I saw the      Let $\mathbf{A}$ be a $n\times n$-matrix, and $p(\lambda)=\det(\lambda \mathbf{I}_n-\mathbf{A})$ the characteristic polynomial of $\mathbf{A}$. Then $p(\mathbf{A})=\mathbf{0}$   and I thought it was trivial since $\det(\mathbf{A}-\mathbf{A})=0$. I checked with  and learned that this didn't work because $\lambda$ is a scalar, and $\det(\mathbf{0})$ is also a scalar, and not a matrix. This brings me to my question:  Are there any other results that seem trivial, but aren't?
30009
In , I asked about proof of statements which are simple but incorrect.   Here, I ask about statements which seems, at a first glance, straightforward, but if we try to write a proof, we can see it's much harder than it looked. So I expect the answers to contain:   the statement; why it looks easy to prove; why actually it isn't.
39836
Let $f(x)=(p_1-x)\cdots (p_n-x)$ $p_1,\ldots, p_n\in \mathbb R$ and let $a,b\in \mathbb R$ such that $a\neq b$  Prove that $\det A={bf(a)-af(b)\over b-a}$ where $A$ is the matrix:  $$\begin{pmatrix}p_1 &amp; a &amp; a &amp; \cdots &amp; a \\ b &amp; p_2 &amp; a &amp; \cdots &amp; a \\ \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\ b &amp; b &amp; b &amp; \cdots &amp; p_n \end{pmatrix}$$  that is the entries $k_{ij}=a$ if $i&lt;j$, $k_{ij}=p_i$ if $i=j$ and $k_{ij}=b$ if $i&gt;j$  I tried to do it by induction over $n$. The base case for $n=2$ is easy $\det(A)=p_1p_2-ba={bf(a)-af(b)\over b-a}$  The induction step is where I don´t know what to do. I tried to solve the dterminant by brute force(applying my induction hypothesis for n and prove it for n+1) but I don´t know how to reduce it. It gets horrible.  I would really appreciate if you can help me with this problem. Any comments, suggestions or hints would be highly appreciated
28416
Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$?     deals with the case that $R$ is a polynomial ring over a field. There it was observed that there is a straightforward proof when $R$ is an integral domain by passing to the fraction field.  In the general case I have neither a proof nor a counterexample. Here are three general observations about properties that a counterexample $M$ (trivial kernel but zero determinant) must satisfy. First, recall that the  $\text{adj}(M)$ of a matrix $M$ is a matrix whose entries are integer polynomials in those of $M$ and which satisfies  $$M \text{adj}(M) = \det(M).$$  If $\det(M) = 0$ and $\text{adj}(M) \neq 0$, then some column of $\text{adj}(M)$ lies in the kernel of $M$. Thus:     If $M$ is a counterexample, then $\text{adj}(M) = 0$.   When $n = 2$, we have $\text{adj}(M) = 0 \Rightarrow M = 0$, so this settles the $2 \times 2$ case.   Second observation: recall that by Cayley-Hamilton $p(M) = 0$ where $p$ is the characteristic polynomial of $M$. Write this as $$M^k q(M) = 0$$  where $q$ has nonzero constant term. If $q(M) \neq 0$, then there exists some $v \in R^n$ such that $w = q(M) v \neq 0$, hence $M^k w = 0$ and one of the vectors $w, Mw, M^2 w,\dots, M^{k-1} w$ necessarily lies in the kernel of $M$. Thus if $M$ is a counterexample we must have $q(M) = 0$ where $q$ has nonzero constant term.  Now for every prime ideal $P$ of $R$, consider the induced action of $M$ on $F^n$, where $F = \overline{ \text{Frac}(R/P) }$. Then $q(\lambda) = 0$ for every eigenvalue $\lambda$ of $M$. Since $\det(M) = 0$, one of these eigenvalues over $F$ is $0$, hence it follows that $q(0) \in P$. Since this is true for all prime ideals, $q(0)$ lies in the intersection of all the prime ideals of $R$, hence     If $M$ is a counterexample and $q$ is defined as above, then $q(0)$ is nilpotent.    This settles the question for reduced rings. Now, $\text{det}(M) = 0$ implies that the constant term of $p$ is equal to zero, and $\text{adj}(M) = 0$ implies that the linear term of $p$ is equal to zero. It follows that if $M$ is a counterexample, then $M^2 \mid p(M)$. When $n = 3$, this implies that $$q(M) = M - \lambda$$  where $\lambda$ is nilpotent, so $M$ is nilpotent and thus must have nontrivial kernel. So this settles the $3 \times 3$ case.   Third observation: if $M$ is a counterexample, then it is a counterexample over the subring of $R$ generated by the entries of $M$, so     We may assume WLOG that $R$ is finitely-generated over $\mathbb{Z}$.
121890
I'd like to know if the following problem of elementary linear algebra is already solved / solvable.  For two (singular) $n\times n$ matrices $P$ and $Q$, if $\det(\lambda P+\mu Q)=0$ for any $\lambda,\mu\in\mathbb{R}$, what are conditions on $P$ and $Q$?
106658
I need to prove the following:     Let $A$ be a square matrix over a field. If the matrix $A$ is invertible, then the minimal polynomial $m_A$ satisfies $m_A(0) \neq 0$.   There is one definition I am unsure of or need help making more clear.  I will proceed with proof by contraposition:  We must show that if $m_A(0) = 0$ then $A$ is not invertible. By definition of minimum polynomial of $A$ we have:  $m_A(x) = x^r - \lambda_{r-1} x^{r-1} - \ldots - \lambda_1 x + \det(A)$. Not sure about the determinant term here  So, $m_A(0) = \det(A) = 0$. We know $\det(A) = 0 \implies A$ is not invertible.
100008
Actually I need to show that $\det(AB) = \det(A)\det(B)$ if $A$ is a singular matrix.  The determinant of $A$ is $0$ if $A$ is singular, so $\det(AB)$ has to be $0$ as well, but I have problems showing that $AB$ is singular if $A$ is singular. How can I show that?
62412
Let $A$ be a $n \times n$ real matrix such that $$ A + A^t = 2I, $$ where $I$ is the $n \times n $ identity matrix. Prove that $\det(A) \geq 1$.  It is obvious that tr$(A) = n$. Furthermore, we also have that $$A - 2I = -A^t, $$ so we get that $$\det(A-2I) = (-1)^n \cdot \det(A).$$ Now, we let $\lambda_1, \lambda_2, \cdots, \lambda_n \in \mathbb{C}$ be the  eigenvalues of $A$, so we get that $$(\lambda_1 - 2)(\lambda_2 - 2) \cdots (\lambda_n - 2) = (-1)^n\lambda_1\lambda_2 \cdots \lambda_n, $$ but I couldn't derive anything about the product $\lambda_1\lambda_2 \cdots \lambda_n = \det(A)$ (the only known thing is $\lambda_1 + \cdots + \lambda_n = n$). Also, I tried the same approach for $2 \times 2$ and $3 \times 3 $ matrices, but it didn't lead to anything (especially since the eigenvalues can be complex numbers).
42917
If $P$ is real orthogonal matrix with $\det P = -1$, prove that $-1$ is an eigenvalue of $P$.    can anyone help me please how can I solve this problem?
40757
If $A \in M_{n\times n}(F)$ is not invertible then the rank of $A$ is less than $n$, thus $\det(A) =0$.    I proved that way, but looks like too simple so I think maybe there is a trick that I missed. Or I proved it right?   Modification: ($\Rightarrow$) If $A$ is invertible, $AB=I$, $\det(AB)=\det(A)\det(B)=1$ so $\det(A) \neq 0$ ($\Leftarrow$) If $\det(A) \neq 0$ , by Cramer's rule, $Ax=b$ has a unique solution and it means $A$ is invertible ($x=A^{-1}b$)    Now I think I complete the proof.
21358
I have a qualifying exam coming up in a couple days and I am just trying to understand some pathological examples I have in my notes. I will list a similar problem which I know the solution to and then the question.  True or False:   let $A,B \in \mathbb{Q}^{n \times n }$  Suppose $xI-A$ and $xI-B$ are equivalent then $\det(xI-A) = \det(xI-B)$.     I wrote in my notes this is true because any two matrices are equivalent in $\mathbb{Q[x]}^{n \times n}$ if and only if they have the same invariant factors.  Since the characteristic polynomial is a product of the invariant factors it follows that $\det(xI-A) = \det(xI-B)$.        True or False:   let $A,B \in \mathbb{Q}^{n \times n }$. If $\det(xI-A) = \det(xI-B)$ (in $\mathbb{Q}[x]$) then $xI-A$ and $xI-B$ are equivalent  in $\mathbb{Q}[x]^{n \times n}$   I think this is false because of the fact that equality of determinants is not strong enough to guarantee all invariant factors are equal but I do not have an example.
60530
I would like a proof of the following result, given on .  For all square matrices $\mathbf{A}$ and column vectors $\mathbf{u}$ and $\mathbf{v}$ over some field $\mathbb{F}$, $$ \det(\mathbf{A}+\mathbf{uv}^\mathrm{T}) = \det(\mathbf{A}) + \mathbf{v}^\mathrm{T}\mathrm{adj}(\mathbf{A})\mathbf{u}, $$ where $\mathrm{adj}(\mathbf{A})$ is the adjugate matrix of $\mathbf{A}$.  Note that $\mathbf{A}$ may be singular. However, the proof given on wikipedia requires that $\mathbf{A}$ is nonsingular.
17749
I read something that suggests that if $I$ is the $n$-by-$n$ identity matrix, $v$ is an $n$-dimensional real column vector with $\|v\| = 1$ (standard Euclidean norm), and $t &gt; 0$, then $$\det(I + t v v^T) = 1 + t$$ Can anyone prove this or provide a reference? More generally, is there also an (easy) formula for calculating $\det(A + wv^T)$ for $v,w \in \mathbb{K}^{d \times 1}$ and some (invertible) matrix $A \in \Bbb{K}^{d \times d}$?
1470
This question aims to create an &quot;&quot;  of numerous questions that ask about determinants of specific matrices (I may have missed a few):              The general question of this type is  Let $A$ be a square matrix of rank$~1$, let $I$ the identity matrix of the same size, and $\lambda$ a scalar. What is the determinant of $A+\lambda I$?  A clearly very closely related question is  What is the characteristic polynomial of a matrix $A$ of rank$~1$?
35912
Is it true that if $A$ is an $m \times n$ matrix and $B$ is an $n \times m$ matrix, with $m &gt; n $ then $\det(AB)=0$?
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.
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?
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.