Hot answers tagged

4

The answer to your initial question is "no", a counterexample is given by $A = 0$, but for example also by $$A = \begin{pmatrix}-1 & 0\\ 0 & 0\end{pmatrix}.$$ The problem with your reasoning is the step "$A ( A + I) = 0 \implies A = 0$ or $A + I = 0$". The ring of square matrices of size at least $2\times 2$ has zero divisors, so ...


3

There are reasons to know in advance that such a decomposition is possible (see Ben's comment). But that's not essential to the argument. A more verbose version of your proof might go as follows: "We make the guess that the inverse $A^{-1}$ has the form $aB + bI$ for some scalars $a,b$, and try to solve for $a$ and $b$. If we succeed, our guess was ...


2

Continuing with $R_{1}\rightarrow R_{1}+\frac{1}{4}R_3$ and $R_{2}\rightarrow R_{2}+\frac{1}{4}R_{3}$ we have $$\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{array}\mid\begin{array}{ccc} 1 & 0 & 1\\ \frac{1}{2} & \frac{1}{2} & 1\\ 2 & 1 & 4 \end{array}\right]$$


2

Building on SV-97's comment, I often find it's a good idea to do it in smaller steps: If you're looking for a homeomorphism $f : X \to Y$, you instead try to find homeomorphisms $f_1 : X\to X_1$, $f_2 : X_1 \to X_2$ until $f_k : X_{k-1} \to Y$ for some $k \in \mathbb{N}$, and then you simply define $f := f_k \circ \ldots \circ f_1$. A nice application of ...


2

Let's say the equation has two solutions for some $b$. That means there exist two different vectors $x_1, x_2$ such that $Ax_1=b$ and $Ax_2=b$. Now, suppose the matrix $A$ has an inverse $A^{-1}$. The property of the inverse matrix must be that if $A$ maps the vector $x$ to $y$, then $A^{-1}$ maps the vector $y$ to $x$. Well, we know that $A$ maps $x_1$ to $...


2

First we use the equation given as $$A^{-1}(A+B)=A^{-1}AB$$ which implies $$I+A^{-1}B=B.$$ Then to find the inverse of $B$ we start by assuming that $B$ is invertible which leads to $$(I+A^{-1}B)B^{-1}=BB^{-1}=I.$$ This gives $$B^{-1}+A^{-1}=I$$ and then we can write $$B^{-1}=I-A^{-1}$$ which we can check satisfies: $$(I-A^{-1})B=B-A^{-1}B=I$$


2

$\psi$ is not a bijection since the range is contained in $\{(x,y): x >0\}$ If you restrict the codomain to this set you can find the inverse as follows: $t=\ln x$ and $s=ye^{-2t}=ye^{-2\ln x}=y/x^{2}$ So $\psi^{-1}(x,y)=(\ln x , yx^{-2})$ on $\{(x,y): x>0, y \in \mathbb R\}$.


1

After you arrive at the following equation $$f'(y)\cdot\frac{dy}{dx} = 1$$ you can just substitute the value of $y$ to arrive at the final answer. This way could save much of your confusion. What you seem to have done is the following: When you say that $$\frac{df}{df^-1}\cdot\frac{dy}{dx} = 1$$ you're absolutely right. But, $f = x$ and $f^-1 = y$. So this ...


1

For (1), we will use the fact that $\frac{x^Ty-1}{x^Ty-1} = 1$ and $x^Ty = y^Tx$. We also use the fact that $xy^Txy^T = (x(y^Tx)y^T) = (y^Tx)xy^T =y^Txxy^T =x^Tyxy^T $. Obser for (1) we have that $$\begin{equation}\begin{split}\left(I-x y^T\right)\left(I-\frac{1}{x^T y- 1} x y^T\right) &= I - \frac{1}{x^T y- 1} x y^T-x y^T + \frac{1}{x^T y- 1} x y^Tx ...


1

A being invertible implies that $A=−I$, so if $A≠−I$, it's non-invertible (note that this is the contrapositive of the first implication). So, you can't say for sure if $A$ is invertible or not unless you know what A is. But you can say that $A$ is invertible and the equation in the title holds iff $A=−I$. ($A$ is invertible $\land$ $A^2 + A = 0$) $\iff A = -...


1

The Wronski's formula \eqref{f(t)g(t)=1=determ} below is a best answer to this question. If $a_0\ne0$ and $$ P(t)=a_0+a_1t+a_2t^2+\dotsm $$ is a formal series, then the coefficients of the reciprocal series $$ \frac{1}{P(t)}=b_0+b_1t+b_2t^2+\dotsm $$ are given by \begin{equation}\label{f(t)g(t)=1=determ}\tag{1} b_r=\frac{(-1)^r}{a_0^{r+1}} \begin{vmatrix} ...


1

The important step is usually not constructing the homeomorphism in the sense you seem to be talking about here; rather, it's determining whether or not such a homeomorphism exists in the first place. Usually, the construction will be obvious if you have a good intuition for why the spaces are homeomorphic.


1

The notation is a little confusing here. Given $f(x) = 3^x$, you're wanting to determine a $y$ such that $f^{-1}(y) = 4$ (replacing the $y$ with $x$ leads to a little confusion since your $x$ variables are initially referred to as elements in the domain of $f$). Now, by assuming invertability of $f$, we have that $y = f(f^{-1}(y)) = f(4) = 3^4$, i.e. $y = 81$...


1

The characteristic polynomial of A is of degree n from which the inverse is found out in the form $A^{-1}$= f(A) for some f. Substituting B-I for A we get a polynomial in B which can be reduced to degree one in B as B is a rank one matrix as observed by Ben. Hence the assumption about the inverse of A.


1

Consider the canonical projection $\pi:\mathbb{Z}\to\mathbb{Z}/5\mathbb{Z}$. The image is a field, and so the element $\pi(2)=2+5\mathbb{Z}$ is invertible. However, $2$ is obviously not invertible in $\mathbb{Z}$.


1

First of all, we need to properly set the domain and co-domain. The domain is $$ D=\{x\in \mathbb{R}: x \ge 0, x \ne 1\} = [0,1[\cup]1,+\infty[ $$ and the co-domain is $D^* =\mathbb{R}$. As you have mentioned, $f:D \to D^*$ is one-to-one, and therefore invertible. An expression for the inverse can be obtained using the method mentioned by @SteveKim: $$ \frac{...


Only top voted, non community-wiki answers of a minimum length are eligible