I have to show that given the matrix A equal to:  $$\pmatrix{a_{11} &amp; \dots &amp; a_{1n} \\ \vdots &amp; \ddots &amp; \vdots \\ a_{n1} &amp; \dots &amp; a_{nn}}$$  Where:  $$ \sum_{j =1}^{n} a_{ij} = u, \forall i \in \{1, \dots, n\} $$  Then $ u $ is a eigenvalue, and then get his corresponding eigenvector....  I tried to show that checking if $ A \vec{v} = u \vec{v} $ or $ \det{(A - uI)} = 0 $, but I couldn't conclude anything....  How can I conclude that $ u $ is a eigenvalue?
30963
Let $A$ be an $n \times n$ matrix.  $i)$Prove that if the sum of each row of $A$ equals $s$, then $s$ is an eigenvalue of $A$.  $ii)$Prove that if the sum of each column of $A$ equals $s$, then $s$ is an eigenvalue of $A$.  I think that being an eigenvalue of $A$ implies that $sv=Av$ for some vector $v$. Furthermore, I know that $[a_i] = s$ if we let $a_i$ denote the i-th row of $A$. However, I do not seem to be able to find a link between these two facts. Could anyone please help me out?
94189
Suppose $A$ is a $n \times n$ constant matrix. How can I prove $\det(e^A) = e^{\displaystyle \sum_{\lambda_i\in\sigma(A)} \lambda_i}$,  where $\sigma(A)$ is the multiset of eigenvalues of $A$?  The following matlab code shows this is true in $n = 3$ case:  A = rand(3) detA = exp(sum(eig(A))) detmA = det(expm(A))
73571
I just wanted to make sure that the following statement is true:  Let $A$ be a normal matrix with eigenvalues $\lambda_1,...,\lambda_n$ and eigenvectors $v_1,...,v_n$. Then $A^*$ has the same eigenvectors with eigenvalues $\bar{\lambda_1},..,\bar{\lambda_n}$, correct?
104885
Let $A = (a_{ij}) \in R^{n\times n}$. Find the determinant if:  $$a_{ij}= |i-j|$$  So we have the symmetric matrix \begin{bmatrix} 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; \dots &amp; n-1 \\ 1 &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; \dots &amp; n-2 \\ 2 &amp; 1 &amp; 0 &amp; 1 &amp; 2 &amp; \dots &amp; n-3 \\ \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ n-1 &amp; n-2 &amp; n-3 &amp; \dots &amp; \dots &amp; \dots &amp; 0 \end{bmatrix}  But i can't find a way to diagonalize the matrix nor find the determinant by Laplace expansion.  Any ideas ??
48334
So $Ax=\lambda x$ where $A$ is the matrix, $\lambda$ and $x$ its eigenvalue and eigenvector resptectively.  Now I have another matrix similar to $A$ i.e. $B=TAT^{-1}$  I'll let a vector $y$ which is an eigenvector of $B$, i.e. $By=\mu y$ where $\mu$ is its ($B$'s) eigenvalue.  I have to now prove that  $By=\mu x$.  Some maths leads me to  $By = TAT^{-1}y$ but here I'm stuck.  answers wants me to assume $y=Tx$ which will essentially solve it but why/how is $y=Tx$?
12774
If the $\det(A)=0$ why does the matrix $A$ have an eigenvector? Explain why there is a basis $B$ in $R^n$ so that the matrix $[A]_B$ has the zero vector as its first column   I know that if the $\det(A)=0$ of a matrix is zero then it is singular and thus not invertible. That in turn means that it doesn't have one solution. Why does that mean that it has an eigenvector?  As for the second I have no idea.
6236
Let A = $\begin{bmatrix}1 &amp; 1 &amp; 4\\0 &amp; 3 &amp; -4\\0&amp;0&amp;-1\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? Determine the eigendecomposition $A = PΛP^{-1}$. A = $\begin{bmatrix}3 &amp; 1 &amp; 0\\1 &amp; 2 &amp;1\\0&amp;1&amp;3\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that orthogonally diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? (Note that A is real symmetric so that you do not have to compute the inverse of P).   My solution attempts:   det(λI - A) = $\begin{vmatrix}λ - 1 &amp; -1 &amp; -4\\0 &amp; λ - 3 &amp;4\\0&amp;0&amp;λ + 1\end{vmatrix}$ = (λ -1) (λ -3) (λ + 1) = 0   $λ_{1}$ = 1, $λ_{2}$ = 3, $λ_{3}$ = -1  ($λ_{1}I - A)x_{1}$ = 0  $\begin{bmatrix}0 &amp; -1 &amp; -4\\0 &amp; -2 &amp;4\\0&amp;0&amp;2\end{bmatrix}$ $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $\begin{bmatrix}0\\0 \\0\end{bmatrix}$   $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{11} \begin{bmatrix}1\\0 \\0\end{bmatrix}$  It follows that $\begin{bmatrix}1\\0 \\0\end{bmatrix}$ is a basis for the eigenspace of A corresponding to $λ_{1}$ = 1.  In a similar way, I found other two basis vectors as $\begin{bmatrix}1/2\\1 \\0\end{bmatrix}$ and $\begin{bmatrix}-2/5\\1 \\1\end{bmatrix}$, and I got $P = \begin{bmatrix}1 &amp; 1/2 &amp; -2/5\\0 &amp; 1 &amp; 1\\0&amp;0&amp;1\end{bmatrix}$. Then my $P^{-1} = \begin{bmatrix}1 &amp; -1/2 &amp; 9/10\\0 &amp; 1 &amp; -1\\0&amp;0&amp;1\end{bmatrix}$. And I said A's left-multiplication by $P^{-1}$ and right-multiplication by $P (P^{-1}AP)$ gives the diagonal matrix Λ that is similar to A. I found Λ as $\begin{bmatrix}1 &amp; 0 &amp; 0\\0 &amp; 3 &amp; 0\\0&amp;0&amp;21/5\end{bmatrix}$. Then I said that if we multiply both sides of the equation $P^{-1}AP = Λ$ by $P$ from the left and by $P^{-1}$ from the right we get $PP^{-1}APP^{-1} = PΛP^{-1} = A$ and that this is the eigendecomposition of A. Here is my first question: Do you think I understood and did correctly what the asker wanted me to do by saying "Determine the eigendecomposition $A = PΛP^{-1}$."?   I started in the same manner and I got three eigenvectors as  $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{12} \begin{bmatrix}0\\1 \\0\end{bmatrix}$,  $\begin{bmatrix}x_{21}\\x_{22}\\x_{23}\end{bmatrix}$ = $x_{21} \begin{bmatrix}1\\1 \\1\end{bmatrix}$ and  $\begin{bmatrix}x_{31}\\x_{32}\\x_{33}\end{bmatrix}$ = $x_{33} \begin{bmatrix}1\\-2 \\1\end{bmatrix}$.   I said A is not diagonalizable since the eigenvectors are not linearly independent.  Here is my second question: What the hell does the sentence "Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors?" mean?
110845
Let 'A' be a $2 \times 3$ matrix where as B be a $3 \times 2$ matrix if $\det(AB) = 4$ the find value of the $\det(BA)$  My attempt: I took A =  $$         \begin{bmatrix}         2 &amp; 0  &amp;0\\         0 &amp; 0  &amp;2\\         \end{bmatrix} $$  B= $$         \begin{bmatrix}         1 &amp; 0  \\         0 &amp; 0  \\         0 &amp; 1  \\         \end{bmatrix} $$  It satisfies given condition and I get $\det(BA)=0$   But I have not proved it  How do I prove that it is always zero (background)I am 12th grader and I know about adjoint,inverse,determinant,rank of a matrix and the other basics.  However I do NOT know about eigenvalues and eigenvectors.
52271
I have stumbled across a very old exam question from my linear algebra course and the solutions are not available. I was wondering if my working/logic is correct and if any improvements can be made.     Let  $\ B=\begin{pmatrix}     1 &amp; -1 &amp; 4 \\     6 &amp; -7 &amp; 2 \\     -3 &amp; 1 &amp; -6       \end{pmatrix}\ $ and $\ \vec{v_1}=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.      Part (i): Calculate $(B+5I)\vec{v_1}$ and $(B+5I)^2\vec{v_1}$.   Using simply matrix multiplication, I calculated $(B+5I)\vec{v_1}=\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $(B+5I)^2\vec{v_1}=\vec{0}$.     Part (ii): Find all eigenvalues and eigenvectors of B.   From our calculations in part (i), we can see that $\ \lambda=-5,-5$ are two of the eigenvalues, with corresponding eigenvector $\ \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and generalised eigenvector $\ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. From the trace of B, the remaining eigenvalue is $\lambda=-2$ with corresponding eigenvector $\begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}$.     Part (iii): Giving reasons, write down a basis for the generalised eigenspace $GE_{-5}$ of B.   I am unsure of this part of the question. If I had to guess, I would say a basis for $GE_{-5}=\left\{\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\right\}$. Is this correct?     Part (iv): Solve the initial value problem $\vec{y'}=B\vec{y}$ where $\ B=\begin{pmatrix}     1 &amp; -1 &amp; 4 \\     6 &amp; -7 &amp; 2 \\     -3 &amp; 1 &amp; -6       \end{pmatrix}\ $ and $\vec{y}(0)=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.   Well, the solution to this IVP will take the form $\vec{y}=e^{tB}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. I noticed that $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ was a generalised eigenvector and hence we can write $\vec{y}=e^{-5t}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}+e^{-5t}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}t$. Thus $\vec{y}=e^{-5t}\begin{pmatrix} 1+t \\ 1+2t \\ -1-t \end{pmatrix}$
55372
Given a matrix $A=(a_{ij})_{n\times n}$, let $C_{i,j}$ be the cofactor in position $(i,j)$. By the determinant formula, we have  $$\det A=\sum_{i=1}^n a_{i,1}C_{i,1}.$$  What about if we take a different column for the cofactors, that is  $$\sum_{i=1}^n a_{i,1}C_{i,2}$$  Must this evaluate to zero?
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.
71234
I have a question to ask down below, that I have been having some trouble with and would like some help and clarification on.  Suppose A  is an $n \times n$ matrix with (not necessarily distinct) eigenvalues $\lambda_{1}, \lambda_{2}, \ldots, \lambda_{n}$. Can it be shown that:     (a) The sum of the main diagonal entries of A, called the trace of A, equals the sum of the eigenvalues of A.   (b) A $- ~ k$ I has the eigenvalues $\lambda_{1}-k, \lambda_{2}-k, \ldots, \lambda_{n}-k$ and the same eigenvectors as A.  Thank You very much.
2421
As a part of an exercise I have to prove the following:  Let $A$ be an $(n \times m)$ matrix. Let $A^T$ be the transposed matrix of $A$. Then $AA^T$ is an $(n \times n)$ matrix and $A^TA$ is an $(m \times m)$ matrix. $AA^T$ then has a total of $n$ eigenvalues and $A^TA$ has a total of $m$ eigenvalues.  What I need to prove is the following:  $AA^T$ has an eigenvalue $\mu \not = 0$ $\Longleftrightarrow$ $A^TA$ has an eigenvalue $\mu \not = 0$  In other words, they have the same non-zero eigenvalues, and if one has more eigenvalues than the other, then these are all equal to $0$.  How can I prove this?  Thanks and regards.
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?
45709
We were asked to prove the following:  Let $ A $ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.   It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.  Thanks
64151
Let $A$ be an $n\times n$ matrix, with $A_{ij}=i+j$.  Find the eigenvalues of $A$.  A student that I tutored asked me this question, and beyond working out that there are 2 nonzero eigenvalues $a+\sqrt{b}$ and $a-\sqrt{b}$ and $0$ with multiplicity $n-2$, I'm at a bit of a loss.
99495
So if I have an $n\times n$ matrix $A$ that is diagonalizable then how do I determine $\mbox{tr}(A)$ and  $\det(A)$ as functions of the eigenvalues of the matrix?   For  $\det(A)$, I know the formula for eigenvalues is $\det(A−tI)$ which would be  $$(-1)^n(t-\lambda_1)\dots (t-\lambda_n)$$ where $\lambda_i$ are the eigenvalues of $A$.   For $\mbox{tr}(A)$ could I also do something similar to that? I'm just not sure how this question works.