Hot answers tagged

5

Yes this is possible even for a (left) module $M$ over a ring $R$ with a multiplicative identity $1$: $$(1+1)(x+y)=x+y+x+y$$ by the left distributive law. But also $$(1+1)(x+y)=2(x+y)=2x+2y=x+x+y+y.$$Cancelling $x$ and $y$ at both sides yields $x+y=y+x$, that is, $M$ must be an abelian group.


5

If $\langle x,y \rangle = \langle Tx,Ty \rangle$, then taking $y=x$ gives $\|x\| = \|Tx\|$. If $\|x\| = \|Tx\|$ for all $x$, then the polarisation identity gives: $\left \langle Tx,Ty \right \rangle=\frac{1}{4}\left \|Tx+Ty \right \|^2-\frac{1}{4}\left \|Tx-Ty \right \|^2 =\frac{1}{4}\left \|T(x+y) \right \|^2-\frac{1}{4}\left \|T(x-y) \right \|^2 ...


4

Write $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$. Then the equation $$ \int_{-1}^1 p(x) \, dx = \int_{-1}^1 (a_0 + a_1 x + a_2 x^2 + a_3 x^3)\, dx = \left[ a_0 x + a_1 \frac{x^2}{2} + a_2 \frac{x^3}{3} + a_3 \frac{x^4}{4} \right]_{x=-1}^{x=1} \\ = 2a_0 + \frac{2}{3} a_2 = 0$$ is a linear equation for $a_0, a_1, a_2, a_3$. Solve the equation and use the ...


3

Using the notation $\;\langle u,v\rangle=u\cdot v\;$ for the inner (scalar) product of two vectors: $$\langle u,\,v-\text{proj}_uv\rangle=\left\langle u,\,v-\frac{\langle u,v\rangle}{||u||}u\right\rangle=\langle u,v\rangle-\frac{\langle u,v\rangle}{||u|||}\langle u,u\rangle=\langle u,v\rangle-\langle u,v\rangle=0$$ because $\;||u||=\langle u,u\rangle\;$


3

Hint: show that all matrices in $\langle A\rangle$ are diagonalizable, and that the only nilpotent diagonalizable matrix is the zero matrix.


2

The second volume of Jacobson's Lectures in abstract algebra (in particular, Chapter VIII on infinite-dimensional vector spaces) could be a good reference.


2

I'm just going to apologize in advance. If there is any lesson to be learned here, it's that clearly stating your goals and cast of characters, as well as good notation both help immensely. SCENE I Enter Vector Space $P$, the space of polynomials with real coefficients and variable $t$, with his servant the Mapmaker. Vector Space $P$: Mapmaker! I ...


2

First of all, I think you need $V,W$ finite dimensional for the statement to be true. For example, if you take $W=\mathbb{R}$ and $V$ any infinite dimensional real vector space, then $Hom(V,W)=Hom(V,\mathbb{R}) = V^*$, which has strictly larger dimension for $V$ infinite dimensional. In your expression, $\beta_i^*(f(\alpha_k))$ $\textbf{is}$ a constant for ...


2

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Hint Prove and use the following lemma Let $D$ be a matrix, $\lambda$ an eigenvalue of $D$, and $$V_{\lambda} = \Set{ v : D v = \lambda v}$$ the corresponding eigenspace. If $A D = DA$, then $$ A V_{\lambda} \subseteq V_{\lambda}. $$


2

Span is clear. For linear independence, use $0$ can be expressed uniquely. $$0 = 0v_1 + \cdots + 0v_n $$ so if $c_1 v_1 + \cdots + c_n v_n = 0$, then $c_i = 0$ for all $i$.


1

If $V$ is defined the way you say with the constraint on $a_2$, how can $(0x^2+0x+0)$ be an element of $V$? So it is not a vector space. If you eliminate the constraint $a_2 \ne 0$ then you get a vector space: it is the space of polinomials of degree less than or equal to 2.


1

If $S_1$ and $S_2$ are dependent, neither $S_1 \cap S_2$ nor $S_1 \setminus S_2$ need to be dependent. If $S_1 \cap S_2 = \emptyset$, then $S_1 \cap S_2$ is independent if you use the convention that $\emptyset$ is independent. If $S_1 \cap S_2$ contains a single vector $v \ne 0$, then $S_1 \cap S_2$ is independent. In particular, a set $(v)$ containing ...


1

$$\begin{align*}&\bullet\;\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\frac{a+d}2&\frac{b-c}2\\-\frac{b-c}2&\frac{a+d}2\end{pmatrix}+\begin{pmatrix}\frac{a-d}2&\frac{b+c}2\\\frac{b+c}2&-\frac{a-d}2\end{pmatrix}\;\implies\;V=W_1+W_2\\{}\\ &\bullet\;\begin{pmatrix}a&b\\c&d\end{pmatrix}\in W_1\cap ...


1

Go through the conditions systematically. I.e. $x_1+x_2 = 0 \Rightarrow x_1 = -x_2$, $x_3+x_4=0 \Rightarrow x_3=-x_4$. From this, note that any vector $u \in U$ can be written in the form $(a,-a, b, -b)$, where $a,b \in \mathbb{R}$. Do the same to write some arbitrary $w \in W$. Then take your arbitrary $u,w$ and compute $u+w$. This should be the conditions ...


1

The differential of the Holder 1-norm (h) of a matrix (Y) is $$ dh = {\rm sign}(Y):dY$$ where the sign function is applied element-wise and the colon represents the Frobenius product. Now substitute $Y=(X-AX)$ $$\eqalign{ dY &= (I-A)\,dX \cr dh &= {\rm sign}(Y):(I-A)\,dX\cr &= (I-A)^T\,{\rm sign}(Y):dX\cr &= (I-A)^T\,{\rm ...


1

Let $\lambda_1, \lambda_2 \in \mathbb{F}$ and assume that $\lambda_1 u + \lambda_2 v = 0_V$. Apply the linear map $T$ to both sides of the equality and use the independence of $T(u)$ and $T(v)$ to deduce that $\lambda_1 = \lambda_2 = 0$, showing the independence of $u$ and $v$.


1

Hint: Because $T(u),T(v)$ are linearly independent, we know that if $c_1,c_2$ are constants such that $$ c_1 T(u) + c_2T(v) = 0 $$ then we must have $c_1 = c_2 = 0$. Now, suppose that $c_1,c_2$ are such that $$ c_1 u + c_2 v = 0 $$ how can we show that $c_1$ and $c_2$ are necessarily zero?


1

For (a), note that $A$ is not diagonalizable, and so $a\not=0$ (otherwise you would have a nontrivial Jordan canonical form). For (b), simply compute. Note that $$ M=\left(\begin{array} (0 & a & b\\ 0 &0 & c\\ 0 & 0 & \lambda_2\end{array}\right)\left(\begin{array} (\lambda_1 & a & b\\ 0 & \lambda_1 & c\\ 0 & 0 ...


1

Since $\dim W_1=1$ and $\dim W_2=2$, if $\mathbb{R}^3=W_1+W_2$ the dimension formula gives $$ \dim(W_1\cap W_2)=\dim\mathbb{R}^3-\dim W_1-\dim W_2=0 $$ So you just need to show that $$ \mathbb{R}^3=\operatorname{Span}\{(1,1,1),(1,0,0),(1,1,0)\} $$ The matrix $$ \begin{bmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 1 & 0 \end{bmatrix} $$ ...


1

In general it is not sufficient to check the characteristic polynomial to make sure that two matrices are similar. In order to be similar, there needs to exist an invertible matrix $P$ such that $A = P^{-1}B P$. If two matrices are similar and one of them is diagonalizible (say, $B=Q^{-1}DQ$), then $A$ is automatically diagonalizible, too (by means of ...


1

The "functional" $$\phi(x):=\int_0^1 t^2\>x(t)\>dt$$ takes as input a function $x:\>t\mapsto x(t)$ and produces as output a number. According to the rules about integrals you have learned in Calculus 101 (and have applied a hundred times ever since) it is obvious that $$\phi(x+y)=\phi(x)+\phi(y), \qquad \phi(\lambda x)=\lambda\>\phi(x)$$ for all ...


1

It is either a basic theorem about direct sums (or an alternative definition of a direct sum) that if $V=U \oplus W$, $\alpha$ is a basis for $U$ and $\beta$ is a basis for $W$, then $\alpha \cup \beta$ (a disjoint union) is a basis for $V$. So you're close. Pick a basis for the image, say $\alpha = \{u_1,\dots,u_r \}$. Then (don't try to "complete" the ...


1

Note that we should integrate with respect to $y$, not $x$. Write $f(x) = x$. Then $$ T(f)(x) = \int_{-1}^1 (x - y)^2 f(y) \, dy - 2f(0)x^2 = \int_{-1}^1 (x - y)^2 y \, dy = \int_{-1}^1 (x^2y - 2xy^2 + y^3) \, dy =\left[ \frac{x^2y^2}{2} - \frac{2xy^3}{3} + \frac{y^4}{4} \right]^{y = 1}_{y = -1} = -\frac{4}{3} x. $$


1

$$c_1(u+v)+c_2(v+w)=0$$ So $$c_1u+(c_1+c_2)v+c_2w=0$$ As $u,v,w$ are linearly independent, $$c_1=0, c_2=0$$ If they were not zero, one could be written in terms of the other.


1

Your proof for the first one is correct. For the second one, look at $\left(\begin{matrix} 0 & 1\\ 0 & 0 \end{matrix}\right)+\left(\begin{matrix} 0 & 0\\ 1 & 0 \end{matrix}\right)$. For the third, look at $\left(\begin{matrix} 1 & 0\\ 0 & 1 \end{matrix}\right)+\left(\begin{matrix} 0 & 1\\ 1 & 0 \end{matrix}\right)$.


1

Hints/Comments: 1. Correct 2. Also correct but it would be clearer if you gave an explicit counterexample. Can you think of two matrices $A$ and $B$ that are not symmetric but for which $A+B$ is symmetric? Don't overthink this one! 3. Can you think of two invertible matrices $A$ and $B$ whose sum is not invertible? Try simple examples in which the ...



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