Reputation
67,838
Next tag badge:
99/100 score
37/20 answers
Badges
3 59 141
Newest
 Nice Answer
Impact
~1.4m people reached

13h
comment Is there any known way to sum a subserie (square indices) of geometric series?
There always exists a function (in fact many) that has a given formal series as its Taylor series. There might not however be any nice expression for such a function.
13h
comment Can you show that $3n+1$ is not divisible by $5$ using congruences?
My apologies, that was a wrong edit of mine. But $3n+1$ is not a difference between consecutive cubes either.
13h
revised Can you show that $3n+1$ is not divisible by $5$ using congruences?
added 10 characters in body; edited title
13h
revised Can you show that $3n+1$ is not divisible by $5$ using congruences?
correctly render formulas
23h
comment How to explain polynomial coefficients by minimezed Error function?
You have already explained what $\bf w$ is and how to find it. So what is the question?
23h
revised How to explain polynomial coefficients by minimezed Error function?
edited tags; edited tags; edited tags
2d
revised If $A$ is normal and upper triangular then it is diagonal
edited title
2d
comment $im(A)=im(AA^T)$
Concerning $\rk(A)=\rk(A^T)$, see this question and also this question asking for an intuitive explanation.
Apr
28
revised Determining if $\def\rk{\operatorname{rank}}\rk(A)$ is always equivalent to $\rk(A^t*A)$
added 70 characters in body
Apr
28
answered Determining if $\def\rk{\operatorname{rank}}\rk(A)$ is always equivalent to $\rk(A^t*A)$
Apr
28
revised Determining if $\def\rk{\operatorname{rank}}\rk(A)$ is always equivalent to $\rk(A^t*A)$
added 54 characters in body
Apr
28
comment $im(A)=im(AA^T)$
OK, so here a bit more detail. First, by definition of rank (dimension of the image), having $\im(A)=\im(AA^T)$ will imply that $A$ and $AA^T$ have the same rank, but the latter also implies $\im(A)=\im(AA^T)$, since $\im(AA^T)$ is then subspace of $\im(A)$ of the same dimension, which forces it to be all of $\im(A)$. So proving $\im(A)=\im(AA^T)$ is equivalent to proving $\def\rk{\operatorname{rank}}\rk(AA^T)=\rk(A)$. Second it is well known that $\rk(A)=\rk(A^T)$, so it is also equivalent to proving $\rk(AA^T)=\rk(A^T)$. But that is done in the linked question (swapping names of $A,A^T$).
Apr
27
comment Prove rank $A^TA$ = rank $A$ for any $A_{m \times n}$
Or $A^T=(1~~\mathbf i)$.
Apr
27
comment $im(A)=im(AA^T)$
Also note the result is not true in the stated generality: it is true for real matrices (or matrices over an ordered field), but not for complex matrices (take $A=(1~~\mathbf i)$).
Apr
27
comment $im(A)=im(AA^T)$
Marked as duplicate because $A$ and $A^T$ have the same rank (so it makes no difference to say the rank of a product of two transposed matrices equals the rank of the first or of the second), and the inclusion $\def\im{\operatorname{im}}\im(AA^T)\subseteq\im(A)$ is immediate, so equality of their ranks suffices.
Apr
27
comment Prove rank $A^TA$ = rank $A$ for any $A_{m \times n}$
I cannot decipher what is said here, but it must be wrong since it never uses that the matrices are over $\Bbb R$ (or more generally an ordered field) rather than for instance over $\Bbb C$ where the result is not true.
Apr
27
revised If $F$ is closed subset of $R^n$ and $x \in R^n, $ is $x+F$ still closed?
deleted 3 characters in body
Apr
27
answered If $F$ is closed subset of $R^n$ and $x \in R^n, $ is $x+F$ still closed?
Apr
27
comment Complement of rationals has empty interior
And a set being dense means that its complement has empty interior.
Apr
27
answered Complement of rationals has empty interior