# Humanity

 Sep 13 comment Eigenvalues of $AB$ from eigenvalues of $A$ and $B$If $A$ and $B$ commute, you can diagonalize them simultaneously and the relation is clear. If they don't commute, you can still deduce in this case that the spectral radius $\rho(AB)$ is not greater than $\rho(A)\rho(B)$ but that's about it. Sep 12 comment for $p \in [1,\infty)$ the space $l^p$ is separableHint: sequences with finite support are dense in $\ell^p$. Sep 11 comment Banach Algebra-MathematicsWell, actually, this should be: $x_ny_n-x_py_p=(x_n-x_p)y_n+x_p(y_n-y_p)$. And you do have to prove that a Cauchy sequence is bounded (I can't see a way around that). Take $\epsilon =1$ etc... Sep 11 comment Banach Algebra-MathematicsNow just prove or use that a Cauchy sequence is bounded to handle that $\|y_n\|$. Sep 6 comment Geometric intuition for the tensor product of vector spacesGood question, I look forward to the answers. If $W^*$ is the dual of $W$, I always feel more comfortable seeing $V\otimes W^*$ as the vector space of linear maps $L(W,V)$, and in particular the tensors $v\otimes w^*$ as the rank $1$ maps $w^*(\cdot)v$. Unfortunately this requires an identification of $W$ with $W^*$ to describe $V\otimes W$. But at least the identification $V\otimes W^*\simeq L(W,V)\simeq L(V^*,W^*)\simeq W^*\otimes V^{**}\simeq W^*\otimes V$ is intrinsic via the transpose and the canonical identification $V\simeq V^{**}$. Sep 4 comment Example of matrices that do not satisfy the submultiplicative propertyTake any matrix norm such that $\|I_n\|=1$. Then consider the new matrix norm $\|A\|_0=\frac{1}{2}\|A\|$ and $A=B=I_n$. But note that every induced matrix norm, and every Schatten matrix norm is submultiplicative. Sep 4 comment A basic question on diagonalizabilityIf it were diagonalizable, it would be $T=0$. Why? Sep 2 awarded Enthusiast Aug 17 awarded Nice Answer Aug 17 revised Why is $\dim(V/\bigcap_{i=1}^n\ker(f_i))\leq n$ for linear functionals $f_i$?added 516 characters in body Aug 17 answered Why is $\dim(V/\bigcap_{i=1}^n\ker(f_i))\leq n$ for linear functionals $f_i$? Aug 17 answered Show that $T(V)$ is finite-dimensional? Aug 17 awarded Commentator Aug 15 comment Prove that the rank of the block matrix is rank $A$ + rank $B$@CnR 2. The rank of a (square or rectangular) matrix is not affected by multiplication by an invertible matrix, left or right. For a linear algebraic approach of this, in this case, this is right multiplication. The rank of $M$ is the dimension of the range, that is the vector space $M(K^{2n})$. Now if $S$ is invertible, you have $S(K^{2n})=K^{2n}$ (ie $S$ is surjective). So $MS(K^{2n})=M(K^{2n})$. That is the range of $MS$, whence its rank, is the same as that of $M$. Aug 15 comment Prove that the rank of the block matrix is rank $A$ + rank $B$@CnR 1. The determinants of $S$ and $T$ is $1$ (they are both triangular with $1$ on the diagonal). It is also easy to find the inverses if you prefer: just replace $B$ by $-B$, and you get the inverse of $T$. Do the same trick with $S$. See transvection matrices for more about this type of matrices. Aug 15 revised Prove that the rank of the block matrix is rank $A$ + rank $B$added 2 characters in body Aug 15 answered Prove that the rank of the block matrix is rank $A$ + rank $B$ Aug 15 revised Prove that the rank of the block matrix is rank $A$ + rank $B$This needed to be TeXified. Aug 14 comment $A$ - positive definite then $A + A^{-1} -2I$ - semi-positive definiteA product of positive semidefinite matrices need not be positive semidefinite. What makes things work in this approach is the supplementary equation $=A^{-1/2}(A-I)^2A^{-1/2}$. And now it is clear, as $P^*BP$ is positive for any $P$, and any positive $B$. Which is clearly what you had in mind. Just answering @Davide Giraudo's comment. Aug 14 answered $A$ - positive definite then $A + A^{-1} -2I$ - semi-positive definite