Question: Let $\mathbb{F}$ a commutative field and $ a,b,c,d,e,f\in\mathbb{F} $ be scalars and suppose that $ A $ and $ B $ are the following matrices:  $$ A = \begin{bmatrix} a &amp; b \\ 0 &amp; c \\ \end{bmatrix} $$ and $$ B = \begin{bmatrix} d &amp; e \\ 0 &amp; f \\ \end{bmatrix} $$   Prove that $ AB = BA $ if and only if  $ \det\begin{pmatrix} \begin{bmatrix} b &amp; a - c \\ e &amp; d - f \\ \end{bmatrix} \end{pmatrix} = 0 $   My steps:  $$ \det\begin{pmatrix} \begin{bmatrix} b &amp; a - c \\ e &amp; d - f \\ \end{bmatrix} \end{pmatrix} = 0 $$ $$ b(d - f) - e(a - c) = 0 $$ $$ bd - bf - ea + ec = 0 $$ $$ bd - bf = ea - ec $$ At this point I considered that one of the following things must be true for the solution to hold: $$ b = e = 0 $$ which would mean that $ A $ and $ B $ are diagonal matrices meaning $ AB = BA $   Another condition for the solution to hold would be $$ b = e, a = d, f = c $$ which would mean that $ A = B $ hence $ AB = BA $ must be true.  Would this be the right approach to the problem or is there a point that I am missing?
122569
Show that the matrices    $$A=\begin{bmatrix} a &amp; b \\ 0 &amp; c \end{bmatrix} \text{and B = $\begin{bmatrix} d &amp; e \\ 0 &amp; f \end{bmatrix}$}$$    commute if and only if    $$\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0.$$     My approach:  For $AB=BA$, we must have  $AB=\begin{bmatrix} ad &amp; ae+bf \\\ 0 &amp; cf\end{bmatrix}$ and $BA=\begin{bmatrix} ad &amp; bd+ce \\\ 0 &amp; cf\end{bmatrix}$ where $ae+bf=bd+ce$. We need to show that if $ae+bf=bd+ce$ then $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0.$   Since $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0$, we see that: $$b(d-f)-e(a-c)=0 \iff bd-bf-ae+ce=0 \iff bd+ce=ae+bf.$$   Thus, if  $AB=BA$ then $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0.$   (Showing the converse)  If $\begin{vmatrix} b &amp; a-c \\ e &amp; d-f\end{vmatrix}=0$ then we have $bd+ce=ae+bf$ and $AB=BA$ if and only if $ae+bf=bd+ce$, which we have.    Is this a proper approach am I missing something here?
85458
So the question asks: Let A be a $2\times 2$ matrix, and suppose that there exist scalars $λ_1,λ_2$ in$\mathbb{R}$ and nonzero vectors $v_1,v_2$ in $\mathbb{R}$ such that $Av_1=λ_1v_1, Av_2=λ_2v_2$, and $λ_1$ does not equal to $λ_2$.Prove that the vectors $v_1,v_2$ form a basis of $R^2$.   So so far I got:   As the eigenvector $v_1,v_2$ are corresponding to two different eigen values$λ_1,λ_2$ respectfully.   Therefore the eigenvectors $v_1,v_2$ are linearly independent.   And since $v_1,v_2$ spans A in $\mathbb{R}$.   Hence,  $v_1,v_2$ forms a basis of $\mathbb{R}$.   Does this proof look right? Do I need to write a linear combination of $v_1,v_2$ to show they span A?
49764
The Problem Let $A$ be the matrix $\bigl(\begin{smallmatrix}a&amp;b\\c&amp;d\end{smallmatrix} \bigr)$, where no one of $a,b,c,d$ is $0$. Let $B$ be a $2\times 2$ matrix such that $AB+BA=\bigl(\begin{smallmatrix} 0&amp;0\\ 0&amp;0 \end{smallmatrix} \bigr)$. Show that either   $a+d=0$, in which case the general solution for $B$ depends on 2 parameters, or $ad-bc=0$, in which case the general solution for $B$ depends on one parameter.   (this is question 22 of the last matrix exercise of Further Pure Mathematics by Bostock et al.)  Comments Writing $B=\bigl(\begin{smallmatrix} e&amp;f\\ g&amp;h \end{smallmatrix} \bigr)$ and multiplying out I get that    $(a+d)(f+g)+(b+c)(e+h)=0$ $ae+bg+cf+dh=0$   but I am unable to get the required restrictions on $a,b,c,d$. Is there a quick way of doing the problem that doesn't require manual computation? I thought of considering invertible and non-invertible cases but couldn't get anywhere. Help would be much appreciated.
38731
If $A$ and $B$ are two $4\times3$ matrices such that $A\mathbb{v}=B\mathbb{v}$ for all $\mathbb{v}\in\mathbb{R}^3$, then $A=B$?  If it's not true, can you give me an example of it?  Thank you.
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?
85957
I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please. $$\left|\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3$$    Second question, what is the easiest way to argue that $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent in $\mathbb{Q}$?    Motivation: Prove that $\mathbb{Q}[\sqrt[3]{2}] = \{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field.   Proof: Since $\mathbb{Q}[\sqrt[3]{2}] \subset \mathbb{R}$, we prove $\mathbb{Q}[\sqrt[3]{2}]$ is a subfield of $(\mathbb{R},+,\cdot)$    $\forall (a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)\in \mathbb{Q}[\sqrt[3]{2}]\backslash\{0\}.$ We want to find $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)$ such that    $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)(d+e\sqrt[3]{2}+f(\sqrt[3]{2})^2) =$ $ (ad+2ec+2bf)+(ae+bd+2cf)\sqrt[3]{2}+(af+cd+be)(\sqrt[3]{2})^2 = 1$    Since $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent (?) over $\mathbb{Q}$, we show there is unique solution to: $$\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix} \cdot \left[\begin{array}{l l} d\\e\\f \end{array}\right] = \left[\begin{array}{l l} 1\\0\\0 \end{array}\right] $$ Which is equivalent in showing the determinant is non-zero $$\left|\begin{bmatrix} a &amp; 2c &amp; 2b\\b &amp; a &amp; 2c\\ c &amp; b &amp; a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3=(?)$$    By subfield test, 1)2)3)4) is enough to say that $(\mathbb{Q}[\sqrt[3]{2}],+,\cdot)$ is a subfield of $(\mathbb{R},+,\cdot)$ therefore a field.  EDIT: If you have shorter way that prove the proposition without touching my 2 questions, that is even better.
105685
This question came from . As the OP hasn't edited his question and I really want the answer, I'm adding my thoughts.      Let $A, B$ be two real $n\times n$ matrices that commute and $\det(A + B)\ge 0$.   Prove or disprove : $\forall k\in \mathbb{N}^*$   $$\det(A^k + B^k) \geq 0$$   What I think about it I think we can consider two cases : k even and k odd. For k even, the result is pretty easy considering $$\det(A^{2k}+B^{2k})=\det(A^{k}+iB^{k}){\det(A^{k}-iB^{k})}=\det(A^{k}+iB^{k})\overline {\det(A^{k}+iB^{k})}\ge 0$$ For k odd, I thought about this equality but I don't know if it can help $$A^{2k+1}+B^{2k+1}=(A+B)\left(\sum_{i=0}^{2k} (-1)^i A^{2k-i}B^{i}\right)$$ so $$\det(A^{2k+1}+B^{2k+1})\ge 0 \iff \det\left(\sum_{i=0}^{2k} (-1)^i A^{2k-i}B^{i}\right)\ge 0$$
33605
\begin{array}{rrrrr|r}     b &amp; a &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; b &amp; a &amp; \cdot \cdot \cdot &amp; a \\     a &amp; a &amp; b &amp; \cdot \cdot \cdot &amp; a \\     \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ \cdot &amp; \cdot &amp; \cdot &amp; \space &amp; \cdot\\ a &amp; a &amp; a &amp; \cdot \cdot  \cdot &amp; b   \end{array}  I have the above matrix $A\in M_{n\times n}(F)$ where $F$ is a field and $n\geq1$, $a,b\in F$.  I'm trying to find out how to use row operations to make it into an upper triangular matrix in order to figure out the determinant. But I'm not sure how I would approach it.
124076
Let $V$ be the vector space over $\mathbb{R}$ consisting of all polynomials of degree $\leqslant n$. If $f,g\in V$, let $$\langle f,g \rangle = \int_\limits{0}^{1}f(t)g(t)dt$$ Find the matrix of this scalar product with respect to the basis $\{1,t,\dots,t^n\}$.  What does it mean to find a matrix of the scalar product? Thanks in advance!
3798
Consider the transpose inverse automorphism on $GL_n(\mathbb F)$ where $n\geq2$ and $|\mathbb F|&gt;2$. (i.e. $\mathbb F$ is a field, possibly infinite, with three or more elements). I want to show this automorphism is not inner. I was told to consider $\det(BAB^{-1}) = \det(A)$ and $\det(\,^TA^{-1})=\det(A)^{-1}$ and derive a contradiction.  However, in some fields (where the result holds) I fail to see the contradiction. What about $\mathbb F_3$? An element of $GL_2(\mathbb F_3)$ either has determinant $1$ or $2$, both of which are their own multiplicative inverses in $\mathbb F_3$. I'm not sure if my knowledge of fields and determinants is flawed or the hint is.
32114
Let $k$ and $n$ be positive integers and let $F$ be a field. For matrices $A,B \in M_{k\times n} (F)$, show that the rank of $A+B$ is no more than the sum of the ranks of $A$ and $B$ I believe this question is addressed , but to be honest I don't quite understand the explanations given. Can someone possibly help with a more detailed explanation?
87274
In a course of group theory an exercise is left with no indications :      Let $A$ be an abelian group of finite type and $a_1,\dots ,a_n$ a $\mathbb{Z}$-basis of $A$. Let $f : A \to A$ be a group homomorphism. Les $M=(m_{ij})$ be the matrix (invertible) defined by $\displaystyle f(a_j)=\sum_{i=1}^n m_{ij}a_i$ with $\det (M)\neq 0$. Show that $[A:\operatorname{Im}(f)]=|\det M |$, i.e the index of $\operatorname{Im}(f)$ in $A$ as a subgroup is precisely $\det (M)$.   I am not able to prove this fact, and need a bit of help. In fact I do not see where I can make $\det (M)$ appear. Can you give me some hints ?   Thank you
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.
67342
Let $b\in\mathbb{F}$ then {$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$:$x_3=5x_4+b$} is a subspace iff $b=0$. This is from axler's Linear algebra done right. $\mathbb{F}$ stands for a field either the real numbers or complex numbers. I proved the reverse direction but now I have to show the foward direction. This is what I have so far.      ($\rightarrow$) Assume that $b\in\mathbb{F}$ and let T={$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$: $x_3=5x_4+b$} where $T$ is a subspace of $\mathbb{F}^4$. Suppose not that $b\neq 0$. Since  $T$ is a subspace by assumption that means that $0\in T$ That is $(x_1,x_2,x_3,x_4)=(0,0,0,0)$ This means that $x_3=0$ and $x_4=0$. If $x_3=0$ then we have $0=5x_3+b \implies \frac{5}{b}=x_4$ but since $b\neq 0$ we have that $x_4\neq 0$. Suppose that $x_4=0$. Again we have $x_3=5(0)+b=b$ Since $b\neq 0$ this means $x_3\neq 0$. Thus a contradiction. Would this be right?
80341
Suppose $A$ and $B$ are two commuting square matrices over an algebraically closed field. It is true that the spectrum of $A+B$ is contained in the set $\{\lambda_1+\lambda_2:\lambda_1 \in \sigma(A), \lambda_2 \in \sigma(B)\}$ where $\sigma(A)$ denotes the spectrum of $A$?  If so, can you provide a reference for this? can more be said?  Thanks
10907
So here's one that I can't quite crack:     Let $A\in M_n(\mathbb{F})$ be an invertible matrix with integer eigenvalues. Its minimal   polynomial is $m_A(\lambda)=\lambda^k+b_{k-1}\lambda^{k-1}+...+b_0$. Prove that $detA$ is divisible by $b_0$.   I'm afraid I don't quite see the connection between the minimal polynomial and the determinant. All I know is that for an invertible matrix $b_0 \neq 0$, but I can't find a way to relate that to the determinant.  Any tips and guidance would be much appreciated! Thanks in advance.
60640
Let $\mathbf A, \mathbf B$ be Hermitian matrices of the same size. What is the characterization of $\mathbf A, \mathbf B$ such that $p(\lambda)=\det(\mathbf A-\lambda\mathbf B)=0$ has only real roots?   If $\mathbf B$ is positive definite (I corrected this), it is easy to see $p(\lambda)$ has only real roots.