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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 $$ p_i\circ (S-T)=p_i\circ S - p_i\circ T = T_i-T_i=0$$
Apr
24
answered Prove there exists a unique map T
Apr
24
revised Prove that matrix is symmetric and positive definite given the fact that $A+iB$ is.
added 5 characters in body; edited title
Apr
24
reviewed Edit Show the inverse of a bijective linear operator is continuous iff $\inf_{\|x\|=1} \|Tx\|>0$
Apr
24
reviewed Approve Coordinate Geometry Help (circles + trigonometry)