Tag Info

Hot answers tagged

5

HINT: Let $v_1$, $v_2$ be column vectors of $I$ (identity matrix).


4

Yes, we need an infinite-dimensional vector space. An interesting example is: $V$ the space of continuous functions $[0,1]\to\mathbb R$ and $f$ integration $f(g)(x)=\int_0^xg(t)\,\mathrm dt$. This is not surjective because $f(g)(0)=0$ for all $g$


4

I'm going to start with your example and work towards a more abstract notion of structure throughout this writing. So let's see, the bijection you give is a function $f:A\rightarrow B$. But all we have are the sets $A,B$. No other information is given. So what does the bijection encode? Well, both sets have $3$ elements. Perhaps that is what we should look ...


3

I don't think the following could be caracterized as "nasty determinant calculation". I don't know how one can prove the equality without indulging in some computation. Let $r=\operatorname{rank}(A)$ From a well-known theorem, derive that there exists $P,Q$ invertible $m\times m$ and $n \times n$ matrices such that $$A=P\begin{bmatrix}I_r& 0\\ 0 ...


3

$3$ vectors are, and only are linearly independent, if only $0u+0v+0w=0$(null vector) stands, and you can't make the null vector from any of their other combinations. Since $2u+3v-w=(8,6,-4)+(6,-18,21)-(14,-12,17)=(0,0,0)$, they are linearly dependent. Your approach is totally correct, your book must be false.


2

Note that all sets are subsets of vector spaces. Thus you have to check whether they are subspaces: Is the zero element of the underlying vector space also in the subset? Take two arbitrary elements from the subset. Is their sum also in the subset? Take an element $x$ of the subset and a number $\lambda$ of the underlying field (in most cases either ...


2

You want to see whether the sets are subspaces of the given vector spaces. The first necessary condition to check is whether the zero vector belongs to the set: if not, we're done because the set is not a subspace. Note that this is not sufficient, so if the zero vector is in the set we need to do other checks. The zero vector doesn't belong to the set in ...


2

In finite dimensions we have that bijectivity $\Leftrightarrow$ injectivity $\Leftrightarrow$ surjectivity. Hence we have to come up with an infinite-dimensional example. The idea is to pick a basis $v_i, i\in \mathbb N$, and shift every basis vector $v_i \mapsto v_{i+1}$. We can do that not hitting the first basis vector only because we have infinitely many ...


2

Hint: Do you know how to find a basis for the null space of a given matrix? If so, it is a fact that $C(A)^\perp = N(A^T)$ for any matrix $A$. That is, take the transpose of your matrix and find a basis for its null space, and the result will give you what you are looking for. For troubleshooting purposes, I have gone through some of the process below. ...


2

Set $X={A^{-1}}^T B$ where I assume $A$ and $B$ are $n\times n$ positive definite matrices. Then, $X$ has eigenvalues $\lambda_1,\ldots,\lambda_n>0$, and the trace $\text{tr}(X)=\sum_j\lambda_j$. The constraint $A_{ii}=1$ (added after my original answer) have very little impact on the eigenvalues, and so I'll ignore them. Since ...


2

As it is, your equations are not linearly independent, since you have $(2)+(3)=(1)$. This means that you can pick arbitrarily one of the variables $h$, $i$ or $n$, and only then will the other two follow by your system of equations. Notice that you can derive an additional constraint on your known parameters, namely that $f+g=e$.


2

For a route different to induction, express the matrix as $$A=aI_3+G$$ where $I_3$ is the identity matrix and $G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ A binomial expansion will yield $$\begin{align}A^k&=(aI_3+G)^k\\&={k \choose 0}a^kI_3+{k \choose 1}a^{k-1}I_3G+{k \choose ...


1

If the (linear) map $f$ is not injective, its kernel is not 0, i. e. there is an element $v=(a,b,c)\ne (0,0,0)$ such that $f(v)=0$. As $f$ is linear, for all $\lambda\in\mathbf R$, $$f(\lambda v)=\lambda f(v)=\lambda0=0.$$ Hence the whole line directed by $v$ maps to $0$. Now there might be two linearly independent vectors that map to $0$. In such a case, ...


1

If a linear map $T: \Bbb V \to \Bbb W$ is not injective, there is some nonzero vector $v \in \Bbb V$ such that $T v = 0$. Then, $T$ maps the line $\langle v \rangle$ to the origin. It is possible for a linear map $T$ to map a $k$-plane to an $l$-plane for any $l \leq k$, and so can map a plane to a point, or a plane to a line, or a line to a line (or all ...


1

They are linearly independent since the minor is non-zero, i.e : $$\text{det}\begin{pmatrix} 4 & -1 \\ -4 & 10 \end{pmatrix}=36\neq 0$$


1

Sometimes it is solvable - and has infinitely many solutions - but sometimes it is not. It depends on the values of $e,f$ and $g$. In particular, if there is a solution then $h-i+i-n+n-h=f+g-e=0$. In that case, $h$ can be anything, say $=x$, then $n=x-e$ and $i=x-f$


1

Note that if $Av = \lambda v$, then $(A+cI)v = (\lambda+c)v$. Also, $\chi_{A+cI}(x) = \det (xI-A-cI) = \det( (x-c)I -A) = \chi_A(x-c)$.


1

You use the Gram-Schmidt process. The Gram-Schmidt process takes a set of vectors and produces from them a set of orthogonal vectors which span the same space. It is based on projections -- which I'll assume you already are familiar with. Let's say that we want to orthogonalize the set $\{u_1, u_2, u_3\}$. So we want a set of at most $3$ vectors $\{v_1, ...


1

i think you can do this with cayley-hamilton theorem which says that a matrix satisfies its characteristic equation. that is $$A^2 -(a+b)A + abI = 0 $$ or $$A^2 = (a+b)A - abI$$ we can use this to write every power of $A$ as a lineay combination of $A, I$. we will the division algorithm to find the remainder. suppose $$x^n = q(x)(x-a)(x-b) + \alpha x + ...


1

We have $$A = X \begin{bmatrix}a & 0\\0 & b\end{bmatrix} X^{-1}$$ This means $$A^n = X \begin{bmatrix}a^n & 0\\0 & b^n\end{bmatrix} X^{-1}$$ We have $$A-bI = X \begin{bmatrix}a-b & 0\\0 & 0\end{bmatrix} X^{-1} \text{ and }A-aI = X \begin{bmatrix}0 & 0\\0 & b-a\end{bmatrix} X^{-1}$$ Hence, $$\dfrac{a^n}{a-b}\left(A-bI\right) = ...


1

I got $$x_1=-7-5x_3$$ $$x_2=8-6x_3$$ and $$x_3=h-11$$ for every real $$h$$ there is a uniquely determined solution.


1

The system matrix is $$A = \begin{bmatrix} 3 & 4 & -8\\ -6 & -5 & 0\\ 1 & 1 & -1 \end{bmatrix}$$ We have $\det(A) = -1$, which means the matrix is invertible and hence the system is consistent for all values of $h$.


1

Just follow your nose: Since $x=Ax+x-Ax$, take $u=Ax$ and $v=x-Ax$, and you only need to see that $v$ satisfies the required property: $Av=A(x-Ax)=Ax-A^{2}x=Ax-Ax=0$. Have you studied projections yet?


1


1

Hint: In terms of inner product, the projection of vector $X=(x,y,z)$ onto vector $u=(a,b,c)$ is simply: $$p_u(X)=\frac{\langle X,u\rangle}{\langle u,u\rangle}\, u$$ For the second question, two strategies for the projection onto a plane: If you have the equation of the plane, as is the case, you also have a normal vector of the plane: $n=(1,-2,3)$. ...


1

i think it is easier to find the projection on to the line $u = (1, -2, 3)^\top$ that is orthogonal to the plane and then subtract from the identity to get the projection onto the plane. the projection matrix onto the line $u$ is $$uu^\top/(u^\top u) = \frac1{14}\pmatrix{1&-2&3\\-2&4&-6\\3&-6&9}.$$ therefore the projection matrix on ...


1

If $U_1$ and $U_2$ are subspaces of the vector space $V$, then $U_1\cup U_2$ is a subspace if and only if either $U_1\subseteq U_2$ or $U_2\subseteq U_1$ Proof. Suppose $U_2\not\subseteq U_1$ and let $y\in U_2$, $y\notin U_1$. Let $x\in U_1$; if $U_1\cup U_2$ is a subspace, then $x+y\in U_1\cup U_2$. However $x+y\in U_1$ implies $y\in U_1$, which is ...


1

Think of a simple counter-example: $$X = \pmatrix{1&0\\0&0} \in \mathbb R^{2\times 2}; \qquad y = \pmatrix{1\\1}$$ We have $C(X) = \{(x,0) \mid x\in\mathbb R\}$ and $C(X)^\perp = \{(0,x) \mid x\in\mathbb R\}$


1

Call your vectors $v_1,v_2$. Pick two vectors $v_3,v_4$ such that $\{ v_1,\dots,v_4 \}$ is a linearly independent set. Pick linearly independent images $w_1,w_2$ for them. (The linear independence ensures that the kernel contains only your given vectors.) Then you want $A$ such that $$A \begin{bmatrix} v_1 & v_2 & v_3 & v_4 \end{bmatrix} = ...


1

the eigenvalues $\pmatrix{0&1&1\\1&0&1\\1&1&0} = uu^\top - I, \text{ where } u = (1,1,1)^\top$ are $2, -1, -1$ and the corresponding eigenvectors are $u, u^\perp.$ what you have listed are two linearly independent vectorsin $u^\perp.$ the matrix you have is an example of a rank one perturbation of the identity matrix. their ...



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