Marc van Leeuwen
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 12h answered How can we know if the minimal polynomial of a matrix has no multiple products? 13h comment How can we know if the minimal polynomial of a matrix has no multiple products? @kuhaku To be precise, a square-free polynomial is one that is not divisible by the square of any non-constant polynomial. (More generally in a commutative ring square-free means not divisible by the square of any non-invertible element.) For polynomials there is a practical test that does not require factoring: $P$ is square-free iff $\gcd(P,P')=1$ where $P$ is the (formal) derivative of $P$. 15h comment $-1$ as the only negative prime. @user254665: That kind of historic reconstruction to explain the present is usually extremely inexact, and it is certainly in this case. I don't want to go into linguistic issues (the English language did not even exist when this first came up), but the notion of prime number arose many many centuries before negative numbers were conceived. And the terms invertible and decomposable are of much more recent coinage. 15h comment Is it accurate to say that multiplication of two integers yields an integer? This question is confusingly formulated, and it is not clear to me at all what is being asked. When $n$ is an integer, then neither $n+\sqrt3$ nor $n-\sqrt3$ are ever integers, not even approximately. If you were asking whether multiplying non-integers can sometimes give an integer the answer is yes, but that is not what you state as question. Also, calling $n^2-3=(n+\sqrt3)(n-\sqrt3)$ a factorisation is misleading, since that term usually means decomposition as a product of prime numbers (necessarily integers); for instance for $n=5$ the factorization of $n^2-3$ is different: $22=2\times11$. 15h comment Meaning of the phrase “span of vectors contains” @Dan Since you do know the definition of linear combination, the definition of the span of a certain list of vectors is simply the set of all their linear combinations. Showing that $(d,e,f)$ belongs to that span is therefore the same as showing that $(d,e,f)$ is (=can be written as) such a linear combination. 16h comment Meaning of the phrase “span of vectors contains” Did they give you a definition of the span of a sequence of vectors? (The span is a certain, usually infinite, set of vectors). The definition should answer your question. 1d comment Determinant of a rank $1$ update of a scalar matrix, or characteristic polynomial of a rank $1$ matrix @kuhaku: Yes, the set of roots of the characteristic polynomial gives you all the eigenvalues. 2d comment True/false questions about minimal and characteristic polynomials of a matrix That $(X+77)(X-1432)$ annihilates $B$ follows from the fact that $(B+77I)(B-1432I)$ acts as $0$ on each of a basis of eigenvectors for$~A$. 2d comment True/false questions about minimal and characteristic polynomials of a matrix $P[x]$ means the result of substituting $x$ for $X$ into the polynomial $P$. For instance, with $P=X^5+6X^3+X^2-1$ and $x=-2$ one has $(X^5+6X^3+X^2-1)[-2]=(-2)^5+6(-2)^3+(-2)^2-1=-77$, which is the scalar by which $B=P[A]$ acts on eigenvectors for $\lambda=-2$ of $A$. Also "multiple" root is the opposite of "simple" root. I used it at some point instead of "double" because I did not know yet (and did not care) whether it might actually be a triple root (it might happen that $P[-2]=P[4]$ for a certain $P$). Feb 5 comment What is the square root of $i$? @gebra done.... Feb 5 answered What is the square root of $i$? Feb 5 comment What is the square root of $i$? I don't understand your logic at all. You are looking for real $a,b$ that satisfy $(2ab - 1)i=b^2-a^2$, and although you do observe that this is only possible when both sides are $0$ (since LHS is imaginary and RHS is real), you decide to forget about solving for $a,b$, and rather do a division that does not give you any concrete answer, and besides that must work out as $\frac00$ whenever $a,b$ do solve the equation? Feb 5 comment Finding a matrix given eigenvalues and eigenvectors. Note that $A=PDP^T$ is not the correct formula (unless $P$ is an orthogonal matrix and therefore $P^T=P^{-1}$, but this does not follow from your description of $P$), and $A=P^TDP$ is even less so. Feb 5 comment Why doesn't Cantor's diagonalization work on integers? If you try this for the standard enumeration of the natural numbers $0,1,2,3,4,\ldots$ (trying to show it cannot exist), then you will rapidly notice that most numbers on the list do not even reach the diagonal. This is not the essential problem (which is that your "extra natural number" is not one), but you should have tried this obvious example and notice the difficulty. A similar argument does show that there are uncountably many $10$-adic integers. Feb 5 revised Why doesn't Cantor's diagonalization work on integers? edited tags Feb 5 comment True/false questions about minimal and characteristic polynomials of a matrix Yes, $P[A]$ acts on the $2$-dimensional eigenspace of$~A$ for $\lambda=-2$ by the scalar $P[-2]=-77$ and on the $1$-dimensional eigenspace of$~A$ for $\lambda=4$ by the scalar $P[4]=1423$, so $f_B=(X+77)^2(X-1423)$. Feb 5 comment True/false questions about minimal and characteristic polynomials of a matrix You did not compute $f_A$ correctly. Feb 5 answered True/false questions about minimal and characteristic polynomials of a matrix Feb 4 revised Understanding the method to find Eigenvectors deleted 15 characters in body Feb 4 answered Understanding the method to find Eigenvectors