HK Lee
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 13h revised $V=ker\ (T)\oplus im\ (T)$ if $T$ is a self-adjoint edited title 13h answered $V=ker\ (T)\oplus im\ (T)$ if $T$ is a self-adjoint 16h revised Prove or disprove the inequality edited tags 1d comment Show that $g(a) = g(b) = 0,\ \int_a^b f(x)g(x)dx=0$ implies $f(x)=0$ It is an assumption of $f$. If $f$ satisfies such assumption, we must show that $f=0$. That is, when $f$ satisfies the assumption, if $f$ is not $0$, we complete the proof by showing a contradiction. 1d revised Prove boundedness of the matrix series added 31 characters in body 1d revised Proving $AD_1A^{-1}=D_2$ added 91 characters in body 1d comment Proving $AD_1A^{-1}=D_2$ I just enumerated properties but it is not crucial to here. I editted. $A_1DA_1$ is diagonal when we compute directly. 1d revised Proving $AD_1A^{-1}=D_2$ deleted 99 characters in body 1d revised Injectivity of $T:C[0,1]\rightarrow C[0,1]$ where $T(x)(t):=\int_0^t x(s)ds$ added 11 characters in body 1d revised A continuous function $f$ is differentiable when so is $|f|$. edited body; edited title 1d revised Proving $AD_1A^{-1}=D_2$ added 22 characters in body 1d comment Proving $AD_1A^{-1}=D_2$ $ADA^{-1}=A_1\cdots A_m D A_m\cdots A_1$ hence if $A_mDA_m$ is diagonal, then after $m$-steps we have a diagonal $ADA^{-1}$ 1d answered Proving $AD_1A^{-1}=D_2$ 1d revised A continuous function $f$ is differentiable when so is $|f|$. added 39 characters in body 1d answered Is the completion of a separable normed linear space is also separable? 2d comment Prove there exists a unique map T It is an assumption of $S$ : $p_i\circ S=T_i$ Hence $$p_i\circ (S-T)=p_i\circ S - p_i\circ T = T_i-T_i=0$$ Apr24 answered Prove there exists a unique map T Apr24 revised Prove that matrix is symmetric and positive definite given the fact that $A+iB$ is. added 5 characters in body; edited title Apr24 reviewed Edit Show the inverse of a bijective linear operator is continuous iff $\inf_{\|x\|=1} \|Tx\|>0$ Apr24 reviewed Approve Coordinate Geometry Help (circles + trigonometry)