain: $$ (t-1)I_n=(tI_n-A)(\frac{B}{(t-1)^2}+\frac{C}{t-1}+D+\frac{(t-1)E}{t-2}) $$ Then take limits when $t\rightarrow 1$: $$0=\lim_{t\rightarrow1}(I_n-A)(\frac{B}{(t-1)^2}+\frac{C}{t-1}+D) \\(A-I_n)D=\lim_{t\rightarrow1}(I_n-A)(\frac{B}{(t-1)^2}+\frac{C}{t-1}) $$ This is really close to the required form but I do not know how to continue, I saw some answer concluding that: $$ \lim_{t\rightarrow1}(I_n-A)(\frac{B}{(t-1)^2}+\frac{C}{t-1})=C $$ I don't know if this is correct and it makes no sense to me. Thanks so much for your answers in advance!
28073
Let $$M = \begin{bmatrix} -7 &amp; 8 \\ -8 &amp; -7 \end{bmatrix}.$$ Find formulas for the entries of $M^n$ where $n$ is a positive integer.  (Your formulas should not contain complex numbers.)  Your answer should be in the form of a matrix.  I diagonalized to the form $M = P D P^{-1}$  and $M^n = P D^n P^{-1}$ where $P$ is my matrix of eigenvectors and $D$ is my matrix of eigenvalues.  My final answer after diagonalization was $$M^n = \begin{bmatrix} .5((-7+8i)^n+(-7-8i)^n) &amp; (i/2)(-(-7+8i)^n+(-7-8i)^n) \\ (.5/i)(-(-7+8i)^n+(-7-8i)^n) &amp; .5((-7+8i)^n+(-7-8i)^n) \end{bmatrix}$$  I can't seem to find an answer not in terms of complex numbers.  Can someone show me what I'm missing?
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 .
32992
I'm self studying Steven Roman's Advanced Linear Algebra, and this is problem 10 of Chapter 8.      Let $A,B\in M_2(\mathbb{C})$, $A^2=B^3=I$, $ABA=B^{-1}$, but $A\neq I$ and $B\neq I$. If $C\in M_2(\mathbb{C})$ commutes with $A$ and $B$, then $C=rI$ for some $r\in\mathbb{C}$.   Is there a way to solve this without writing out arbitrary matrices and attempting to solve a huge system of equations? The only thing I observe is that $A=A^{-1}$, so $B\sim B^{-1}$, so $B$ and $B^{-1}$ have the same characteristic polynomial. I'm stymied trying to show $C$ is diagonal, let alone a multiple of $I$. Thanks for any ideas.  I should add that I know that the center $Z(M_n(\mathbb{C}))$ consists of scalar multiples of $I$, but I don't see any reason to assume or prove $C$ commutes with everything.
11069
\begin{pmatrix} 2na &amp; -a &amp; -a &amp; -a &amp; -a &amp;  -a&amp; -a\\   -a&amp; a+b &amp; 0 &amp; 0 &amp; -b &amp; 0 &amp; 0\\   -a&amp; 0 &amp; a+b &amp; 0 &amp; 0 &amp; -b &amp;0 \\   -a&amp; 0 &amp; 0 &amp; a+b &amp; 0 &amp;  0&amp;-b \\   -a&amp; -b &amp; 0 &amp; 0 &amp; a+b &amp; 0 &amp; 0\\   -a&amp;  0&amp;-b  &amp; 0 &amp; 0 &amp; a+b &amp;0 \\   -a&amp;  0&amp;  0&amp;-b  &amp; 0 &amp; 0 &amp; a+b \end{pmatrix}   I represent the matrix when $n=3$. There are block matrices $(a+b)I_3$ and $(-b)I_3$ above. I think the characteristic polynomial of this matrix for any natural number $n$ is calculated efficiently.  The rank is $2n$, so the characteristic polynomial should zero as the constant part. But I cannot calculate the determinant and I can't get zero when I put $t=0$ when I calculate $\text{char}(M)(t)$ for any $n$. Please give me a favor.
121270
$$B = \begin{bmatrix} 0 &amp; A^* \\ A&amp;0 \end{bmatrix}$$  I think that $\det(B) = \det(A) * \det(A^*)$ and probably eigen values just get squared. What is the right answer?  EDIT: $\operatorname{rank}(B) = 2 * \operatorname{rank}(A)$, so there are twice more eigenvalue or their multiples. And the additional eigenvalues are just negatives of the eigenvalues of $A$ (Thanks to Simon).  Still can't figure out.
86559
I would be very grateful if someone can help me with the following assignment:  I'm given an $n\times n$ matrix $A$.  If $A \neq I,0$ and $A = A^2$, I need to prove that $\lambda=0$ and $\lambda=1$ are $A$'s eigenvalues and that they are $A$'s only eigenvalues.
10950
I have the following $n\times n$ matrix:  $$A=\begin{bmatrix} a &amp; b &amp; \ldots &amp; b\\ b &amp; a &amp; \ldots &amp; b\\ \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\ b &amp; b &amp; \ldots &amp; a\end{bmatrix}$$  where $0 &lt; b &lt; a$.  I am interested in the expression for the determinant $\det[A]$ in terms of $a$, $b$ and $n$.  This seems like a trivial problem, as the matrix $A$ has such a nice structure, but my linear algebra skills are pretty rusty and I can't figure it out.  Any help would be appreciated.
61681
Let $Z_n$ be the set of all $n \times n$ matrices that commute with all $n \times n $ matrices. Show that $$Z_n = \{\lambda I_n \ | \  \lambda \in \mathbb R\}$$  ($I_n$ is the $n \times n$ identity matrix)  I don't know how to use $E_{ij}$ (matrix with $1$ in $(i,j)$ and $0$ elsewhere) and the elementary matrix $P_{ij}$ to prove this question. Can anyone explain it please?
42734
Let $A$ be an $n \times n$ matrix and let $\Lambda$ be an $n \times n$ diagonal matrix. Is it always the case that $A\Lambda = \Lambda A$? If not, when is it the case that $A \Lambda = \Lambda A$?  If we restrict the diagonal entries of $\Lambda$ to being the equal (i.e. $\Lambda = \text{drag}(a, a, \dots, a)$), then it is clear that $A\Lambda = AaI = aIA = \Lambda A$. However, I can't seem to come up with an argument for the general case.
79300
The question is motivated from the following problem:     Let $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times 2$ matrix. If $A=A^{-1}$, then the trace of $A$ is   $$ (A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2$$   Since $A=A^{-1}$, $A^2=I$. If the converse of  is true, then $\lambda^2=1$ and thus $\lambda=\pm1$. And then $\rm{trace}(A)=1+(-1)=0$.   Here are my questions:   Is  $C$ the answer to the quoted problem?  Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix $A$, if $p(A)=0$, then $p(\lambda)$ is the characteristic polynomial of the matrix $A$" true? If it is not, then what's the right method to solve the problem above?
34553
Title looks a little bit twisted. What I want to say is the following:  $A\in\mathbb{R}^{n\times n}$, polynomial of matrix $A$: $P(A)=\displaystyle \sum_{k=0}^{n} c_k A^k$. $\lambda(A)$ is the set of eigenvalues of $A$.   So, we want to prove $\lambda(P(A))=P(\lambda(A))$  So far, I can prove it when $A$ is diagonalisable, but when $A$ is not diagonalisable, it seems we can't just simply use Jordan normal form to prove it. (Or this statement won't hold when it's not diagonalisable?)  Can anyone suggest me something and help me out? Thanks!
101417
Can anyone help me to understand what is suppose to do in this question?  "A matrix that is similar to the identity matrix" I should say something about this but I do not understand what is meant to do.  I know that two n x n matrices A and B are similar if $B=P^{-1}AP $.   They also have some properties such as:  The same:  Rank; Characteristic equation;  Determinant; Trace; Eigenvalues etc.   However I do not understand the question and consequently I do not know how to start.  Can anyone help me on this?  Thanks
18758
Question:     Let $A$ be a $3 \times 3$ Matrix such that $[-3,4,1]$   is the eigenvector corresponding to eigenvalue $3$, and $[6,-3,2]$ is an eigenvector corresponding to the eigenvalue $2$. If $v$ = $[0,5,4]$, compute $A^{2}v$.      Hint: $[6,-3,2]+2[-3,4,1]$ = $v$   Where I'm stuck:  I know that we can  solve A with the equation:   A = $P^{-1}DP$, where $P$ is the matrix consisting of the eigenvectors and $D$ is the matrix consisting of the eigenvalues.  However, since $A$ is a $3$ x $3$, we should have a $3 \times 3$ matrix $D$ and $ 3 \times 3 $ matrix $P$ as well, consisting of the eigenvalues and eigenvectors respectively - and in this case, we only have two eigenvalues and two eigenvectors. Where can I obtain the other one of each?  Any help will be very appreciated.  Thank you.
87268
Our teacher didn't explain us how to find it so I've had to look up a bit by myself.  I have this matrix :  $$A = \begin{pmatrix} 9 &amp; 4 &amp; 5 \\ -4 &amp; 0 &amp; -3 \\ -6 &amp; -4 &amp; -2 \end{pmatrix}$$  I've found its characteristic polynomial $p_A(\lambda) = -(\lambda -3)(\lambda - 2)^2$.  I've found its eigenspaces : $E_2 = span(\begin{pmatrix}-2 &amp; 1 &amp; 2\end{pmatrix}^T)$, $E_3 = span(\begin{pmatrix}-3 &amp; 2 &amp; 2\end{pmatrix}^T)$.  And I've found the generalized eigenspace of $\lambda = 2$ :  $\ker((A-2I_3)^2) = span(\begin{pmatrix}0 &amp; 1 &amp; 0\end{pmatrix}^T, \begin{pmatrix}1 &amp; 0 &amp; -1\end{pmatrix}^T)$  But I don't know what to do now. I've tried to see what result I would get with $$M = \begin{pmatrix} -3 &amp; -2 &amp; 0 \\ 2 &amp; 1 &amp; 1 \\ 2 &amp; 2 &amp; 0 \end{pmatrix}.$$  But $$M^{-1}AM = \begin{pmatrix} 3 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; -2 \\ 0 &amp; 0 &amp; 2 \end{pmatrix}.$$  So there it is. Any help ?
97035
Given the matrices $A,B,C\in M_n(\mathbb R)$ which satisfy the condiditions:  $A+B+C+\lambda ΑΒ=0 \quad(1)$  $Α+Β+C+\lambda BC=0 \quad(2)$  $A+B+C+\lambda CA=0 \quad(3)$  for some $\lambda\neq0$.  (a) Prove that $I+\lambda Α$, $Ι+\lambda Β$, $I+\lambda C$ are invertible and $AB=BC=CA$.  (b) Prove that $A=B=C$.  For (a) i understand that the matrices are invertible, because we have non zero eigenvalue. In addition we have AB=BC=CA, because the (1)-(2) and the (2)-(3) give us these equalities. I have been stucked in (b). I tried to use the equality $AB=BC=CA$ and the fact that $I+\lambda Α,Ι+\lambda Β,Ι+\lambda C$ are invertible, but i can't prove that $A=B=C$. I thought that A, B and C are null matrix, but i don't find a way to prove that. I want help for (b).
94994
Let $M$ be a (real) $n \times n$ matrix. For $1 \leq i, j \leq n$ we denote by $M_{ij}$ the $(n-1) \times (n-1)$ matrix that we get when the $i$th row and $j$th column of $M$ are removed. Now, consider fixed $i$ and $j$ with $i\neq j$. Let $N$ be the $(n-2) \times (n-2)$ matrix that we get when removing both the $i$th and $j$th row and the $i$th and $j$th column from $M$. Then the following identity holds: $$ \det M \det N = \det M_{ii}\det M_{jj} - \det M_{ij} \det M_{ji}. $$  We were able to prove this looking at all different terms that can occur on both sides when evaluating the determinant as a sum of $n!$ terms (for instance, terms containing $a_{ji}a_{ij}$ are counted both $1$ time in the LHS and $1$ time in the RHS).  We are looking for a slick proof that does not involve writing out the determinant.  Any suggestions or approaches to this problem are appreciated!