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5

The functions $\cos x$ and $\sin x$ are linearly independent. There are various ways to see this: I give a method which picks up on your own working. First let $$a\cos x+b\sin x=0\ ,$$ and make sure you understand that this is an equality of functions: that is, it means that the LHS is zero for all values of $x$. Taking $x=\pi/4$, as you have done, shows ...


5

Why not $a(\mathbf{0})=a(\mathbf{0}+\mathbf{0})=a(\mathbf{0})+a(\mathbf{0})$; now add the inverse of $a(\mathbf{0})$ to both sides?


3

$$\operatorname{rank}(AB)\ge \operatorname{rank}(A^{-1}AB)$$


3

Assuming $A$ is an $m$ by $n$ matrix, to say that $A$ has rank $n$ is equivalent to saying that the $n$ columns of $A$ are linearly independent, which in turn is equivalent to saying that $A$ has kernel $\{0\}$. This means $A$ is one-to-one. Hence in particular $A$ is one-to-one on the range of $B$, a space of dimension $rank(B)$. Hence the range of $AB$ ...


2

Since you only need a subset for this example- we can "cheat" a bit. We shall define our subset $U$ to be the union of $x$ axis, and the $y$ axis, i.e. $U$= $\{(x,0):x \in \mathbb{R}\} \cup \{(0,y):y \in \mathbb{R}\}$


2

$J$ has eigenvalues $0$ and $101$, with multiplicity $100$ and $1$ respectively. Hence its characteristic polynomial is $p_J(x)=x^{100}(x-101)$. The characteristic polynomial of $J-I$ is $$P_{J-I}(x)=P_J(x+1)=(x+1)^{100}(x-100)$$ Then $$Det(J-I)=(-1)^{101}P_{J-I}(0)=100$$


1

Yes, a vector space is a subspace of itself. Just like a set can be a subset of itself (but not a proper subset).


1

If $A$ is invertible, then $\det A \neq 0$ and if $B$ is non-invertible, then $\det B = 0$. But $\det AB = \det A \det B = \alpha \cdot 0 = 0$ for some $\alpha \neq 0$, so $AB$ is non-invertible.


1

Even though one may say $x_1$ is not contributing anything but this very fact makes it a free variable. So in your case both $x_1$ and $x_4$ are free variables. The remaining are pivot variables. So what you are doing is correct. You can have $x_1=1$ and $x_4=0$ for the second special solution. Note: Another way to think about it is that $Ax$ is also the ...


1

The operator is $L(x) = (x_1 \cos \theta-x_2 \sin \theta , x_1 \sin \theta+ x_2 \cos \theta) = Q x$, where $Q = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$. Since $Q^T Q = I$, we have $\|L(x)\|^2 = \langle Qx , Q x \rangle = \langle Q^T Qx , x \rangle = \langle x , x \rangle = \|x\|^2$, hence $L$ is an ...


1

The equation $0=-3x+6$ gives $x=2$ which is the equation of a vertical line.


1

If $c_1,\ldots c_n$ are the entries on the diagonal of an upper triangular matrix $A$, then the characteristic polynomial is $p(x) = (x-c_1) \cdots (x-c_n)$. $A-c_iI$ is upper triangular with the $i$th diagonal entry being zero. See what happens when you then compute $p(A)=(A-c_1 I) \cdots (A-c_nI)$.


1

In general, the length of a curve $\gamma:[a,b]\to\mathbb{R}^n$ is given by $$l(\gamma)=\int_a^b\|\dot{\gamma}(t)\|dt.$$ Given a linear $T:\mathbb{R}^n\to\mathbb{R}^m$, we compose and get $$l(T\circ\gamma)=\int_a^b\|T(\dot{\gamma}(t))\|dt.$$ I don't think one can hope for something much better in general. For example, the ellipse can be obtained by a very ...


1

The difference is that with volume we have $\text{vol}(Tx_1,\dots,Tx_n)=|\det A|\,\text{vol}(x_1,\dots,x_n)$ for parallelepipeds formed by $n$ vectors, by multiplicativity of determinants. But with length $l(Tx)$ is not any characteristic of $T$ times $l(x)$. The best we can have is the inequality $l(Tx)\leq\|T\|\,l(x)$. Both the equality and the inequality ...


1

Using abstract index notation: There are two types of indices here: I will use $i,j,\ldots$ for indices relative to the domain (which appears to be $\mathbb{R}^2$) and $A,B,\ldots$ for indices relative to the co-domain (basically from the number of equations) which is $\mathbb{R}^N$. We can write $\mathbf{c}$ as the rank 4 tensor in index notation $$ ...


1

Any matrix is similar to its Jordan Canonical form. The matrices implementing the "similarity" need not be unitary.



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